# Chapter – 5 : Simultaneous Linear Equations | Chapter Solution Class 9

Ganit Prakash 2023 Simultaneous Linear Equations Solution Class 9 Maths. Chapter 5 – Simultaneous Linear Equations is provided here with simple step-by-step explanations. These solutions for Simultaneous Linear Equations are extremely popular among class 9 students. The Simultaneous Linear Equation solution is handy for quickly completing your homework and preparing for exams.

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Simultaneous Linear Equation (5th Chapter)

## Let us work out – 5.1

Let me frame the simultaneous equations in each of the following cases and see whether they can be solved or not.

Question 1

The sum of my elder sister’s present age and my father’s age is 55 years. By calculating, I observed that after 16 years, my father’s age will be two times more than my sister’s.

1. Let me draw the graph after framing simultaneous equations.
2. With the help of a graph, let me find out whether the general solution of two equations can be determined.
3. Let me write the present age of my elder sister and father from the graph.

Solution:

Let the age of my father be ‘x’

and the age of my sister be ‘y’

According to the 1st condition of the problem,

x + y = 55

x = 55 – y —- (i)

 x 30 25 20 y 25 30 35

according to 2nd problem,

y + 16 = 2(x + 16)

or, y + 16 = 2x + 32

or, 2x = y – 16

or, x = \frac{y-16}{2}\\

If y = 10, x = \frac{10-16}{2}=\frac{-6}{2}=-3\\

If y = 20, x = \frac{20-16}{2}=\frac{4}{2}=2\\

If y = 30, x = \frac{30-16}{2}=\frac{14}{2}=7\\

 x -3 -2 7 y 10 20 30

(b) The equation will have a general solution.

(c) The age of the sister = 13 years.

The age of father  = 42 years.

Question 2

Mita bought 3 pens and 4 pencils at Rs. 42 from the shop of Jadabkaku. I bought 9 pens and 1 dozen pencils at the same rate to give gifts to my friend at Rs. 126.

1. Let me draw the graph after forming the simultaneous equations.
2. With the help of a graph, let me find whether the general solution of two equations can be determined.
3. Let e write the price of 1 pen and 1 pencil separately from the graph.

Solution:

Let the cost of 1 pen be Rs. x

and the cost of 1  pencil be Rs. y

According to the 1st condition of the problem,

3x + 4y = 42, or, 3x = 42 – 4y —- (i)

∴ x=\frac{42-4 y}{2}\\

If y = 0, x = \frac{42-4 \times 0}{3}=\frac{42-0}{3}=\frac{42}{3}=14\\

If y = 3, x = \frac{42-4 \times 3}{3}=\frac{42-12}{3}=\frac{30}{3}=10\\

If y = 6, x = \frac{42-4 \times 6}{3}=\frac{42-24}{3}=\frac{18}{3}=6\\

 x 14 10 6 y 0 3 6

According to the 2nd condition of the problem,

9x + 12y = 126, or, 3(3x + 4y) = 126

or, 3x + 4y = 42

Which gives the previous equation. So we can get only one St. line in the graph paper. So, in this case, the equation will have infinite numbers of general solutions.

The equation will have infinite solutions hence if the cost of 1 pen is Rs. 10, the Cost of 1 pencil is Rs.3, again, If the cost of one pen is Rs.6, the Cost of 1 pencil is Rs.6.

Question 2

Today we shall draw the pictures we like in our school. For this, I bought 2 pieces of art paper and 5 sketch pens at Rs.16. But Dola bought 4 art paper and 10 sketch pens at the same rate and from the same shop.

1. Let me form simultaneous equations and draw the graph.
2. Let me see whether the general solution of the equations can be found out from the graph.
3. Let me whether I can get the price of 1 art paper and 1 sketch pen.

Solution:

Let the cost of art papers be ‘x’

and the cost of sketch pens be ‘y’

According to the 1st condition of the problem,

2x + 5y = 16

or, 2x = 16 – 5y

∴ x = \frac{16-5y}{2}\\

If y = 0, x = \frac{16-5 \times 0}{2}=\frac{16-0}{2}=\frac{16}{2}=8 \\

If y = 2, x = \frac{16-5 \times 2}{2}=\frac{16-10}{2}=\frac{6}{2}=3 \\

If y = 4, x = \frac{16-5 \times 4}{2}=\frac{16-20}{2}=\frac{-4}{2}=-2\\

 x 4 -6 -16 y 0 4 8

according to 2nd condition of the problem,

4x + 10 y = 16

or, 4x = 16 – 10y

∴ \therefore x=\frac{16-10 y}{4}

If y = 0, x = \frac{16-10 \times 0}{4}=\frac{16-0}{4}=\frac{16}{4}=4\\

If y = 4, x = \frac{16-10 \times 4}{4}=\frac{16-40}{4}=\frac{-24}{4}=-6\\

If y = 8, x = \frac{16-10 \times 8}{4}=\frac{16-80}{4}=\frac{-64}{4}=-16\\

(b) The equation will have no general solution.

(c) We can not differentiate the value of one art paper and one sketch pen.

## Let us work out 5.2

Question 1

Let us draw the graph of the following equations and write whether are solvable or not and if they are solvable, let us write that particular solution or 3 solutions if they have an infinite number of solutions:

Question 1 (a)

2x + 3y – 7 = 0

3x + 2y – 8 = 0

Solution:

2x + 3y – 7 = 0 — (i)

From equation (i)

or, 2x + 3y -7 = 0

or, 2x = 7-3y

3x + 2y – 8 = 0 – (ii)

From equation (ii)

or, 3x + 2y – 8 = 0

or, 3x = 8 – 2y

∴ x = \frac{7-3y}{2}

 x 2 -1 -4 y 1 3 5

∴ x = \frac{8-2y}{3}

 x 2 0 -2 y 1 4 7

The point of intersecting the equations (i) and (ii) is P(2, 1)

∴ x = 2, y = 1.

Question 1 (b)

4x – y = 11

– 8x + 2y = -22

Solution:

4x – y = 11 —- (i)

From equation (i)

or, 4x – y = 11

or, 4x = 11 + y

-8x + 2y= -22 – (ii)

From equation (ii)

or, -2(4x – 2y) = -22

or, 4x – y = 11

\therefore x = \frac{11+y}{4}
 x 6 4 5 y 1 5 9

Which is the same as equation (i)

 x 3 4 5 y 1 5 9

The given equations are solvable, and have an infinite number of solutions i.e x = 3, y = 1; x = 4, y = 5; x = 5, y = 9……………….

Question 1 (c)

7x+3y= 42

21x + 9y= 42

Solution:

(c) 7x + 3y = 42 – (i)

From equation (i)

or, 7x + 3y = 4

or, 7x = 42 – 3y

21x + 9y= 42

From equation (ii)

21x + 9y = 42

or, 3(7x + 3y) = 42

\therefore x = \frac{42-3y}{7}
 x 6 3 0 y 0 7 14
or, 7x + 3y = 14

or. 7x = 14 – 3y

= \frac{14 - 3y}{7}

The given lines are parallel to each other and inconsistent

\frac{a_1}{a_2} ≠ \frac{b_1}{b_2} ≠ \frac{c_1}{c_2} \\

Question 1 (d)

5x + y = 13

5x+5y = 12

Solution:

5x + y = 13 – (i)

From equation (i)

5x + y = 13

5x = 13 – y

\therefore x = \frac{13 - y}{5}

 x 2 1 0 y 3 8 13

5x + 5y = 12 – (ii)

From equation (ii)

or, 5x = 12 – 5y

\therefore x = \frac{12-5y}{5}

 x \frac{12}{5} \frac{7}{5} \frac{2}{5} y 0 1 2

Solving equations (i) and (ii), we get x = \frac{53}{20}, y = -\frac{1}{4} which is the intersecting point.

Question 2

By comparing the coefficients of the same variables and constants of the following pairs of equations. Let us write whether the pair of equations is solvable or not and check them by drawing the graphs of the equations.

Question 2 (a)

x + 5y = 7

x + 5y = 20

Solution:

x + 5y = 7 – (i)

x + 5y = 20 – (ii)

\therefore \frac{1}{1}=\frac{5}{5} \neq \frac{7}{20}

The two equations (i) and (ii) are in consistent i.e. they are unsolvable.

I draw the graph of two equations (i) and (ii)

x + 5y = 7 – (i)

x = 7 – 5y

 x 7 2 -3 y 0 1 2

and x + 5y = 20 – (ii)

x = 20 – 5y

 x 20 15 10 y 0 1 2

we are observing that the graphs of two equations (i) and (ii) are respectively \overleftrightarrow{A B} and \overleftrightarrow{\mathrm{CD}}.which are parallel to each other from the graph we get are not generally solvable i.e. they have no general solution.

Question 2 (b)

2x + y = 8

2y – 3x = -5

Solution:

2x + y = 8

2y – 3x = -5

i. e. 2x + y = 8 – (i)

-3x + 2y = -5 (ii)

\therefore \frac{2}{-3} \neq \frac{1}{2} \neq \frac{8}{-5}

∴ The two equations (i) and (ii) are in consistent i.e. they are solvable.

I draw the graph of two equations (i) and (ii)

2x + y = 8 – (i)

or, 2x = 8 – y

\therefore x=\frac{8-y}{2}

 x 4 3 2 y 0 2 4

and -3x + 2y = -5 – (ii)

or, -3x = -5 – 2y

or, -3x = -(5 + 2y)

\therefore x=\frac{5+2 y}{3}

 x 3 1 5 y 2 -1 5

we are observing that the graphs of two equations (i) and (ii) are respectively \overleftrightarrow{A B} and \overleftrightarrow{\mathrm{CD}}, each are intersect each other at the point P.

∴ From the graph, we get the two equations (i) and (ii) are generally solvable.

Question 2 (c)

5x + 8y = 14

15x + 24y = 42

Solution:

5x + 8y = 14

15x + 24y = 42

\therefore \frac{5}{15}=\frac{8}{24}=\frac{14}{42}

∴ From the above ratios of the co-efficients, we are observing that the equations (i) and (ii) are generally solvable, but we shall get infinite/ unlimited number of general solutions.

I draw the graph of two equations (i) and (ii)

5x + 8y = 14 – (i)

or, 5x = 14 – 8y

\therefore x=\frac{14-8 y}{5}
 x 6 -2 -10 y -2 3 8

and 15x + 24y = 42 – (ii)

or, -15x = 42 – 24y

or, ^{5} 15 x=3(14-8 y)\\ \therefore x=\frac{14-8 y}{5}

which is the same as equation (i)

 x 6 -2 -10 y -2 3 8

we are observing that the two graphs of two equations (i) and (ii) overlapping to each other become a straight line  \overleftrightarrow{A B}.

∴ From the graph, we get that one equation (i) is generally solvable and has unlimited/infinite solutions.

Question 2 (d)

3x + 2y = 6

12x + 8y = 24

Solution:

3x + 2y = 6

12x + 8y = 24

\therefore \frac{3}{12}=\frac{2}{8}=\frac{6}{24}

From the above ratios of the coefficients, we are observing that equations (i) and (ii) are generally solvable, but we shall get an infinite/ unlimited number of general solutions.

I draw the graph of two equations (i) and (ii)

3x + 2y = 6 – (i)

or, 3x = 6 – 2y

\therefore x=\frac{6-2 y}{3}
 x 2 0 -2 y 0 3 6

and, 12x + 8y = 24 – (ii)

or, (3x + 2y) = 24

or, 3x + 2y = 6

which is the same as equation (i)

 x 2 0 -2 y 0 3 6

we are observing that the two graphs of two equations (i) and (ii) overlapping to each other become a straight line \overleftrightarrow{A B}.

∴ From the graph, we get that one equation (i) is generally solvable and has unlimited/infinite solutions.

Question 3

Let us determine the relations of the same variable and constants of the following pairs of equations and write whether the graphs of the equations will be parallel or intersecting or overlapping.

Question 3 (a)

5x+3y=11

2x – 7y = -12

Solution:

5x+3y=11 – (i)

2x – 7y = -12 – (ii)

\therefore \frac{5}{2} \neq \frac{3}{-7} \neq \frac{11}{-12}

The graphs of the two equations (i) and (ii) will be two intersecting straight lines.

Question 3 (b)

6x – 8 y = 2

3x – 4y = 1

Solution:

6x – 8 y = 2
3x – 4y = 1

\therefore \frac{6}{3}=\frac{-8}{-4}=\frac{2}{1}

The graphs of the two linear equations in two variables will be one overlapping to each other and one straight line will be obtained.

Question 3 (c)

8x – 7y = 0

8x – 7y = 56

Solution:

8x – 7y = 0 – (i)

8x – 7y = 56 – (ii)

\therefore \frac{8}{8}=\frac{-7}{-7} \neq \frac{0}{56}

The graphs of the two equations (i) and (ii) will be parallel to each other.

Question 3 (d)

4x – 3y = 6

4y – 5x= -7

Solution:

4x – 3y = 6 – (i)

4y – 5x = -7 – (ii)

\text { i. e. } 4x – 3y = 6 –\text { (i) }

-5x + 4y = -7 –\text { (ii) }

\therefore \frac{4}{-5} \neq \frac{-3}{4} \neq \frac{6}{-7}

The graphs of the two equations (i) and (ii) will be two intersecting lines.

Question 4

Let’s solve the following pairs of equations by drawing graphs which are solvable and write 3 solutions which have an infinite number of solutions.

Question 4 (a)

4x + 3y = 20

8x + 6y = 40

Solution:

4x + 3y = 20 — (i)

8x+6y=40 – (ii)

From equation (i)

4x + 3y = 20

or, 4x = 20 – 3y

∴ x= \frac{20-3 y}{4}

 x 5 2 -1 y 0 4 8

\therefore \frac{4}{8}=\frac{3}{6}=\frac{20}{40}

From equation (ii)

8x + 6y = 40

or, 2(4x – 3y) = 40

∴ 4x + 3y = 20

which is the same as equation (i)

 x 5 2 -1 y 0 4 8

From the above ratios of the coefficients, we are observing that equations (i) and (ii) are generally solvable, but we shall get infinite/ unlimited general solutions. I draw the graph of two equations (i) and (ii).

We are observing that the two graphs of the two equations (i) and (ii) overlapping to each other becomes a straight line \overleftrightarrow{A B} and \overleftrightarrow{\mathrm{CD}}.

∴ From the graph, we got that one équation (i) is generally solvable have infinite solutions.

Question 4 (b)

4x + 3y = 20

12x + 9y = 20

Solution:

4x + 3y = 20 – (i)

12x + 9y = 20 – (ii)

From equation (i)

4x + 3y = 20

or, 4x = 20 – 3y

∴ x = \frac{20-3 y}{4}

 x 5 2 -1 y 0 4 8

From equation (ii)

12x + 9y = 20

or, 12x = 20 – 9y

∴ x = \frac{20-9y}{12}

 x \frac{5}{3} \frac{11}{12} \frac{1}{6} y 0 1 2

\therefore \frac{4}{12}=\frac{3}{9} \neq \frac{20}{20}

From the above ratios of the coefficients, we are observing that the equations (i) and (ii) are inconsistent i.e. they are unsolved.

I draw the graph of two equations (i) and (ii).

We are observing that the two graphs of the two equations(i) and (ii) are respectively \overleftrightarrow{A B} and \overleftrightarrow{C D} which are parallel to each other.

Question 4 (c)

4x + 3y = 20

\frac{3 x}{4}-\frac{y}{8}=1

Solution:

4x + 3y = 20 – (i)

\frac{3 x}{4}-\frac{y}{8}=1-(ii)

From equation (i)

4x + 3y = 20

or, 4x = 20 – 3y

x=\frac{20-3 y}{4}

 x 5 2 -1 y 0 4 8

\therefore \frac{4}{\frac{3}{4}} \neq \frac{3}{-\frac{1}{8}} \neq \frac{20}{1}

From equation (ii)

\frac{3 x}{4}-\frac{y}{8}=1

or, \frac{6 x-y}{8}=1

or, 6x – y=8

or, 6x = 8 + y

∴ x = \frac{8-y}{6}

 x 2 3 4 y 4 10 16

From the above ratios of the coefficients, we are observing that the equations (i) and (ii) are inconsistent i.e. they are solvable. I draw the graph of two equations (i) and (ii).

We are observing that the two graphs of the two equations (i) and (ii) are respectively \overleftrightarrow{A B} and \overleftrightarrow{\mathrm{CD}} which are intersect each other at the point P.

∴ From the graph, we get the two equations (i) and (ii) are generally solvable.

Question 4 (d)

p – q = 3

\frac{p}{3}+\frac{q}{2}=6

Solution:

p – q = 3 – (i)

\frac{p}{3}-\frac{q}{2}=6 – (ii)

From equation (i)

p – q = 3

or, p = 3 + q

 x 3 6 9 y 0 3 6

\therefore \frac{1}{\frac{1}{3}} \neq \frac{-1}{-\frac{1}{2}} \neq \frac{3}{6}

From equation (ii)

\frac{p}{3}+\frac{q}{2}=6

or, \frac{2 p+3 q}{6}=6

or, 2p + 3q = 36

or, 2p = 36 – 3q

\therefore p=\frac{36-3 q}{2}

 x 15 12 9 y 2 4 6

From the above ratios of the coefficients, we are observing that the equations (i) and (ii) are inconsistent i.e. they are solvable. I draw the graph of two equations (i) and (ii).

We are observing that the two graphs of the two equations (i) and (ii) are respectively \overleftrightarrow{\mathrm{AB}} and \stackrel{\leftrightarrow}{\mathrm{CD}} which are intersecting each other at the point P.

\therefore\quad From the graph, we get the two equations (i) and (ii) are generally solvable.

Question 4 (e)

p – q =3

\frac{p}{5}-\frac{q}{5}=3

Solution:

p – q = 3 – (i)

\frac{p}{5}-\frac{q}{5}=3-\text{(ii)}

From equation (i)

p – q = 3

or, p = 3 + q

\therefore \frac{1}{\frac{1}{5}}=\frac{-1}{-\frac{1}{5}} \neq \frac{3}{3}

 x 3 9 6 y 0 6 3

From equation (ii)

\frac{p}{5}-\frac{q}{5}=3

or, \frac{p-q}{5}=3

or, p – q =  15

or, p = 15 + q

 x 15 10 20 y 2 -5 5

From the above ratios of the co-efficients, we are observing . that the equations (i) and (ii) are inconsistent i.e. they are unsolved. I draw the graph of two equations (i) and (ii). We are observing that the two graphs of the two equations(i) and (ii) are respectively \overleftrightarrow{\mathrm{AB}}and \overleftrightarrow{\mathrm{CD}}which are parallel to each other.

Question 4 (f)

p – q = 3

8p – 8q = 5

Solution:

p – q = 3 – (i)

8p – 8q = 5 – (ii)

From equation (i)

p – q = 3

or, p = 3 + q

 x 3 9 6 y 0 6 3

From equation (ii)

8p – 8q = 5

or, 8p = 5 + 8q

\therefore p =\frac{5+8 q}{8}

\therefore \frac{1}{8}=\frac{-1}{-8} \neq \frac{8}{5}

 x \frac{5}{8} \frac{13}{5} \frac{21}{5} y 0 1 2

From the above ratios of the coefficients, we are observing that the equations (i) and (ii) are inconsistent i.e. they are unsolved. I draw the graph of two equations (i) and (ii). We are observing that the two graphs of the two equations (i) and (ii) are respectively \overleftrightarrow{A B} and \overleftrightarrow{\mathrm{CD}} which are parallel to each other.

Question 5

Tathagata has written a linear equation in two variables x + y = 5. I write another linear equation in two variables, so that, the graphs of the two equations will be:

(a) parallel to each other (b) intersecting (c) overlapping

Solution:

(a) x + y = 5 – (i)

The equation of a straight line parallel to the graph of the equation (i) will be

2x + 2y = 14\left[\text { since } \frac{1}{2}=\frac{1}{2} \neq \frac{5}{14}\right]

(b) The equation of a straight line intersecting to the graph of the equation  x + y = 5 will be

3x + 2y = 6\left[\text { since } \frac{1}{3} \neq \frac{1}{2}\right]

(c) The equation of a straight line overlapping to the graph of the equation x + y = 5 will be

5x + 5y = 25\left[\because \frac{1}{5}=\frac{1}{5}=\frac{5}{25}\right]

## Let us work out 5.3

Question 1

Let us solve the following linear equations in two variables by elimination method and check them graphically:

Question 1 (a)

8x+ 5y = 11 = 0

3x – 4y – 10 = 0

Solution:

8x + 5y – 11 = 0

3x – 4y – 10 = 0

i. e. 8x + 5y = 11 — (i)

3x – 4y = 10 — (ii)

Multiplying equation (i) by 4 and equation (ii) by 5, we get,

32x + \cancel{20y} = 44\\ 15x - \cancel{20y} = 50

or, 47x = 94

or, x = \frac{{\cancel{94}}^2}{{\cancel{47}}^1}

∴ x = 2

Now putting the value of x in equation (i), we get

8x + 5y = 11

or, 8 × 2 + 5y = 11

or, 16 + 5y = 11

or, 5y = 11 – 16

or, 5y = -5

or, y = \frac{-5}{5} \therefore \quad y = -1

\left.\begin{array}{l}\therefore x=2 \\ y=-1\end{array}\right\} Ans.

By graphically,

From equation (i)

8x + 5y = 11

or, 8x = 11 – 5y

or, x = \frac{11-5y}{8}

 x 2 -3 -8 y -1 7 15

From equation (ii)

3x – 4y = 10

or, 3x = 10 + 4y

or, x = \frac{10+4y}{3}

 x 2 6 10 y -1 2 5

We observe from the graph, x = 2, y = – 1

Question 1 (b)

2x + 3y – 7 = 0

3x + 2y – 8 = 0

Solution:

2x + 3y – 7 = 0

3x + 2y – 8 = 0

i. e. 2x + 3y = 7 — (i)

3x + 2y = 8 — (ii)

Multiplying equation (i) by 2 and equation (ii) by 3 , we get.

(By subtracting)

\cancel{-}5 x=\cancel{-}10

or, x = \frac{\cancel{10}^2} {\cancel{5}} \therefore \quad x = 2

Now putting the value of x in equation (i),

\text {2x + 3y = 7}

\text {or, 2 × 2 + 3y = 7}

\text {or, 4 + 3y=7}

\text {or, 3y = 7 - 4}

\text {or, 3y = 3 }

or, y = \frac{\cancel{3}^1}{{\cancel{3}}^1}

∴ y = 1

\left.\begin{array}{l} x=2 \\ y=1\end{array}\right\} Ans.

By graphically,

From equation (i)

2x + 3y = 7

or, 2x = 7 – 3y

or, x = \frac{7-3 y}{2}

 x 2 -1 -4 y 1 3 5

From equation (ii)

3x + 2y = 8

or, 2x = 7 – 3y

or, 3x = 8 – 2y

or, x =\frac{7-3 y}{2}

or, x =\frac{8-2 y}{3}

 x 2 0 -2 y 1 4 7

We observe from the graph, x = 2, y = 1

Question 2

To eliminate y, what number is multiplied with the equation 7x – 5y + 2 = 0 and then add to the equation 2x + 15y + 3 = 0.

Solution:

We have to multiply the equation 7x – 5y + 2 = 0 by 3. and the equation 2x + 15y + 3 = 0 by 1, when we added these equations then y is eliminated.

Question 3

Let us write the least natural number by which we can multiply both the equations 4x – 3y = 16, 6x + 5y = 62 and get the equal co-efficients of x.

Solution:

Multiplying 4x – 3y = 16 by 3 and 6x + 5y = 62 by 2.

Question 4

Let us solve the following linear equations in two variables by elimination method:

Question 4 (i)

3x + 2y = 6 — (i)

2x – 3y = 17 — (ii)

Solution:

3x + 2y = 6 — (i)

2x – 3y = 17 — (ii)

Multiplying equation (i) by 3 and equation (ii) by 2 , we get,

9x+\cancel{6y}=18 \\4 x-\cancel{6y}=34

or, 13x = 52

or, x = \frac{\cancel{52}^4}{\cancel{13}} \therefore x = 4

Now putting the value of x in equation (i), we get

3x + 2y = 6

or, 3 × 4 + 2y = 6

or, 12 + 2y = 6

or, 2y = 6 – 12

or, 2y = – 6

or, y = \frac{-6}{2} \therefore \quad y = – 3

The required solution:

\left.\begin{array}{l}\therefore x=4 \\y=-3\end{array}\right\}Ans.

Question 4 (ii)

2x + 3y = 32

11y – 9x = 3

Solution:

i. e. 2x + 3y = 32 — (i)

11y – 9x = 3 — (ii)

Multiplying equation (i) by 11 and equation (ii) by 3 , we get.

(By subtracting)

49x = 343

or, x =\frac{\cancel{343}^{\cancel{49}^{7}}}{\cancel{49}_{\cancel{7}_{1}}}

∴ x= 7

Now putting the value of x in equation (i)

2x + 3y = 32

or, 2 × 7 + 3y = 32

or, 14 + 3y = 32

or, 3y = 32 – 14

or, 3y = 18

or, y = \frac{\cancel{18}^{6}}{\cancel{3}} \therefore \quad y = 6

The required solution is

\left.\begin{array}{l}\therefore x=7 \\y=6\end{array}\right\}Ans.

Question 4 (iii)

x + y = 48

x + 4 = \frac{5}{2} (y+4)

Solution:

x + y = 48 – (i)

x + 4 = \frac{5}{2} (y+4) - (ii)

from equation (ii)

x + 4 = \frac{5}{2} (y+4)

or, 2x – 5y = 20 – 8

or, 2x – 5y = 12 — (iii)

∴ x + y = 48 – (i)

2x – 5y = 12 – (iii)

Multiplying equation (ii) by 5 and equation (iii) by 1, we get.

5x + \cancel{5y} = 240 \\ 2x - \cancel{5y} = 12

or, 7x = 252

or, x = \frac{\cancel{252}^{36}}{\cancel{7}}

x = 36

Now putting the value of x in equation (i)

x + y = 48

or, 36 + y = 48

or, y = 48 – 36

y = 12

The required solution is

\left.\begin{array}{l}\therefore x=36 \\y=12\end{array}\right\}Ans.

Question 4 (iv)

\frac{x}{2} + \frac{y}{3} = 8

\frac{5x}{4} - 3y = -3

Solution:

\frac{x}{2} + \frac{y}{3} = 8  - (i)

\frac{5x}{4} - 3y = -3 - (ii)

from equation (i)

\frac{x}{2} + \frac{y}{3} = 8

or, \frac{3x +2y}{6} = 8

or, 3x + 2y = 48 – (iii)

∴ 3x + 2y = 48 (iii)

5x – 12y = -12 – (iv)

from equation (ii)

\frac{5x}{4} - 3y = -3

or, \frac{5x +12y}{4} = -3

or, 5x – 12y = -12 – (iv)

Multiplying equation (iii) by 6 and equation (iv) by 1.

18x + \cancel{12y} = 288 \\ 5x-\cancel{12y} =-12

or, 23x = 276

or, x = \frac{\cancel{276}^{12}}{\cancel{23}}

∴ x = 12

Now putting the value of x in equation (iii).

3x + 2y = 48

or, 3 × 12 + 2y = 48

or, 36 + 2y = 48

or, 2y = 48 – 36

or, 2y = 12

∴ y = \frac{\cancel{12}^6}{\cancel{2}} = 6

The required solution is

\left.\begin{array}{l}\therefore x=12 \\y=6\end{array}\right\}Ans.

Question 4 (v)

3 x-\frac{2}{y}=5 \\

x+\frac{4}{y}=4

Solution:

3 x-\frac{2}{y}=5 — (i)

x+\frac{4}{y}=4 — (ii)

Multiplying equation (i) by 2 and equation (ii) by 1, we get.

6 x-\cancel{\frac{4}{y}}=10 \\ \underline{x+\cancel{\frac{4}{y}}=4}

7x = 14

or, x = \frac{\cancel{14}^{2}}{\cancel7}

Now putting the value of x in equation (i)

3x-\frac{2}{y}=5

or, 3 \times 2-\frac{2}{y}=5

or, 6-\frac{2}{y}=5

or, -\frac{2}{y}=5-6

or, \cancel{-}\frac{2}{y}= \cancel{-}1

The required solution is

\left.\begin{array}{r}\therefore x=2 \\ y=2\end{array}\right\}Ans.

Question 4 (vi)

\frac{x}{2}+\frac{y}{3}=1

\frac{x}{3}+\frac{y}{2}=1

Solution:

i. e. \frac{x}{2}+\frac{y}{3}=1 — (i)

\frac{x}{3}+\frac{y}{2}=1 — (ii)

from equation (i)

\frac{x}{2}+\frac{y}{3}=1

or, \frac{3 x+2 y}{6}=1

or, 3x + 2y = 6 — (iii)

\therefore \quad 3x + 2y = 6 — (iii)

2x + 3y = 6 — (iv)

from equation (ii)

\frac{x}{3}+\frac{y}{2}=1

or, \frac{2 x+3 y}{6}=1

or, 2x + 3y = 6 — (iv)

Multiplying equation (iii) by 3 and equation (iv) by 2 , we get

9x+\cancel{6 y}=18 \\4x+\cancel{6 y}=12 \\(-)(-) \quad(-)

(By subtracting)

or, 5x = 6

Now putting the value of x in equation (i)

3x + 2y = 6

or, 3 \times \frac{6}{5}+2 y=6

or, \frac{18}{5}+2 y=6

or, 2 y=6-\frac{18}{5}

or, 2 y=\frac{30-18}{5}

or, \cancel{2y}=\frac{\cancel{12}^{6}}{5}

The required solution is

\left.\begin{array}{r}\therefore x=\frac{6}{5} \\ y=\frac{6}{5}\end{array}\right\} Ans.

Question 4 (vii)

\frac{x+y}{2}+\frac{3 x-5 y}{4}=2 \\

\frac{x}{14}+\frac{y}{18}=1

Solution:

\frac{x+y}{2}+\frac{3 x-5 y}{4}=2 — (i)

\frac{x}{14}+\frac{y}{18}=1 — (ii)

from equation (i)

\frac{x+y}{2}+\frac{3 x-5 y}{4}=2

or, \frac{2 x+2 y+3 x-5 y}{4}=2

or, 5x – 3y = 8 — (iii)

from equation (ii)

\frac{x}{14}+\frac{y}{18}=1

or, \frac{9 x+7 y}{126}=1

\therefore \quad 9x + 7y = 126 — (iv)

\therefore \quad 3 x+2 y=6 — (iii)

\quad 2 x+3 y=6 — (iv)

Multiplying equation (iii) by 7 and equation (iv) by 3, we get

35 x-\cancel{21 y}=56 \\\underline{27 x+\cancel{21 y}=378}

or, 62x = 434

or, x =\frac{\cancel{434}^{\cancel{277}^{7}}}{\cancel{62}_{\cancel{31}}}

Now putting the value of x  in equation (i)

5x – 3y = 8

or, 5 \times 7-3 y=8

or, 35 – 3y = 8

or, -3y = 8 – 35

or, \cancel{-} 3y = \cancel{-} 27

or, y=\frac{\cancel{27}^{9}}{\cancel{3}}

The required solution is

\left.\begin{array}{r}\therefore x=7 \\ y=9\end{array}\right\} Ans.

Question 4 (viii)

\frac{x y}{x+y}=\frac{1}{5} \\

\frac{x y}{x-y}=\frac{1}{9}

Solution:

\frac{x y}{x+y}=\frac{1}{5} — (i)

\frac{x y}{x-y}=\frac{1}{9}— (ii)

or, 2\cancel{x} = 14\cancel{x}y

or, y =\frac{\cancel{2}^{1}}{\cancel{14}_{7}}

Now putting the value of y in equation (i)

x + y = 5xy

or, x+\frac{1}{7}=5 x \times \frac{1}{7}

or, \frac{7 x+1}{\cancel{7}}=\frac{5 x}{\cancel{7}}

or, 7x + 1 = 5x

or, 7x – 5x = -1

or, 2x = -1

The required solution is

\quad \left.\begin{array}{r}\therefore x=-\frac{1}{2} \\y=\frac{1}{7}\end{array}\right\} \text { Ans. }

Question 4 (ix)

\frac{1}{x-1}+\frac{1}{y-2}=3 \\

\frac{2}{x-1}+\frac{3}{y-2}=5

Solution:

\frac{1}{x-1}+\frac{1}{y-2}=3 — (i)

\frac{2}{x-1}+\frac{3}{y-2}=5 — (ii)

Multiplying equation (i) by 3 and equation (ii) by 1, we get

\frac{3}{x-1}+\cancel{\frac{3}{y-2}}=9 \\ \underline{{_{(\text-)}}\frac{2}{x-1}+{_{(\text-)}}\cancel{\frac{3}{y-2}}={_{(\text-)}}5}

(By Subtracting)

\frac{1}{x-1} = 4

or, 4x – 4 = 1

or, 4x = 1 + 4

or, 4x = 5

Now putting the value of x in equation (i)

\frac{1}{x-1}+\frac{1}{y-2}=3

or, \frac{1}{\frac{5}{4}-1}+\frac{1}{y-2}=3

or, \frac{1}{\frac{5-4}{4}}+\frac{1}{y-2}=3

or, \frac{1}{\frac{1}{4}}+\frac{1}{y-2}=3

or, 4+\frac{1}{y-2}=3

or, \frac{1}{y-2}=3-4

or, \frac{1}{y-2}=-1

or, -y + 2 = 1

or, -y = 1 – 2

or, \cancel{-} y = \cancel{-} 1

The required solution is

\left.\begin{array}{r}\therefore x=\frac{5}{4} \\ y=1\end{array}\right\}Ans.

Question 4 (x)

\frac{14}{x+y}+\frac{3}{x-y}=5

\frac{21}{x+y}-\frac{1}{x-y}=2

Solution:

\frac{14}{x+y}+\frac{3}{x-y}=5 — (i)

\frac{21}{x+y}-\frac{1}{x-y}=2 — (ii)

Let x + y = a, x – y = b

Then,

\frac{14}{a}+\frac{3}{b}=5-(\text{i}) \\\frac{21}{a}-\frac{1}{b}=2- (\text{ii})

Multiplying equation (i) by 1 and equation (ii) by 3, we get

\frac{14}{a}+\cancel{\frac{3}{b}}=5\\

\frac{63}{a}-\cancel{\frac{3}{b}}=6

\frac{77}{a}=11

or, 11a = 77

or, a =\frac{\cancel{77}^{7}}{\cancel{11}}

Now putting the value of a in equation (i)

\frac{14}{a}+\frac{3}{b}=5

or, \frac{\cancel{14}^{2}}{\cancel{7}}+\frac{3}{b}=5

or, 2+\frac{3}{b}=5

or, \frac{3}{b}=5-2

or, \frac{3}{b}=3

or, 3b = 3

or, b=\frac{\cancel{3}^{1}}{\cancel{3}}

\therefore \quad x+y=a=7-\text { (iii) } \\\underline{x-y=b=1-\text { (iv) }}

or, 2x = 8

or, x =\frac{\cancel{8}^{4}}{\cancel{2}}

Now putting the value of x  in equation (iii)

x + y = 7

or, 4 + y = 7

or, y = 7 – 4

The required solution is

\left.\begin{array}{r}\therefore x=4 \\y=3\end{array}\right\} \text { Ans. }

Question 4 (xi)

\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} \\

\frac{x+y}{3}-\frac{x-y}{2}+\frac{5}{6}=0

Solution:

\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} — (i)

\frac{x+y}{3}-\frac{x-y}{2}+\frac{5}{6}=0 — (ii)

From equation (i),

\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} \\

\text { or, } \frac{4 x+4 y-5 x+5 y}{20}=\frac{7}{20} \\

\text { or, } \frac{-x+9 y}{\cancel{20}}=\frac{7}{\cancel{20}}

or, – x + 9y = 7 — (iii)

From equation (ii),

\frac{x+y}{3}-\frac{x-y}{2}+\frac{5}{6}=0\\

or, \frac{2 x+2 y-3 x+3 y}{6}=-\frac{5}{6}\\

or,\frac{-x+5 y}{\cancel{6}}=\frac{-5}{\cancel{6}}\\

\therefore \quad-x+5 y=-5 -\text{ (iv) } \\

\therefore \quad\cancel{-x}+9 y=7-\text { (iii) } \\ \underline{{_{(\text+)}}\cancel{-x}+{_{(\text-)}}5y={_{(\text+)}}-5-\text { (iv) }}

(By Subtracting)

or, 4 y=12\\

or, y=\frac{12^{3}}{4}\\

\therefore \mathrm{y}=3\\

Now putting the value of y in equation (iii)

-x+9 y=7\\

or, -x+9 \times 3=7\\

or, -x+27=7\\

or, -x=7-27\\

or, - x= -20\\

\therefore \mathrm{x}=20\\

The required solution is

\left.\begin{array}{r}\therefore x=20 \\y=3\end{array}\right\}Ans.\\

Question 4 (xii)

x+y=a+b

a x-b y=a2 – b2

Solution:

x + y = a + b — (i)

a x-b y=a^{2}-b^{2} — (ii)

Multiplying equation (i) by b and equation (ii) by 1, we get.

b x+\cancel{b y}=a b+\cancel{b^{2}} \\\underline{a x-\cancel{b y}=a^{2}-\cancel{b^{2}}}

or, b x+a x=a b+a^{2}

or, x(b+a)=a(b+a)

\therefore \mathrm{x}=\mathrm{a}

Now putting the value of x in equation (i)

x+y=a+b

or, \cancel {a}+y=\cancel{a}+b

The required solution is

\left.\begin{array}{r}\therefore x=a \\y=b\end{array}\right\} \text { Ans. }

Question 4 (xiii)

\frac{x+a}{a}=\frac{y+b}{b} \\

ax – by=a2 – b2

Solution:

\frac{x+a}{a}=\frac{y+b}{b} \\

ax – by=a2 – b2

or, bx + ab = ay + ab

or, bx – a y = 0 – (i)

ax – by=a2 – b2

Multiplying equation (i) by b and equation (ii) by a, we get

b^{2} x-\cancel{a b y}=0 \\\underline{_{(\text-)}{a^{2} x-{_{(\text+)}}\cancel{aby}={_{(\text-)}}a\left(a^{2}-b^{2}\right)}}

(By subtracting)

b^{2} x-a^{2} x=-a\left(a^{2}-b^{2}\right)

or, \cancel{-} x\left(a^{2}-b^{2}\right)= \cancel{-} a\left(a^{2}-b^{2}\right)

\therefore \mathrm{x}=\mathrm{a}

Now putting the value of x in equation (i)

bx – ay=0

or, b \cdot a-a y=0

or, ab – ay=0

or, \cancel{-a}y= \cancel{-a}b

∴ y = b

The required solution is

\left.\begin{array}{l}\therefore x=a \\y=b\end{array}\right\}Ans.

Question 4 (xiv)

ax + by = c

a2 x – b2 y = c2

Solution:

ax + by = c — (i)

a2 x – b2 y = c2 — (ii)

Multiplying equation (i) by b and equation (ii) by 1, we get.

a b x+\cancel{b^{2}y}=b c \\\underline{{_{(\text-)}}{a^{2} x+{_{(\text+)}}\cancel{b^{2} y}={_{(\text-)}}c^{2}}}

(By subtracting)

or, a bx – a2x = bc – c2

or, ax(b – a) = c(b – c)

∴ x = \frac{c(b-c)}{a(b-a)}

Now putting the value of x in equation (i)

ax + b y = c

or, \quad \cancel{a \cdot} \frac{c(b-c)}{\cancel{a \cdot}(b-a)}+b y=c\\

or, \frac{b c-c^{2}}{b-a}+b y=c \\

or, b y=c-\frac{b c-c^{2}}{b-a}\\

or, b y=\frac{\cancel{b c}-a c-\cancel{bc}+c^{2}}{b-a}\\

or, b y=\frac{c^{2}-a c}{b-a}\\

∴ y = \frac{c(c-a)}{b(b-a)}\\

The required solution is

\left.\begin{array}{r}\therefore x=\frac{c(b-c)}{a(b-a)} \\ y=\frac{c(c-a)}{b(b-a)}\end{array}\right\} Ans.

Question 4 (xv)

a x + by = 1

b x+a y=\frac{(a+b)^{2}}{a^{2}+b^{2}}-1

Solution:

a x+b y=1 — (i)

b x+a y=\frac{(a+b)^{2}}{a^{2}+b^{2}}-1 — (ii)

\therefore b x+a y=\frac{(a+b)^{2}}{a^{2}+b^{2}}-1 \\

or, b x+a y=\frac{\cancel{a^{2}}+2 a b+\cancel{b^{2}}-\cancel{a^{2}}-\cancel{b^{2}}}{a^{2}+b^{2}}\\

or, b x+a y=\frac{2 a b}{a^{2}+b^{2}}-\text { (ii) }\\

Multiplying equation (i) by a and equation (ii) by b, we get

\begin{array}{l}a^{2} x+\cancel{a b y}=a \\{_{(\text-)}}\underline{b^{2} x+{_{(\text-)}}\cancel{a b y}={_{(\text-)}}\frac{2 a b}{a^{2}+b^{2}}}\end{array}

(By subtracting)

a^{2} x-b^{2} x=a-\frac{2 a b}{a^{2}+b^{2}}\\

or, x\left(a^{2}-b^{2}\right)=\frac{a^{3}+a b^{2}-2 a b^{2}}{a^{2}+b^{2}}\\

or, x\left(a^{2}-b^{2}\right)=\frac{a^{3}-a b^{2}}{a^{2}+b^{2}}\\

or, x\left(a^{2}-b^{2}\right)=\frac{a\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}}\\

\therefore \mathrm{x}=\frac{\mathrm{a}}{\mathrm{a}^{2}+\mathrm{b}^{2}}\\

Now putting the value of x in equation (i)

a x+b y=1

or, \frac{ a\cdot a}{a^{2}+b^{2}}+b y=1\\

or, \frac{a^{2}}{a^{2}+b^{2}}+b y=1 \\

or, b y=1-\frac{a^{2}}{a^{2}+b^{2}} \\

or, b y=\frac{a^{2}+b^{2}-x^{2}}{a^{2}+b^{2}}\\

or, b y=\frac{b^{2}}{a^{2}+b^{2}} \\

The required solution is

\left.\begin{array}{r}\therefore x=\frac{a}{a^2+b^2} \\ y=\frac{b}{a^2+b^2}\end{array}\right\} Ans.

Question 4 (xvi)

(7x – y – 6)2 + (14x + 2y – 16)2 =0

Solution:

(7 x-y-6)^{2}+(14 x+2 y-16)^{2}=0

If the sum of two positive quantities is zero, then each quantity will be zero.

7x – y – 6 = 0

or, 7x – y = 6 – (i)

Also, 14x + 2y = 16 – (ii)

Multiplying equation (i) by 2 and equation (ii) by 1, we get.

14 x-\cancel{2 y}=12 \\\underline{14 x+\cancel{2 y}=16}

or, 28 x = 28

∴ x = 1

Now putting the value of x in equation (i)

7x – y = 6

or, 7 × 1 – y = 6

or, 7 – y = 6

or, – y = 6 – 7

or, \cancel{-}y= \cancel{-}1\\[/katex]

∴ y = 1

The required solution is

\left.\begin{array}{l}x=1 \\ y=1\end{array}\right\} \text { Ans. }

## Let us work out 5.4

Question 1

Let us express the variable x of the equation \frac{x}{3}+\frac{y}{2}=8 in terms of the variable y.

Solution:

The given equation is

\frac{x}{3}+\frac{y}{2}=8

or, \frac{x}{3}=8-\frac{y}{2}

or, \frac{x}{3}=\frac{16-y}{2}

or, x=\frac{3}{2}(16-y) \text { (Ans.) }

Question 2

Let us express the variable y of the equation \frac{2}{x}+\frac{7}{y}=1 in terms of the variable x.

Solution:

The given equation is

\frac{2}{x}+\frac{7}{y}=1

or, \frac{7}{y}=1-\frac{2}{x}\\

or, \frac{7}{y}=\frac{x-2}{x}\\

or, \frac{y}{7}=\frac{x}{x-2}\\

∴ y = \frac{7 x}{x-2}

Question 3

Let us solve the following equations by comparison method and check whether the solutions satisfy the equations.

Question 3 (a)

2(x – y) = 3

5x + 8y = 14

Solution:

2(x – y) = 3 — (i)

5x + 8y = 14 — (ii)

Taking equation (i)

2(x – y) = 3

or, 2x -2y = 3

or, 2x = 3 + 2y

x = \frac{3+2 y}{2}-\text { (iii) }

Taking equation (ii)

5x + 8y = 14

or, 5x = 14 – 8y

or, x = \frac{14-8 y}{2}-\text { (iv) }

Comparing equations (iii) and (iv)

or, \frac{3+2 y}{2}=\frac{14-8 y}{5}\\

or, 15 + 10y = 28 – 16y

or, 10y + 16y = 28 – 15

or, 26 y = 13

or, y = \frac{\cancel{13}^{1}}{\cancel{26}^{2}} \\

∴ y = \frac{1}{2}\\

Now putting the value of y in equation (iii),

∴ x = \frac{3+2 y}{2} \\

=\frac{3+\cancel{2} \times \frac{1}{\cancel{2}}}{2} \\

=\frac{3+1}{2}\\

=\frac{\cancel{4}^{2}}{\cancel{2}^{1}} \\

∴ x = 2

The required solution is

\left.\begin{array}{l}x=2 \\y=\frac{1}{2}\end{array}\right\} \text {}

Question 3 (b)

5 x-\frac{2}{y}=3\\

Solution:

5 x-\frac{2}{y}=3 — (ii)

Taking equation (i)

or, 2 x=5-\frac{3}{y}\\

or, 2 x=\frac{5 y-3}{y}\\

x=\frac{5 y-3}{2 y}-\text { (iii) }\\

Taking equation (ii)

5 x-\frac{2}{y}=3 \\

5 x=3+\frac{2}{y} \\

or, 5 x=\frac{3 y+2}{y} \\

\therefore x=\frac{3 y+2}{5 y}-(iv)

Comparing equations (iii) and (iv)

or, \frac{5 y-3}{2}=\frac{3 y+2}{5}[\because \quad y \neq 0]\\

or, \frac{5 y-3}{2}=\frac{3 y+2}{5}\\

or, 25y – 15 = 6y + 4

or, 25y – 6y = 4 + 15

or, 19y = 19

or, y=\frac{\cancel{19}^{1}}{\cancel{19}}\\

Now putting the value of y in equation (iii),

∴ x = \frac{5 y-3}{2 y} \\

=\frac{5 \times 1-3}{2 \times 1}

=\frac{5-3}{2} \\

=\frac{\cancel{2}^{1}}{\cancel{2}} \\

∴ x = 1

∴ The required solution is x = 1, y = 1

Question 3 (c)

\frac{x}{3}+\frac{y}{2}=1\\

Solution:

\frac{x}{3}+\frac{y}{2}=1 — (ii)

Taking equation (i)

or, \frac{x}{2}=1-\frac{y}{3}

or, \frac{x}{2}=\frac{3-y}{3}\\

\therefore x=\frac{6-2 y}{3} — (iii)

Taking equation (ii)

\frac{x}{3}+\frac{y}{2}=1

or, \frac{x}{3}=1-\frac{y}{2} \\

or, \frac{x}{3}=\frac{2-y}{2}\\

\therefore \quad x=\frac{6-3 y}{2} — (iv)

Comparing equations (iii) and (iv)

or, \frac{6-2 y}{3}=\frac{6-3 y}{2}\\

or, 12 – 4y = 18 – 9y

or, -4y + 9y = 18 – 12

or, 5y = 6

∴ y = \frac{6}{5}\\

Now putting the value of y in equation (iii),

x = \frac{6-2 y}{3}

=\frac{6-2 \times \frac{6}{5}}{3} \\

=\frac{6-\frac{12}{5}}{3} \\

=\frac{\frac{30-12}{5}}{3} \\

=\frac{\frac{18}{5}}{3} \\

=\frac{\cancel{18}^6}{5} \times \frac{1}{\cancel{3}} \\

=\frac{6}{5} \\

∴ x = \frac{6}{5}

∴ The required solution is x =\frac{6}{5}, y=\frac{6}{5} (Ans.)

Question 3 (d)

4x – 3y = 18

4y – 5x = -7

Solution:

4x – 3y = 18 — (i)

4y – 5x = – 7 — (ii)

Taking equation (i)

4 x-3 y=18

or, 4 x=18+3 y

\therefore x=\frac{18+3 y}{4}-\text { (iii) }

Taking equation (ii)

4y – 5x = -7

or, -5x = -7 – 4y

or, \cancel{-}5 x=\cancel{-}(4 y+7) \\

∴ x = \frac{4 y+7}{5}-\text { (iv) }

Comparing equations (iii) and (iv)

or, \frac{18+3 y}{4}=\frac{4 y+7}{5}\\

or, 16y + 28 = 90 + 15y

or, 16y – 15y = 90 – 28

∴ y = 62

Now putting the value of y in equation (iii),

x = \frac{18+3 y}{4} \\

=\frac{18+3 \times 62}{4} \\

=\frac{18+186}{4}\\

=\frac{\cancel{204}^{51}}{\cancel{4}} \\

= 51

∴ The required solution is

\left.\begin{array}{l}x=51 \\y=62\end{array}\right\}

Question 4

Let us solve the equations  2 x+y=8  and 2 y-3 x=-5  by comparison method and justify them by solving graphically.

Solution:

2 x + y = 8 — (i)

2y – 3x = -5 — (ii)

Taking equation (i)

2x + y = 8

or,  2x = 8 – y

∴ x = \frac{8-y}{2}-\text { (iii) }

Taking equation (ii)

2y – 3x = -5

or, -3x = -5 – 2y

or, \cancel{-}3 x={\cancel-}(2 y+5) \\

∴ x = \frac{2 y+5}{3}-\text { (iv) }

Comparing equations (iii) and (iv)

\frac{8-y}{2}=\frac{2 y+5}{3}\\

or, 4y + 10 = 24 – 3 y

or, 4y + 3y=24 – 10

or, 7y = 14

or, y = \frac{\cancel{14}^{2}}{\cancel{7}}\\

or, y = 2

Now putting the value of y in equation (iii),

x =\frac{8-y}{2} \\

=\frac{8-2}{2} \\

=\frac{6^{3}}{7} =3 \\

∴ x = 3

The required solution is

x = 3, y = 2

By graphically

From equation (iii),

x = \frac{8-y}{2}

 x 4 3 2 y 0 2 4

From equation (iv),

x = \frac{2 y+5}{3}

 x 3 5 7 y 2 5 8

From that graph, we see that x = 3, and y = 2 which satisfies the above answer.

Question 5

Let us solve the following equations in two variables by comparison method :

Question 5 (i)

3x – 2y = 3

7x + 3y = 43

Solution:

3x – 2y=3 — (i)

7x + 3y = 43 — (ii)

Taking equation (i)

3x – 2y = 2

or, 3 x = 2 + 2y

or, x = \frac{2+2 y}{3} — (iii)

Taking equation (ii)

7x + 3y = 43

or, 7x = 43 – 3y

or, x = \frac{43-3 y}{7} — (iv)

Comparing equations (iii) and (iv)

or, \frac{2+2 y}{3}=\frac{43-3 y}{7}\\

or, 14 + 14y = 129 – 9y

or, 23y = 115

or, y = \frac{\cancel{115}^{5}}{\cancel{23}}\\

∴ y = 5

Now putting the value of y  in equation (iii),

x = \frac{2+2 y}{3}\\

=\frac{2+2 \times 5}{3}\\

=\frac{2+10}{3}\\

= \frac{\cancel{12}^{4}}{\cancel{3}}=4\\

∴ x = 4

The required solution is

x=4, y=5

Question 5 (ii)

2x – 3y = 8

\frac{x+y}{x-y}=\frac{7}{3}\\

Solution:

2x – 3y = 8 —(i)

\frac{x+y}{x-y}=\frac{7}{3} — (\text{ii})

Taking equation (i)

2x – 3y = 8

or, 2x = 8 + 3 y

∴ x = \frac{8+3 y}{2} – (iii)

Taking equation (ii)

\frac{x+y}{x-y}=\frac{7}{3}\\

or, 7x – 7y = 3x + 3y

or, 7x – 3x = 3y + 7y

or, 4x = 10y

or, x = \frac{^{5} \cancel{10} y}{\cancel{4}^{2}}\\

∴ x = \frac{5 y}{2}-\text { (iv) }\\

Comparing equations (iii) and (iv),

or, \frac{8+3 y}{\cancel{2}}=\frac{5 y}{\cancel{2}} \\

or, 8 + 3y = 5 y

or, 5y – 3y = 8

or, 2y=8

or, y = \frac{\cancel{8}^{4}}{\cancel{2}} \\

∴ y = 4

Now putting the value of y in equation (iii),

x = x=\frac{8+3 y}{2} \\

= \frac{8+3 \times 4}{2}\\

= \frac{8+12}{2}\\

= \frac{\cancel{20}^{10}}{\cancel{2}}=10 \\

∴ x = 10

The required solution is

x = 10; y = 4

Question 5 (iii)

\frac{1}{3}(x-y) =\frac{1}{4}(y-1)

\frac{1}{7}(4 x-5 y) = x - 7

Solution:

\frac{1}{3}(x-y) =\frac{1}{4}(y-1)

\frac{1}{7}(4 x-5 y)=x-7

Taking equation (i)

\frac{1}{3}(x-y)=\frac{1}{4}(y-1)\\

or, 4(x – y) = 3 (y – 1)

or, 4x – 4y = 3y – 3

or, 4x = 3y – 3 + 4y

or, 4x = 7y – 3

∴ x = \frac{7 y-3}{4}-\text { (iii) }\\

Taking equation (ii)

\frac{1}{7}(4x - 5y) = x - 7\\

or, 7(x – 7) = 4x – 5y

or, 7x – 49 = 4x – 5y

or, 7x – 4x = 49 – 5y

or, 3x = 49 – 5y

∴ x = \frac{49-5 y}{3}-\text { (iv) }\\

Comparing equations (iii) and (iv)

\frac{7 y-3}{4}=\frac{49-5 y}{3}\\

or, 21y – 9 = 196 – 20y

or, 21y + 20y = 196 + 9

or, 41y = 205

or, y = \frac{\cancel{205}^{5}}{\cancel{41}} \\

y = 5

Now putting the value of y in equation (iii),

x = \frac{7 y-3}{4} \\

= \frac{7 \times 5-3}{4} \\

= \frac{35-3}{4} \\

=\frac{\cancel{32}^{8}}{\cancel{4}} \\

∴ x = 8

The required solution is

x = 8, y = 5

Question 5 (iv)

\frac{x+1}{y+1}=\frac{4}{5} \\

\frac{x-5}{y-5}=\frac{1}{2}\\

Solution:

\frac{x+1}{y+1}=\frac{4}{5}-\text { (i) }

\frac{x-5}{y-5}=\frac{1}{2}-\text { (ii) }

Taking equation (i)

\frac{x+1}{y+1}=\frac{4}{5}\\

or, 5x + 5 = 4y + 4

or, 5x = 4y + 4 – 5

or, 5x = 4y – 1

\therefore \quad x=\frac{4 y-1}{5}-\text { (iii) }\\

Taking equation (ii)

\frac{x-5}{y-5}=\frac{1}{2}\\

or, 2x – 10 = y – 5

or, 2x = y – 5 + 10

or, 2x = y + 5

∴ x = \frac{y+5}{2} - (\text{iv})\\

Comparing equations (iii) and (iv)

\frac{4 y-1}{5}=\frac{y+5}{2}\\

or, 8y – 2 = 5y + 25

or, 8y – 5y = 25 + 2

or, 3y = 27

or, y = \frac{\cancel{27}^{9}}{\cancel{3}}\\

∴ y = 9

Now putting the value of y in equation (iii),

x = \frac{4y-1}{5} \\

or, x = \frac{4 \times 9-1}{5} \\

= \frac{36-1}{5} \\

= \frac{\cancel{35}^{7}}{\cancel{5}} \\

∴ x = 7

The required solution is

x = 7, y = 9

Question 5 (v)

x+y=11

y+2=\frac{1}{8}(10 y+x)

Solution:

x + y = 11 — (i)

y + 2 = \frac{1}{8}(10 y+x) — (ii)

Taking equation (i)

x+y=11

x = 11 – y — (iii)

Taking equation (ii)

y + 2 = \frac{1}{8}(10 y+x)\\

or, 10y + x = 8(y + 2)

or, 10y + x = 8y + 16

or, x = 8y + 16 – 10y

∴ x = 16 – 2y — (iv)

Comparing equations (iii) and (iv)

11 – y = 16 – 2y

or, -y + 2y = 16-11

or, y = 5

Now putting the value of y in equation (iii),

∴ x = 11 – y

= 11 – 5

= 6

∴ x = 6

The required solution is

x = 6, y = 5

Question 5 (vi)

\frac{x}{3}+\frac{y}{4}=1 \\

2x+4y=11

Solution:

\frac{x}{3}+\frac{y}{4}=1 — (i)

2x + 4y = 11 — (i)

Taking equation (i)

\frac{x}{3}+\frac{y}{4}=1\\

or, \frac{x}{3}=1-\frac{y}{4}\\

or, \frac{x}{3}=\frac{4-y}{4}\\

∴ x= \frac{12-3 \mathrm{y}}{4}\\

Tạking equation (ii)

2x + 4y = 11

or, 2x = 11 – 4 y

∴ x = \frac{11-4 y}{2}-\text { (iv) }

Comparing equations (iii) and (iv)

\frac{12-3 y}{4}=\frac{11-4 y}{2}\\

or, 4 – 6y = 44 – 16 y

or, 16y – 6y = 44 – 24

or, 10y = 20

or, y=\frac{\cancel{20}^{2}}{\cancel{10}} \\

∴ y = 2

Now putting the value of y in equation (iii)

x = \frac{12-3 y}{4}\\

= \frac{12-3 \times 2}{4}\\

= \frac{12-6}{4}\\

= \frac{\cancel{6}^{3}}{\cancel{4}^{2}}\\

= \frac{3}{2} \\

∴ x = \frac{3}{2}\\

The required solution is

x =\frac{3}{2}, y=2

Question 5 (vii)

x + \frac{2}{y}=7 \\

2x – \frac{6}{y}=9

Solution:

x+\frac{2}{y}=7-(i) \\

2 x+\frac{6}{y}=9-\text { (ii) }

Taking equation (i)

x + \frac{2}{y}=7\\

or, x = 7 – \frac{2}{y}\\

∴ x = \frac{7 y-2}{y} \\ – (iii)

Taking equation (ii)

2x – \frac{6}{y}=9\\

or, 2x = 9+\frac{6}{y}\\

or, 2x =\frac{9 y+6}{y}\\

∴ x=\frac{9 y+6}{2 y} – (iv)

Comparing equations (iii) and (iv)

\frac{7 y-2}{\cancel{y}}=\frac{9 y+6}{2 \cancel{y}}[\because y \neq 0]\\

or, \frac{7 y-2}{1}=\frac{9 y+6}{2}

or, 14y – 4 = 9y + 6

or, 14y – 9y = 6 + 4

or, 5 y=10

or, y=\frac{10^{2}}{7}\\

∴ y = 2

Now putting the value of y in equation (iii),

x=\frac{7 y-2}{y} \\

=\frac{7 \times 2-2}{2} \\

=\frac{14-2}{2} \\=\frac{\cancel{12}^{6}}{\cancel{2}} =6 \\

∴ x = 6

The required solution is

x = 6, y = 2

(viii) \frac{1}{x}+\frac{1}{y}=\frac{5}{6} \\

\frac{1}{x}-\frac{1}{y}=\frac{1}{6}

Solution :

\frac{1}{x}+\frac{1}{y}=\frac{5}{6} — (\text{i})

\frac{1}{x}-\frac{1}{y}=\frac{1}{6} — (\text{ii})

Taking equation (i)

\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\\

or, \frac{1}{x}=\frac{5}{6}-\frac{1}{y}\\

or, \frac{1}{x}=\frac{5 y-6}{6 y}\\

\therefore \quad x=\frac{6 y}{5 y-6} - (\text{iii})

Taking equation (ii)

\frac{1}{x}-\frac{1}{y}=\frac{1}{6}\\

or, \frac{1}{x}=\frac{1}{6}+\frac{1}{y}\\

or, \frac{1}{x}=\frac{y+6}{6 y}\\

\therefore \quad x=\frac{6 y}{y+6}-\text { (iv) }

Compairing equations (iii) and (iv)

\frac{6 \cancel{y}}{5 y-6}=\frac{6 \cancel{y}}{y+6}[\because y \neq 0]\\

or, 5 y-6=y+6\\

or, 5 y-y=6+6\\

or, 4 y=12\\

or, y=\frac{\cancel{12}^{3}}{\cancel{4}}\\

Now putting the value of y in equation (iii),

x=\frac{6 y}{5 y-3} \\

=\frac{6 \times 3}{5 \times 3-3} \\

=\frac{18}{15-3} \\

=\frac{\cancel{18}^{3}}{\cancel{12}}

=\frac{3}{2} \\

The required solution is

x=\frac{3}{2}, y=3 \text { (Ans.) }

(ix)\frac{x+y}{x y}=2 \\

\frac{x-y}{x y}=1

Solution :

\frac{x+y}{x y}=2 — (\text{i})

\frac{x-y}{x y}=1 —(\text{ii})

Taking equation (i)

\frac{x+y}{x y}=2

or, 2 x y=x+y\\

or, 2 x y-x=y\\

or, x(2 y-1)=y\\

\therefore \quad x=\frac{\mathrm{y}}{2 \mathrm{y}-1}-\text { (iii) }\\

Taking equation (ii)

\frac{x-y}{x y}=1\\

or, x-y=x y\\

or, x-x y=y \\

or, x(1-y)=y\\

\therefore \quad x=\frac{y}{1-y}-\text { (iv) }\\

Compairing equations (iii) and (iv)

\frac{\cancel{y}}{2 y-1}=\frac{\cancel{y}}{1-y}[\because y \neq 0]\\

or, \frac{1}{2 y-1}=\frac{1}{1-y}\\

or, 2 y-1=1-y\\

or, 2 y+y=1+1\\

or, 3 y=2\\

Now putting the value of y in equation (iii),

x=\frac{y}{2 y-1} \\

=\frac{\frac{2}{3}}{2 \times \frac{2}{3}-1}\\

=\frac{\frac{2}{3}}{\frac{4-3}{3}} \\

=\frac{\frac{2}{3}}{\frac{1}{3}}

=\frac{2}{\cancel{3}} \times \frac{\cancel{3}}{1}=2 \\

The required solution is x=2, y=\frac{2}{3} \text { (Ans.) }

(x) \frac{x+y}{5}+\frac{x-y}{4}=5 \\

\frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}

Solution:

\frac{x+y}{5}+\frac{x-y}{4}=5 \\ — (\text{i})

\frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5} — (\text{ii})

Taking equation (i)

\frac{x+y}{5}+\frac{x-y}{4}=5\\

or, \frac{4 x+4 y+5 x-5 y}{20}=5 \\

or, \frac{9 x-y}{20}=5 \\

or, 9 x-y=100\\

or, 9 x=100+y\\

Taking equation (ii)

\frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}\\

or, \frac{5 x+5 y+4 x-4 y}{\cancel{20}_{4}}=\frac{29}{\cancel{5}}\\

or, 9 x+y=116\\

or, 9 x=116-y \\

\therefore \quad x=\frac{116-y}{9}-\text { (iv) }\\

Compairing equations (iii) and (iv)

\frac{100+y}{\cancel{9}}=\frac{106-y}{\cancel{9}}\\

or, 100+y=116-y\\

or, y+y=116-100\\

or, 2 y=16\\

or, y=\frac{\cancel{16}^{8}}{2} \\

Now putting the value of y in equation (iii),

x=\frac{100+y}{9} \\

=\frac{100+8}{9}

=\frac{\cancel{108}^{12}}{\cancel{9}}=12 \\

The required solution is

\left.\begin{array}{l}x=12 \\y=8\end{array}\right\}(Ans.)

(xi) \frac{4}{x}-\frac{y}{2}=-1 \\

\frac{8}{4}+2 y=10

Solution:

\frac{4}{x}-\frac{y}{2}=-1 — (\text{i})

\frac{8}{4}+2 y=10 — (\text{ii})

Taking equation (i)

\frac{4}{x}-\frac{y}{2}=-1\\

or, \frac{4}{x}=\frac{y}{2}-1\\

or, \frac{4}{x}=\frac{y-2}{2}\\

or, x(y-2)=8\\

Taking equation (ii)

\frac{8}{x}+2 y=10\\

or, \frac{8}{x}=10-2 y\\

or, \frac{x}{8}=\frac{1}{10-2 y}\\

\therefore \quad x=\frac{8}{10-2 y} - (\text{iv})\\

Compairing equations (iii) and (iv)

\frac{\cancel{8}}{y-2}=\frac{\cancel{8}}{10-2 y}\\

or, y-2=10-2 y\\

or, y+2 y=10+2 \\

or, 3 y=12 \\

or, y=\frac{\cancel{12}{4}}{\cancel{3}}\\

Now putting the value of y in equation (iii),

=\frac{8}{4-2} \\

=\frac{\cancel{8}^{4}}{\cancel{2}} \\

The required solution is

x=4, y=4 (Ans.)

(xii) 2-2(3 x-y)=10(4-y)-5 x=4 x(y-x)

Solution:

2-2(3 x-y)=4(y-x)-\text { (i) } \\

10(4-y)-5 x=4(y-x)-\text { (ii) }

Taking equation (i)

2-2(3 x-y)=4(y-x)\\

or, 2-6 x+2 y=4 y-4 x\\

or, -6 x+4 x=4 y-2 y-2\\

or, -2 x=2 y-2\\

or, +2 x=+2(1-y)\\

Taking equation (ii)

10(4-y)-5 x=4(y-x) \\

or, 40-10 y-5 x=4 y-4 x \\

or, -5 x+4 x=4 y+10 y-40 \\

or, -x=14-40 \\

\therefore \quad x=40-14 y-\text { (iv) }

Compairing equations (iii) and (iv)

1-y=40-14 y \\

or, -y+14 y=40-1 \\

or, 13 y=39 \\

or, y=\frac{\cancel{39}^{3}}{\cancel{13}} \\

Now putting the value of y in equation (iii),

=1-3 \\

=-2 \\

The required solution is

x=-2, y=3 (Ans.)

## Let us work out 5.5

1. Let us express x of the equation \frac{2}{x}+\frac{3}{y}=1 in term of the variable y.

Solution:

\frac{2}{x}+\frac{3}{y}=1 \\

or, \frac{2}{x}=1-\frac{3}{y} \\

or, \frac{2}{x}=\frac{3-y}{y}\\

or, \frac{x}{2}=\frac{y}{y-3}\\

\therefore \quad x=\frac{2 y}{y-3} \text { (Ans.) }\\

2. Let us write the value of x by putting \frac{7-4 x}{-5} instead of y in the equation 2 x+3 y=9

Solution: We have

2 x+3 y=9\\

or, 2 x+3 \cdot\left(\frac{7-4 x}{-5}\right)=9\left[\because y=\frac{7-4 x}{-5}\right]\\

or, 2 x-\frac{21-12 x}{5}=9\\

or, \frac{10 x-21+12 x}{5}=9\\

or, \frac{22 x-21}{5}=\frac{9}{1}\\

or, 22 x-21=45\\

or, 22 x=45+21\\

or, 22 x=66\\

or, x=\frac{\cancel{66}^{3}}{\cancel{22}}\\

3. Let us the following equations in two variables by substitution method and check them graphically.

2 x+4 y=0\\

Solution : (a)

The given equations are

3 x-y=7-\text { (i) }

2 x+4 y=0-\text { (ii) }

Taking equation (i)

3 x-y=7\\

or, 3 x=7+y\\

\therefore \quad x=\frac{7+y}{3}-\text { (iii) }\\

Putting the value of x is equation (ii)

2 x+4 y=0\\

or, 2 \cdot\left(\frac{7+y}{3}\right)+4 y=0\\

or, \frac{14+2 y}{3}+4 y=0\\

or, \frac{14+2 y+12 y}{3}=0\\

or, \frac{14+14 y}{3}=0\\

or, 14+14 y=0\\

or, 14 y=-14\\

or, y=\frac{-\cancel{14}}{\cancel{14}}\\

Now putting the value of y in equation (iii),

or, x =\frac{7+(-1)}{3} \\

=\frac{7-1}{3}

=\frac{\cancel{6}^{2}}{\cancel{3}}=2 \\

By graphically,

Taking equation (i) Taking equation (ii)

3x - 7 = 7\\

or, 3 x=7+y\\

 x 2 3 4 y -1 2 5

The required solution is

\left.\begin{array}{c}\therefore x=2 \\y=-1\end{array}\right\} \text { Ans }

Taking equation (i) Taking equation (ii)

2 x+4 y=0\\

or, 2 x=-4 y\\

or, x=\frac{\cancel{-4}^{2} y}{\cancel{2}}\\

\therefore x=-2 y\\

 x 2 0 -8 y -1 0 4

Solution : (b)

\frac{x}{2}+\frac{y}{3}=2-(\text{i})

\frac{x}{4}+\frac{y}{2}=2-(\text{ii})

Taking equation (i)

\frac{x}{2}+\frac{y}{3}=2\\

or, \frac{x}{2}=2-\frac{y}{3}\\

or, \frac{x}{2}=\frac{6-y}{3}\\

or, x=\frac{2(6-y)}{3}\\

\therefore \quad x=\frac{12-2 y}{3}-\text { (iii) }\\

Now putting the value of x in equation (ii),

\frac{x}{4}+\frac{y}{2}=2\\

or, \frac{1}{4} \cdot\left(\frac{12-2 y}{3}\right)+\frac{y}{2}=2\\

or, \frac{12-2 y}{12}+\frac{y}{2}=2\\

or, \frac{12-2 y+6 y}{12}=2\\

or, \frac{12+4 y}{12}=2\\

or, 12+4 y=24\\

or, 4 y=24-12\\

or, 4 y=12 \\

or, y=\frac{\cancel{12}^{3}}{\cancel{4}}\\

\therefore \quad y=3-\text { (iv) }\\

Putting the value of y in equation (iii),

x =\frac{12-2 y}{3} \\

=\frac{12-2 \times 3}{3} \\

=\frac{12-6}{3}=\frac{\cancel{6}^{2}}{\cancel{3}}=2 \\

The required solution is

x=2, y=3 (Ans.)

By graphically,

Taking equation (i)

\frac{x}{2}+\frac{y}{3}=2\\

or, \frac{3 x+2 y}{6}=2\\

or, 3 x+2 y=12\\

 x 4 2 0 y 0 3 6

Taking equation (ii)

\frac{x}{4}+\frac{y}{2}=2\\

or, \frac{x+2 y}{4}=2\\

or, x+2 y=8\\

 x 8 2 -4 y 0 3 6

4. Let us solve the following equations in two variables by substitution method and check whether the solutions satisfy the equations.

(a)2 x+\frac{3}{y}=1 \\

5 x-\frac{2}{y}=\frac{11}{12}\\

(b) \frac{2}{x}+\frac{3}{y}=2\\

\frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\\

(c)\frac{x+y}{x y}=3 \\

\frac{x-y}{x y}=1\\

(d) \frac{x+y}{x-y}=\frac{7}{3}\\

x+y=\frac{7}{10}\\

Solution : (a)

The given equations are

2 x+\frac{3}{y}=1-\text {(i) } \\

5 x-\frac{2}{y}=\frac{11}{12}-\text { (ii) }\\

Taking equation (i)

2 x+\frac{3}{y}=1\\

or, 2 x=1-\frac{3}{y} \\

or, 2 x=\frac{y-3}{y}\\

\therefore \quad x=\frac{y-3}{2 y}-\text { (iii) }\\

Now putting the value of x in equation (ii)

5 x-\frac{2}{y}=\frac{11}{12}\\

or, 5\left(\frac{y-3}{2 y}\right)-\frac{2}{y}=\frac{11}{12}\\

or, \frac{5 y-15}{2 y}-\frac{2}{y}=\frac{11}{12}\\

or, \frac{5 y-15-4}{2 y}=\frac{11}{12}\\

or, \frac{5 y-19}{2 y}=\frac{11}{12}\\

or, 60 y-228=22 y\\

or, 60 y-22 y=228\\

or, 38 y=228\\

or, y=\frac{\cancel{278}^{6}}{\cancel{38}} \\

Now putting the value of y in equation (iii),

x=\frac{y-3}{2 y} \\

=\frac{6-3}{2 \times 6}=\frac{\cancel{3}}{\cancel{12}^4} =\frac{1}{4} \\

x=\frac{1}{4}\\

The required solution is

The required solution is

x=\frac{1}{4}, y=6 \text { (Ans.) }\\

Solution : (b)

The given equations are

\frac{2}{x}+\frac{3}{y}=2-\text { (i) } \\

\frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}-\text { (ii) } \\

Taking equation (i)

\frac{2}{x}+\frac{3}{y}=2\\

or, \frac{2}{x}=2-\frac{3}{y}\\

or, \frac{2}{x}=\frac{2 y-3}{y}\\

or, \frac{x}{2}=\frac{y}{2y-3}\\

Now putting the value of x in equation (ii),

\frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\\

or, \frac{5}{\frac{2 y}{2 y-3}}+\frac{10}{y}=\frac{35}{6}\\

or, \frac{5(2 y-3)}{2 y}+\frac{10}{y}=\frac{35}{6}\\

or, \frac{10 y-15}{2 y}+\frac{10}{y}=\frac{35}{6}\\

or, \frac{10 y-15+20}{2 y}=\frac{35}{6}\\

or, \frac{10 y+5}{2 y}=\frac{35}{6}\\

or, \frac{\cancel{5}(2 y+1)}{\cancel{2y}}=\frac{\cancel{35}^{7}}{\cancel{6}_{3}}\\

or, \frac{2 y+1}{y}=\frac{7}{3}\\

or, 7 y=6 y+3\\

or, 7 y-6 y=3\\

Now putting the value of y in equation (iii),

=\frac{2y}{2 y-3}\\

or, x=\frac{2 \times 3}{2 \times 3-3} \\

=\frac{6}{6-3}=\frac{\cancel{6}^{2}}{\cancel{3}}=2 \\

The required solution is

x=2, y=6 \text { (Ans.) }

Solution : (c)

\frac{x+y}{x y}=3-(i) \\

\frac{x-y}{x y}=3-(ii)\\

Taking equation (i)

\frac{x+y}{x y}=3\\

or, x+y=3 x y\\

or, x-3 x y=-y\\

or, x(1-3 y)=-y \\

\therefore \quad x=\frac{-y}{1-3 y}-\text { (iii) }\\

Now putting the value of x in equation (ii),

\frac{x-y}{x y}=1\\

or, x-y=x y\\

or, \frac{-y}{1-3 y}-y=\frac{-y}{1-3 y} \cdot y\\

or, \frac{-1}{1-3 y}-1=\frac{-1}{1-3 y} \cdot y\\

or, \frac{-1-1+3 y}{1-3 y}=\frac{-y}{1-3 y}\\

or, -2+3 y=-y\\

or, 3 y+y=2\\

or, 4 y=2\\

or, y=\frac{\cancel{2}^{1}}{\cancel{4}_{2}}=\frac{1}{2}\\

Now putting the value of y in equation (iii),

x=\frac{-y}{1-3 y} \\

=\frac{-\frac{1}{2}}{1-3 \times \frac{1}{2}} \\

=\frac{-\frac{1}{2}}{1-\frac{3}{2}}\\

=\frac{-\frac{1}{2}}{\frac{2-3}{2}} \\

=\frac{\cancel{-}\frac{1}{\cancel{2}}}{\frac{\cancel{-}1}{\cancel{2}}}=1 \\

The required solution is

x=1, y=\frac{1}{2} (Ans.)

Solution : (d)

\frac{x+y}{x-y}=\frac{7}{3}-(i) \\

x+y=\frac{7}{10}-(ii)\\

Taking equation (i)

\frac{x+y}{x-y}=\frac{7}{3}\\

or, 7 x-7 y=3 x+3 y\\

or, 7 x-3 x=3 y+7 y\\

or, 4 x=10 y\\

or, x=\frac{\cancel{10}y^{5}}{\cancel{4}_{2}}\\

\therefore \quad x=\frac{5 y}{2}- (\text{iii}) \\

Now putting the value of x in equation (ii),

x+y=\frac{7}{10}\\

or, \frac{5 y}{2}-y=\frac{7}{10}\\

or, \frac{7y}{\cancel{2}}=\frac{7}{\cancel{10}_{5}}\\

or, 5y=1\\

Now putting the value of y in equation (iii),

x =\frac{5 y}{2} \\

=\frac{\cancel{5}}{2} \times \frac{1}{\cancel{5}} \\

=\frac{1}{2} \\

The required solution is

x=\frac{1}{2}, y=\frac{1}{5} (Ans.)\\

5. Let us solve the following equations in two variables by substitution method.

(i) 2(x-y)=3\\

5 x+8 y=14\\

(ii) 2 x+\frac{3}{y}=5\\

5 x-\frac{2}{y}=3\\

(iii) \frac{x}{2}+\frac{y}{3}=1\\

\frac{x}{3}+\frac{y}{2}=1\\

(iv) \frac{x}{3}=\frac{y}{4}\\

7 x-5 y=2\\

(v)\frac{2}{x}+\frac{5}{y}=1 \\

\frac{5}{x}-\frac{2}{y}=\frac{19}{20}\\

(vi) \frac{1}{3}(x-y)=\frac{1}{4}(y-1)\\

\frac{1}{7}(4 x-5 y)=x-7\\

(vii) \frac{x}{14}+\frac{y}{18}=1\\

\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\\

(viii) p(x+y)=q(x-y)=2 p q\\

Solution : (i)

2(x-y)=3-(i)

5 x+8 y=14-(ii)

Taking equation (i),

2(x-y)=3\\

or, 2 x-2 y=3\\

or, 2 x=3+2 y\\

\therefore \quad x=\frac{3+2 y}{2} - (\text{iii})\\

Now putting the value of x in equation (ii),

5 x+8 y=14\\

or, 5 \cdot\left(\frac{3+2 y}{2}\right)+8 y=14\\

or, \frac{15+10 y}{2}+8 y=14\\

or, \frac{15+10 y+16 y}{2}=14\\

or, \frac{15+26 y}{2}=14\\

or, 15+26 y=28\\

or, 26 y=28-15\\

or, 26 y=13\\

Now putting the value of y in equation (iii),

x=\frac{3-2 y}{2} \\

=\frac{3+7 \times \frac{1}{2}}{2} \\

=\frac{3+1}{2} \\

=\frac{\cancel{4}^{2}}{\cancel{2}} \\

The required solution is

x=2, y=\frac{1}{2} \text { (Ans.) }

Solution : (ii)

2 x+\frac{3}{y}=5-(\text {i) }

5 x-\frac{2}{y}=3-\text { (ii) }

Taking equation (i),

2 x+\frac{3}{y}=5\\

or, 2 x=5-\frac{3}{y}

or, 2 x=\frac{5 y-3}{y}\\

Putting the value of x in equation (ii),

5 x-\frac{2}{y}=3\\

or, 5 \cdot\left(\frac{5 y-3}{2 y}\right)-\frac{2}{y}=3\\

or, \frac{25 y-15}{2 y}-\frac{2}{y}=3\\

or, \frac{25 y-15-4}{2 y}=3\\

or,\frac{25 y-19}{2 y}=3\\

or, 25 y-19=6 y \\

or, 25 y-6 y=19 \\

or, 19 y=19 \\

or, y=\frac{\cancel{19}^{1}}{\cancel{19}}\\

Now putting the value of y in equation (iii),

x=\frac{5 y-3}{2 y} \\

=\frac{5 \times 1-3}{2 \times 1}\\

=\frac{5-3}{2}=\frac{\cancel{2}^{1}}{\cancel{2}}=1 \\

The required solution is

x=1, y=1 \text { (Ans.) }\\

Solution : (iii)

\frac{x}{2}+\frac{y}{3}=1 \\

\frac{x}{3}+\frac{y}{2}=1\\

Taking equation (i),

\frac{x}{2}+\frac{y}{3}=1\\

or, \frac{x}{2}=1-\frac{y}{3}\\

or, \frac{x}{2}=\frac{3-y}{3}\\

Putting the value of x in equation (ii),

\frac{x}{3}+\frac{y}{2}=1\\

or, \frac{1}{3}\left(\frac{6-2 y}{3}\right)+\frac{y}{2}=1\\

or, \frac{6-2 y}{9}+\frac{y}{2}=1\\

or, \frac{12-4 y+9 y}{18}=1 \\

or, \frac{12-5 y}{18}=1\\

or, 12-5 y=18\\

or, 5 y=18-12\\

or, 5 y=6\\

Now putting the value of y in equation (iii),

x =\frac{6-2 y}{3} \\

=\frac{6-2 \times \frac{6}{5}}{3} \\

=\frac{6-\frac{12}{5}}{3} \\

=\frac{\frac{30-12}{5}}{3} \\

=\frac{\frac{\cancel{18}^{6}}{5}}{\cancel{3}} \\

The required solution is

x=\frac{6}{5}, y=\frac{6}{5} (Ans.)

Solution : (iv)

\frac{x}{3}=\frac{y}{4}-(i) \\

7 x-5 y=2-(ii)

Taking equation (i),

or, \frac{x}{3}=\frac{y}{4}\\

Putting the value of x in equation (ii),

7 x-5 y=2\\

or, 7 \times \frac{3 y}{4}-5 y=2\\

or, \frac{21 y}{4}-5 y=2\\

or, \frac{21 y-20 y}{4}=2\\

or, \frac{y}{4}=2 \\

Now putting the value of y in equation (iii),

x=\frac{3 y}{4} \\

=\frac{3 \times \cancel{8}^{2}}{\cancel{4}} \\

=6 \\

The required solution is

x=6, y=8 (Ans.)

Solution : (v)

\frac{2}{x}+\frac{5}{y}=1-(i) \\

\frac{3}{x}+\frac{2}{y}=\frac{19}{20}-(ii)

Taking equation (i),

\frac{2}{x}-\frac{5}{5}=1\\

or, \frac{2}{x}=1-\frac{5}{y}\\

or, \frac{2}{x}=\frac{y-5}{y}\\

or, \frac{x}{2}=\frac{y}{y-5}\\

Putting the value of x in equation (ii),

\frac{\frac{3}{2 y}}{y-5}+\frac{2}{y}=\frac{19}{20}\\

or, \frac{3(y-5)}{2 y}+\frac{2}{y}=\frac{19}{20}\\

or, \frac{3 y-15+4}{\cancel{2 y}}=\frac{19}{\cancel{20}_{10}}\\

or, \frac{3 y-11}{y}=\frac{19}{10}\\

or, 30 y-110=19 y\\

or, 30 y-19 y=110 \\

or, 11 y=110\\

or, y=\frac{\cancel{110}^{10}}{\cancel{11}}\\

Now putting the value of y in equation (iii),

x=\frac{2 y}{y-5} \\

=\frac{2 \times 10}{10-5} \\

=\frac{\cancel{20}^{4}}{\cancel{5}}=4 \\

The required solution is

x=4, y=10 \text { (Ans.) }\\

Solution : (vi)

\frac{1}{3}(x-y)=\frac{1}{4}(y-1)-(i)\\

or, \frac{1}{7}(4 x-5 y)=x-7- (ii)\\

Taking equation (ii),

\frac{1}{7}(4 x-5 y)=x-7\\

or, 7 x-49=4 x-5 y\\

or, 7 x-4 x=49-5 y\\

or, 3 x=49-5 y \\

Putting the value of x in equation (i),

\frac{1}{3}(x-y)=\frac{1}{4}(y-1)\\

i. e. \quad 4 x-4 y=3 y-3\\

or, 4\left(\frac{49-5 y}{3}\right)-4 y=3 y-3\\

or, \frac{196-20 y}{3}-4 y=3 y-3\\

or, \frac{196-20 y-12 y}{3}=3 y-3\\

or, \frac{196-32 y}{3}=3 y-3\\

or, 9 y-9=196-32 y\\

or, 9 y+32 y=196+9\\

or, 41 y=205\\

or, y=\frac{\cancel{205}^{5}}{\cancel{41}}\\

Now putting the value of y in equation (iii),

x=\frac{49-5 y}{3} \\

=\frac{49-5 \times 5}{3} \\

=\frac{49-25}{3} \\

=\frac{24^{8}}{7}=8 \\

The required solution is

x=8, y=5 \text { (Ans.) }

Solution: (vii)

\frac{x}{14}-\frac{y}{18}=1\text { (i) } \\

\frac{x+y}{2}+\frac{3 x-5 y}{4}=2-\text { (ii) }

Taking equation (i),

\frac{x}{14}+\frac{y}{18}=1\\

or, \frac{9 x+7 y}{126}=1\\

or, 9 x+7 x=126\\

or, 9 x=126-7 y\\

\therefore \quad x=\frac{126-7 y}{9} - (\text{iii})\\

Taking equation (ii),

\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\\

or, \frac{2 x+2 y+3 x-5 y}{4}=2\\

or, \frac{5 x-3 y}{4}=2\\

\therefore \quad 5 x-3 y=8-\text { (iv) }\\

Putting x=\frac{126-7 y}{9} in equation (iv),

\text { {5} . }\left(\frac{126-7 y}{9}\right)-3 y=8\\

or, \frac{630-35 y}{9}-3 y=8\\

or, \frac{630-35y-27y}{9}=8 \\

or, \frac{630-62 \mathrm{y}}{9}=8\\

or, 72=630-62 y\\

or, 62 y=630-72\\

or, 62 y=558\\

or, y=\frac{\cancel{558}^{9}}{\cancel{62}_{1}}\\

Now putting the value of y in equation (iii),

x=\frac{126-7 y}{9} \\

=\frac{126-7 \times 9}{9} \\

=\frac{126-63}{9} \\

=\frac{\cancel{63}^{7}}{\cancel{9}}=7 \\

The required solution is

x=7, y=9 \text { (Ans:) }

Solution: (viii)

p(x+y)=q(x-y)=2 p q\\

i. e. \quad p(x+y)=2 p q, q(x-y)=2 p q\\

i. e. \quad x+y=2 q-\text { (i) }\\

x-y=2 p-\text { (ii) }\\

Taking equation (i),

or, x+y=2 q

\therefore \quad x=2 q-y-\text { (iii) }

Putting the value of x in equation (ii),

x-y=2 p\\

or, 2 q-y-y=2 p\\

or, 2 q-2 y=2 p\\

or, 2 y=2 q-2 p\\

or, 2 x=2(q-p)\\

Now putting the value of y in equation (iii),

x=2 q-y \\

= 2 q-(q-p) \\

= 2 q-q+p \\

=p+q \\

The required solution is

x=p+q, y=q-p (Ans.)

## Let us work out – 5.6

1. Let us solve the following linear equations in two variables by applying cross-multiplication method :

(1) 8 x+5 y=11 \\

3 x-4 y=10

Solution :

8 x+5 y=11 \\

3 x-4 y=10 \\

i.e. \quad 8 x+5 y-11=0 \\

3 x-4 y-10=0

By cross multipliation,

\frac{x}{(5)(-10)-(-4)(-11)} =\frac{y}{(-11)(3)-(8)(-10)} =\frac{1}{(8)(-4)-(3)(5)}\\

or, \frac{x}{-50-44}=\frac{y}{-33+80} =\frac{1}{-32-15}\\

or, \frac{x}{-94}=\frac{y}{47}=\frac{1}{-47}\\

The required solution is

x=2, y=-1 \text { (Ans.) }

(2) 3 x-4 y=1

4 x=3 y+6

Solution :

3 x-4 y=1 \\

4 x=3 y+6 \\

3 x-4 y-1=0 \\

4 x-3 y-6=0\\

4 x-3 y-6=0\\

By cross multipliation,

or, \frac{x}{(-4)(-6)-(-1)(-3)}=\frac{y}{(-1)(4)-(3)(-6)} \\=\frac{1}{(3)(-3)-(4)(-4)}\\

or, \frac{x}{24-3}=\frac{y}{-4+18}=\frac{1}{-9+16}\\

or, \frac{x}{21}=\frac{y}{14}=\frac{1}{7}\\

The required solution is

x=3, y=2 \text { (Ans.) }

(3) 5 x+3 y=11 \\

2 x-7 y=-12

Solution :

5 x+3 y=11 \\2 x-7 y=-12 \\

\text { i. e. } \quad 5 x+3 y-11=0 \\2 x-7 y+12=0 \\

By cross multipliation,

or,\frac{x}{(3)(12)-(-7)(-11)}=\frac{y}{(-11)(2)-(5)(12)} =\frac{1}{(5)(-7)-(3)(2)}\\

or, \frac{x}{36-77}=\frac{y}{-22-60}=\frac{1}{-35-6}\\

or, \frac{x}{-41}=\frac{y}{-82}=\frac{1}{-41}\\

The required solution is

x=1, y=2 \text { (Ans.) }

(4) 7 x-3 y-31=0 \\

9 x-5 y-41=0

Solution :

7 x-3 y-31=0

9 x-5 y-41=0

By cross multipliation,

or,\frac{x}{(-3)(-41)-(-5)(-31)}=\frac{y}{(-31)(9)-(7)(-41)} =\frac{1}{(7)(-5)-(9)(-3)}\\

or, \frac{x}{123-155}=\frac{y}{-279+287}=\frac{1}{-35+27} \\

or, \frac{x}{-32}=\frac{y}{8}=\frac{1}{-8}\\

The required solution is

x=4, y=-1 \text { (Ans.) }

(5) \frac{x}{6}-\frac{y}{3}=\frac{x}{12}-\frac{2 y}{3}=4

Solution :

\frac{x}{6}-\frac{y}{3}=\frac{x}{12}-\frac{2 y}{3}=4 \\

\therefore \quad \frac{x}{6}-\frac{y}{3}=4, \frac{x}{12}-\frac{2 y}{3}=4 \\

or, \quad \frac{x-2 y}{6}=4, \frac{x-8 y}{12}=4\\

or, x-2y = 24, x-8y=48

\therefore \quad x-2 y-24=0 \\x-8 y-48=0

By cross multipliation,

or,\frac{x}{(-2)(-48)-(-8)(-24)}=\frac{y}{(-24)(1)-(-48)(1)}=\frac{1}{(1)(-8)-(-2)(1)}\\

or, \frac{x}{96-192}=\frac{y}{-24+48}=\frac{1}{-8+2}\\

or, \frac{x}{-96} =\frac{y}{24}=\frac{1}{-6}\\

The required solution is

x=16, y=-4 \text { (Ans.) }

(6) \frac{x}{5}-\frac{y}{3}=\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0

Solution :

\frac{x}{5}+\frac{y}{3}=0 \\

\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0

By cross multiplication,

or,\frac{x}{\frac{1}{3} \times\left(\frac{-3}{20}\right)-\left(-\frac{1}{3}\right) \times 0}=\frac{y}{0 \times \frac{1}{4}-\frac{1}{5}\left(-\frac{3}{20}\right)}=\frac{1}{\left(\frac{1}{5}\right)\left(-\frac{1}{3}\right)-\left(\frac{1}{3}\right)\left(\frac{1}{4}\right)}

or, \frac{x}{-\frac{1}{20}-0}=\frac{y}{0+\frac{3}{100}}=\frac{1}{-\frac{1}{15}-\frac{1}{12}}\\

or, \frac{x}{-\frac{1}{20}}=\frac{y}{\frac{3}{100}}=\frac{1}{\frac{-4-5}{60}}\\

or, \frac{x}{-\frac{1}{20}}=\frac{y}{\frac{3}{100}}=\frac{1}{\frac{-9}{60}}\\

or, -\frac{1}{20} \times \frac{-60^{3}}{9_{3}}, \quad y=\frac{3}{100}\times \frac{60^{4}}{-9}=-\frac{1}{5}\\

The required solution is

x=\frac{1}{3}, y=-\frac{1}{5} (Ans.)

(7) \frac{x+2}{7}+\frac{y-x}{4}=2 x-8 \\

\frac{2 y-3 x}{3}+2 y=3 x+4

Solution :

\frac{x+2}{7}+\frac{y-x}{4}=2 x-8

or, \frac{4 x+8+7 y-7 x}{28}=2 x-8\\

or, \frac{7 y-3 x+8}{28}=2 x-8\\

or, 56 x-224=7 y-3 x+8 \\

or, 56 x-224-7 y+3 x-8=0 \\

or, 59 x-7 y-232=0 \\

Again,\frac{2 y-3 x}{3}+2 y=3 x+4\\

or, \frac{2 y-3 x+6 y}{3}=\frac{3 x+4}{1}\\

or, \frac{8 y-3 x}{3}=\frac{3 x+4}{1}\\

or, 9 x+12=8 y-3 x\\

or, 9 x+12-8 y+3 x=0 \\

or, 12 x-8 y+12=0 \\

12 x-8 y+12=0 \\

By cross multipliation,

or,\frac x{(-7)(12)-(-8)(-232)} =\frac y{(-12)(-232)-(59)(12)} =\frac{1}{(59)(-8)-(12)(-7)}\\

or, \frac{x}{-84-1856}=\frac{y}{-2784-708}=\frac{1}{-472+84}\\

or, \frac{x}{-1940}=\frac{y}{-3492}=\frac{1}{-388}\\

or, x=\frac{-\cancel{1940}^{5}}{-\cancel{388}}, y=\frac{-\cancel{3792}^{9}}{-\cancel{388}}\\

The required solution is

x=5, y=9 \text(Ans.)

(8) x+5 y=36 \\

\frac{x+y}{x-y}=\frac{5}{3}

Solution :

x+5 y=36-(\text {i) } \\

\frac{x+y}{x-y}=\frac{5}{3}-(\text {ii) }\\

Taking equation (ii)

\frac{x+y}{x-y}=\frac{5}{3}\\

or, 5 x-5 y=3 x+3 y\\

or, 5 x-5 y-3 x-3 y=0\\

or, 2 x-8 y=0\\

\therefore \quad x+5 y-36=0 \\2 x+8 y+0=0\\

By cross multipliation,

or, \frac{x}{(5)(0)-(-36)(-8)}=\frac{y}{(-36)(2)-(0)(1)} =\frac{1}{(1)(-8)-(5)(2}\\

or, \frac{x}{0-288}=\frac{y}{-72+0}=\frac{1}{-8-10}\\

or, \frac{x}{-288}=\frac{y}{-72}=\frac{1}{-18}\\

The required solution is

x=16, y=4 \text { (Ans.) }

(9) 13 y-12 y+15=0 \\

8 x-7 y=0

Solution :

13 y-12 y+15=0

8 x-7 y=0

By cross multiplication,

or,\frac{x}{(-12) \times 0-(-7)(15)} =\frac{y}{(15) \times(8)-(13)(0)} =\frac{1}{(13) \times(-7)-(8)(-12)}\\

or, \frac{x}{0+105}=\frac{y}{120-0}=\frac{1}{-91+96}\\

or, \frac{x}{105}=\frac{y}{120}=\frac{1}{5}\\

or, \frac{x}{105}=\frac{y}{120}=\frac{1}{5}\\

The required solution is x=21, y=24 (Ans.)

(10) x+y=2b\\

x-y=2 a

Solution :

x+y=2 b

x-y=2 a \\

\text { i. e. } \quad x+y-2 b=0 \\x-y-2 a=0

By cross multiplication,

or,\frac{x}{(1) \times(-2 a)-(-1)(-2 b)}=\frac{y}{(-2 b) \times(1)-(1)(-2 a)}=\frac{1}{(1) \times(-1)-(-1)(1)}\\

or, \frac{x}{-2 a-2 b}=\frac{y}{-2 b+2 a}=\frac{1}{-1-1}\\

or, \frac{x}{-\cancel{2}(a+b)}=\frac{y}{\cancel{2}(a-b)}=\frac{1}{-\cancel{2}}\\

or, \frac{x}{-(a+b)}=\frac{y}{a-b}=\frac{1}{-1}\\

The required solution is

x=a+b, y=b-a \text { (Ans.) }

(11) x-y=2 a \\

a x+b y=a^{2}+b^{2}

Solution :

x-y=2 a \\a x+b y=a^{2}+b^{2} \\

\text { i. e. } \quad x+y-2 a=0 \\

a x+b y-\left(a^{2}+b^{2}\right)=0\\

By cross multiplication,

or,\frac{x}{(-1)\left\{-\left(a^{2}+b^{2}\right)\right\}-(-2 a)(b)}=\frac{y}{(-2 a)(a)-\left\{-\left(a^{2}+b^{2}\right)\right\}\{1\}}=\frac{1}{(1)(b)-(-1)(a)}\\

or, \frac{x}{a^{2}+b^{2}+2 a b}=\frac{y}{-2 a^{2}+a^{2}+b^{2}}=\frac{1}{b+a}\\

or, \frac{x}{(a+b)^{2}}=\frac{y}{b^{2}-a^{2}}=\frac{1}{a+b}\\

or, \frac{x}{(a+b)^{2}}=\frac{y}{(b+a)(b-a)}=\frac{1}{a+b} \\

or, \frac{x}{a+b}=\frac{y}{b-a}=\frac{1}{1}\\

The required solution is

x=a+b, y=b-a (Ans.)

(12) {x}{a}+\frac{y}{b}=2 \\

a x-b y=a^{2}-b^{2}

Solution:

\frac{x}{a}+\frac{y}{b}=2 \\

\frac{b x+a y}{a b}=2\\

or, \frac{b x+a y}{a b}=2\\

or, b x+a y=2 a b \\

\therefore \quad b x+a y-2 a b=0 \\

\therefore \quad a x-b y-\left(a^{2}-b^{2}\right)=0 \\

\text{Again}, bx+a y-2 a b=0 \\

By cross multiplication,

\frac{x}{a x\left\{-\left(a^{2}-b^{2}\right)\right\}-(-b) \times(-2 a b)}=\frac{y}{(-2 a b) \times a-\left\{-\left(a^{2}-b^{2}\right)\right\} \times b}=\frac{1}{b} \times(-b)-a \times {a}

or, \frac{x}{-a\left(a^{2}-b^{2}\right)-2 a b^{2}}=\frac{y}{-2 a^{2} b+a^{2} b-b^{3}}=\frac{1}{-b^{2}-a^{2}}\\

or, \frac{x}{-a^{3}+a b^{2}-2 a b^{2}}=\frac{y}{-2 a^{2} b+a^{2} b-b^{3}}=\frac{1}{-a^{2}-b^{2}}\\

or, \frac{x}{-a^{3}-a b^{2}}=\frac{y}{a^{2} b-b^{3}}=\frac{1}{-a^{2}-b^{2}}\\

or, \frac{x}{-a\left(a^{2}+b^{2}\right)}=\frac{y}{b\left(a^{2}+b^{2}\right)}=\frac{1}{-\left(a^{2}+b^{2}\right)} \\

\therefore \quad x=\frac{-a\left(a^{2}+b^{2}\right)}{-\left(a^{2}+b^{2}\right)}=a \text { and } y \frac{-b\left(a^{2}+b^{2}\right)}{-\left(a^{2}+b^{2}\right)}=b

The required solution is

x=a, y=b (Ans.)

(13) a x+b y=1 \\

b x+a y=\frac{2 a b}{a^{2}+b^{2}}

Solution :

a x+b y=1 \\

\therefore \quad a x+b y-1=0 \\

b x+a y=\frac{2 a b}{a^{2}+b^{2}} \\

\therefore \quad b x+a y-\frac{2 a b}{a^{2}+b^{2}}=0\\

Again, ax + by -1=0\\

b x+a y-\frac{2 a b}{a^{2}+b^{2}}=0\\

By cross multiplication,

or,\frac{x}{b x\left(-\frac{2 a b}{a^{2}+b^{2}}\right)-a \times(-1)}=\frac{y}{(-1) \times b-\left(-\frac{2 a b}{a^{2}+b^{2}}\right) \times a} = \frac{1}{a \times a - b \times b }\\

or, \frac{x}{\frac{-2 a b^{2}}{a^{2}+b^{2}}+a}=\frac{y}{-b+\frac{2 a b^{2}}{a^{2}+b^{2}}}=\frac{1}{a^{2}-b^{2}}

or, \frac{x}{\frac{-2 a b^{2}+a^{3}+a b^{2}}{a^{2}+b^{2}}}=\frac{y}{\frac{a^{2} b-b^{3}+2 a^{2} b}{a^{2}+b^{2}}}=\frac{1}{a^{2}-b^{2}}\\

or, \frac{x}{\frac{a^{3}-a b^{2}}{a^{2}+b^{2}}}=\frac{y}{\frac{a^{2} b-b^{3}}{a^{2}+b^{2}}}=\frac{1}{a^{2}-b^{2}}\\

or, x=\frac{a^{3}-a b^{2}}{a^{2}+b^{2}} \times \frac{1}{a^{2}-b^{2}}=\frac{a\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}} \times \frac{1}{a^{2}-b^{2}}=\frac{a}{a^{2}+b^{2}}\\

\text{and} \quad y=\frac{a^{2} b-b^{3}}{a^{2}+b^{2}} \times \frac{1}{a^{2}-b^{2}} =\frac{a\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}} \times \frac{1}{a^{2}-b^{2}}=\frac{b}{a^{2}+b^{2}}\\

\therefore \quad The required solution is

x =\frac{a}{a^{2}+b^{2}}, y=\frac{b}{a^{2}+b^{2}} (Ans.)

## Let us work out – 5.7

1. My friend Rita has bought 5 pens and 3 pencils at ₹ 34 from the book-shop nearby our school. But Sumita hs bought 7 pens and 6 pencils at ₹ 53 at the same rate and from the same shop. I write by calculating the price of each pen and pencil by framing simultaneous equations.

Solution :

Let the price of one pen be x.

and Let the price of one pencil be y.

According to 1st problem,

5 x+3 y=34-(i)

According to 2nd problem,

7 x+6 y=53-\text { (ii) }

We have

5 x+3 y=34-(i) \\

7 x+6 y=53-(ii)

Multiplying equation (i) by 2 and equation (ii) by 1,

(By Subtracting)

or, 3 x=15\\

or, x=\frac{\cancel{15}^{5}}{\cancel{3}}\\

Putting the value of x in equation (i),

5 x+3 y=34\\

or, 5 \times 5+3 y=34\\

or, 25+3 y=34\\

or, 3 y=34-25\\

or, 3 y=9\\

or, y=\frac{\cancel9^{{3}}}{\cancel{3}}=3\\

The price of one pen ₹ 5

The price of one pencil ₹ 3 (Ans.)

2. The weight of my friend Ayesha and Refique altogether is 85 kg. The half of weight of Ayesha is equal to the \frac{4}{9} th of weight of Rafique. Let us calculate and write the weight of them separately by forming simultaneous equations.

Solution:

Let weight of Ayesha be x kgs.

and that Rafique be y kgs.

According to 1st condition of the problem,

or,x+y=85 — (i)

According to 2nd condition of the problem,

\frac{x}{2}=\frac{4 y}{9}\\

or, 9 x=8 y\\

\therefore \quad 9 x-8 y=0 — (ii)

Multiplying equation (i) by 9 and equation (ii) by 1,

(By Subtracting)

or, 17 y=765\\

or,y=\frac{\cancel{765}^{45}}{\cancel{17}}\\

or,y=45 \\

Putting the value of y in equation (i),

or, x+45=85\\

or, x=85-45\\

The weight of Ayesha is 40 \mathrm{~kg}.

The weight of Rafique is 45 \mathrm{~kg}. (Ans.)

3. My uncle’s present age is twice of my sister’s age 10 years ago, my uncle’s age was thrice of my sister’s age. Let me calculate and write their present age separately by forming simultaneous equations.

Solution :

Let present age of my uncle’s be x years.

and that of my sister’s be y years.

According to 1st condition of the problem,

x=2 y

or, x-2 y=0 — (i)

According to the 2nd condition of the problem,

(x-10)=3(y-10)

or, x-10=3 y-30

or, x-3 y=-30+10\\

\therefore \quad x-3 y=-20-\text { (ii) }\\

(By Subtracting)

Putting the value of y in equation (i),

x-2 y=0\\

or, x-2 \cdot 20=0\\

or, x=40\\

The present age of my uncle is 40 years.

The present age of my sister is 20 years. (Ans.)

4. Debkumarkaku of our village draws ₹ 590 through cheque from the bank. If he receive 70 notes on total of the notes of ₹ 10 and ₹ 5, then let us calculate and write the number of the notes of ₹ 10 and ₹ 5.

Solution :

Let the no. of ₹ 10 notes be x.

and that of ₹ 5 notes be y.

According to 1st condition of the problem,

10 x+5 y=590\\

or, 5(2 x+y)=590\\

or, 2 x+y=118-(i)\\

According to the 2nd condition of the problem,

x+y=70-(\text {ii) } \\

(By Subtracting)

or, x=48

Putting the value of x in equation (ii),

x+y=70\\

or, 48+y=70 \\

or, y=70-48\\

or, y=22\\

There were 48 notes of ₹ 10

There were 22 notes of ₹ 5 (Ans.)

5. I write a proper fraction in our school-black board whose denominator is greater than its numerator by 5 and when 3 is added to both numerator and denominator it becomes \frac{3}{4}. Let us form the simultaneous equations and by solving write the proper fraction on the black board.

Solution :

Let the proper fraction be \frac{x}{y}.

According to the 1st condition of the problem,

y=x+5

or, x-y=-5—(i)

According to the 2nd condition of the problem,

\frac{x+3}{y+3}=\frac{3}{4}\\

or, 3 y+9=4 x+12\\

or, 4 x-3 y=-12+9\\

\therefore \quad 4 x-3 y=-3-\text { (ii) }\\

Multiplying equation (i) by 4 and equation (ii) by 1,

(By Subtracting)

Putting the value of x in equation (i),

x-y=-5\\

or, x-17=-5\\

or, x=12\\

The proper fraction is \frac{12}{17}(Ans.)

6. Maria has written two numbers in her exercise copy such that the addition of 21 with the first number gives twice the second number. Again, the addition of 12 with the second number gives twice of the first number. Let us calculate and write the two numbers Maria has written.

Solution :

Let the 1st number be x and the 2nd be y.

According to the 1st condition of the problem,

x+21 =2 y \\

According to the 2nd condition of the problem,

2 x=y+12 \\

Multiplying equation (i) by 2 and equation (ii) by 1,

(By Substracting)

Putting the value of y in equation (i),

x-2 y=-21 \\

x-2 \times 18=-21\\

or, x-36=-21\\

The two numbers are 16 and 18. (Ans.)

7. Both of Lalima and Romen clean their garden. If Lalima works for 4 days and Romen works for 3 days, then \frac{2}{3} part of the work is completed. Again, if Lalima works for 3 days and Romen works for 6 days, then \frac{11}{12} part of the work is completed. Let us form the Simultaneous equations and write the number of days required to complete the work separately by Lalima and Romen by calculating the solution.

Solution :

Let Lalima work for x days and Romen work for y.

According to the 1st condition of the problem,

According to the 2nd condition of the problem,

Multiplying equation (i) by 3 and equation (ii) by 4,

(By Substracting)

or, \frac{-y}{15}=-\frac{3}{5}\\

Putting the value of y in equation (i),

\frac{4}{x}+\frac{3}{y}=\frac{2}{3}\\

or, \frac{4}{x}+\frac{3}{9}=\frac{2}{3}\\

or, \frac{4}{x}=\frac{2}{3}-\frac{3}{9}\\

or, \frac{4}{x}=\frac{2}{3}-\frac{1}{3}\\

or, \frac{4}{x}=\frac{1}{3}\\

Lalima needs to work for 12 days and Romen needs to work for 9 days cleen their garden.

8. My mother has prepared two types of sherbat. There is 5 \mathrm{kg} sugar in 100 litre sherbat of the first type and 8 \mathrm{~kg} sugar in 100 litre sherbat of the second type. By mixing, these two types, I will prepare 150 litre sherbat which contains 9 \frac{2}{3} \mathrm{~kg} sugar. Forming simultaneous equation let us calculate how much quantity of two types of sherbat will be mixed to prepare 150 litre sherbat.

Solution :

Let x Its sherbat of the first type be mixed with y lts sherbat of the second type to prepare 150 litres sherbat.

According to the 1st condition of the problem,

x+y=150-\text { (i) }

According to the 2nd condition of the problem,

\frac{5 x}{100}+\frac{8 y}{100}=9 \frac{2}{3}\\

or, \frac{5 x}{100}+\frac{8 y}{100}=\frac{29}{3}\\

or, \frac{5 x+8 y}{100}=\frac{29}{3}\\

\therefore \quad 15 x+24 y=2900-\text { (ii) }\\

Multiplying equation (i) by 15 and equation (ii) with 1,

\cancel{15 x}+15 y=2250 \\ \cancel{15 x}+24 y=2900\\ \underline {\text{(-) \text{ \quad(-)} \text{\quad(-)}}}\\

(By Subtracting)

or, y=\frac{-650}{-9}\\

Putting the value of y in equation (i),

x+y=150\\

or, x+\frac{650}{9}=150\\

or, x=150-\frac{650}{9}\\

or, x=\frac{1350-650}{9}\\

or, x=\frac{700}{9}\\

77 \frac{7}{9} lts of first type sherbat be mixed.

72 \frac{2}{9} lts of second type sherbat to prepare 150 lts of sherbat.

9. Last year Akhilbabu and Chhandadebi were the candidates in Bakultala Gram-Panchayet election. Akhilbabu defeated Chhandadebi by 75 votes. If 20% of the voters who have casted Akhilbabu would have casted to Chhandadebi, then Chhandadebi could be won by 19 votes. By forming simultaneous equations let us solve and find out how many votes each of them had got.

Solution :

Let x no. of votes be casted to Akhilbabu and y no. of votes be casted to Chhandadebi.

According to the 1st condition of the problem,

x-y=75-(i)

According to the 2nd condition of the problem,

\left(y+\frac{x}{5}\right)-\left(x-\frac{x}{5}\right)=19\\

or, \left(\frac{5 y+x}{5}\right)-\left(\frac{5 x-x}{5}\right)=19\\

or, \frac{5 y+x-4 x}{5}=19\\

or, 5 y-3 x=95\\

\therefore \quad 3 x-5 y=-95-\text { (ii) }\\

Multiplying equation (i) by 3 and equation (ii) with 1,

or, 2 y=320\\

or, y=\frac{320}{2} \therefore y=160\\

Putting the value of y in equation (i),

or, x-160=75 \\

or, x=75+160 \\

10. If the length is increased by 2 \mathrm{~m} and breadth is increased by 3 \mathrm{~m}, then the area of rectangular floor of Rafique is incrased by 75 \mathrm{sq}.\mathrm{m}. But if the length is reduced by 2 \mathrm{~m} and breadth incrased by 3 \mathrm{~m}, the area is increased by 15 \mathrm{sq}. \mathrm{m}. By forming simultaneous linear equations, let us determine the length and breadth of the floor.

Solution :

Let the length be \mathrm{xm}.

According to the 1st condition of the problem,

(x+2)(y+3)=x y+75\\

or, x y+3 x+2 y+6=x y+75\\

According to the 2nd condition of the problem,

(x-2)(y+3)=x y+15\\

or, \cancel{x y}+3 x-2 y-6=\cancel{x y}+15\\

or, 3 x-2 y=21- (ii)\\

(By Subtracting)

or, y=\frac{48}{4} \therefore y=12\\

Putting the value of y in equation (i),

or, 3 x+2 \times 12=69\\

or, 3 x+24=69\\

or, 3 x=69-24\\

or, 3 x=45\\

The length and breadth of Rafique’s rectangular for are 15 \mathrm{~m} and 12 \mathrm{~m} respectively.

11. My friend Meri told Ishan, give me \frac{1}{3} \mathrm{rd} of your money, then I shall have ₹ 200. Isha told Meri, give me half of your money, I shall have ₹ 200. Forming simultaneous equations let us calculate how much money does each of them possess.

Solution :

Let Meri posses ₹ x and Isha passes ₹ y.

According to the 1st condition of the problem,

x+\frac{1}{3} y=200\\

or, \frac{3 x+y}{3}=200\\

According to the 2nd condition of the problem,

y+\frac{1x}{2} =200 \\

\therefore \quad x+2 y =400-(\text {ii) }\\

Multiplying equation (ii) by 3 and equation (i) by 1,

(By Subtracting)

or, y=\frac{-600}{-5}\\

Putting the value of y in equation (i),

3 x+y=600\\

or, 3 x+120=600\\

or, 3 x=600-120\\

or, x=\frac{480}{3}\\

Meri possess ₹ 160 and Ishan possess ₹ 120

12. Today my elder brother and some of his friends will go to a fair. So, grandfather divided some money equally among them. We are observing that if the number of friends is less by 2, then each of them would get ₹18. Again, if the number of friends is more by 3, then each of them would get ₹ 12. Let us calculate and write the number of persons went to the fair and how much money did grandfather divide among them in total?

Solution:

Let x be the number of persons and y be the total money.

According to the 1st condition of the problem,

\frac{y}{x - 2} = 18

18x – 36 = y

18x – y = 36 – (i)

According to the 2nd condition of the problem,

\frac{y}{x+3} = 12

or, 12x + 36 = y

\therefore \quad 12x – y = – 36 – (ii)

\therefore \quad 18x - \cancel{y} = 36 \\ 12x - \cancel{y} = - 36 \\ \underline{\text{(-) \text{ \quad(+)} \text{\quad(+)}}\quad}\\ \text{ or }, \quad 6x = 72 (By Substracting)

or, x = \frac{72}{6}

Putting the value of y in equation (i),

18x – y = 36

or, 18.12 – y = 36

or, y = 216 – 36

\therefore y= 180

\therefore x = 12, y = 180

Grandfather divided ₹ 180 among 12 persons.

13. In my elder brother’s bag there are ₹ 350 with the coins of ₹ 1 and 50 paise together. My sister has put out \frac{1}{3} part of 50 paise coins from the bag and in that place equal umber of coins of ₹ 1 she has put into the bag and now the total amount of money in the bag is ₹ 400. Let us calculate and write the original number of coins of ₹ 1 and 50 paise kept at first separately in my brother’s bag.

Solution :

Letx be the original number of coins of ₹ 1 and y be the original number of coins of 50 paise.

According to the 1st condition of the problem,

x + \frac{y}{2} = 350

\therefore \quad 2x + y = 700

According to the 2nd condition of the problem,

\frac{1}{2}\left(y - \frac{y}{3}\right) + \left(x + \frac{y}{3}\right) = 400

or, \frac{1}{2}\left(\frac{3 y - y}{3}\right) + \left(\frac{3x - y}{3}\right) = 100

or, \frac{1}{2} \cdot \frac{2 y}{3} + \frac{3x + y}{3} = 400

or, \frac{y + 3x + y}{3} = 400

or, \frac{3x + 2 y}{3} = 400

\therefore \quad 3x + 2 y = 1200 – (ii)

\therefore \quad 2x + y = 700 – (i)

3x + 2 y = 1200 –

Multiplying equation (i) by 2 and equation (ii) by 1 , we get

4x + \cancel{2 y} = 1400 \\ 3x + \cancel{2y} = 1200 \\ \underline{\text{(-) \text{ \quad(-)} \text{\quad(-)}}\quad}\\ \quad x = 200 (By Subtracting)

Now putting the value ofx in equation (i),

2x + y = 700

or, 2 × 200 + y = 700

or, 400 + y = 700

or, y = 700 – 400

\therefore \quad original number of coins of ₹ 1 = 200

original number of coins of ₹ 50 paise = 300 (Ans.)

14. Today, we will go to my maternal uncle’s house, so a motor car sets out from our house towards my maternal uncle’s house at a uniform speed. If the speed of the car would be increased by 9 km/hr. then the time required to cover this path would be less by 3 hours. Again, if the speed would be decreased by 6km/hr. then 3 hours more time would be required to cover this path. Let us calculate and write the distance between our house and my maternal uncle’s house and the speed of the car.

Solution:

Let x be the distance and y be the speed of the car.

According to the 1st condition of the problem,

\frac{x}{y + 9} = \frac{x}{y} – 3

or, \frac{x}{y + 9} - \frac{x}{y} = – 3

or, x\left(\frac{1}{y + 9} - \frac{1}{y}\right) = – 3

or, x\left(\frac{x}{y + 9} - \frac{1}{y}\right) = – 3

or, x \cdot\left(\frac{y - (y + 9)}{y(y + 9)}\right) = – 3

or, x \cdot\left(\frac{\cancel{y} - \cancel{y} - 9}{y(y + 9)}\right) = – 3

or, x \cdot \frac{ \cancel{-} \cancel{9}^{3}}{y(y + 9)} = \cancel{-} \cancel{3}^{1}

or, \frac{3x}{y(y + 9)} = 1

or, 3x = y(y + 9)

\therefore\quad x = \frac{y(y + 9)}{3} – (i)

According to the 2nd condition of the problem,

\frac{x}{y - 6} = \frac{x}{y} + 3

or, \frac{x}{y - 6} - \frac{x}{y} = 3

or, x\left(\frac{x}{y - 6} - \frac{1}{y}\right) = 3

or, x \cdot\left(\frac{y - (y - 6)}{y(y - 6)}\right) = 3

or, x \cdot\left(\frac{\cancel{y} - \cancel{y} + 6}{y(y - 6)}\right) = – 3

or, \frac{2x}{y(y - 6)} = 1

\therefore \quad 2x = y(y – 6)

x = \frac{y(y - 6)}{2x} – (ii)

Comparing equation (i) and (ii),

\frac{\cancel{y}(y + 9)}{3} = \frac{\cancel{y}(y - 6)}{2}

or, 3y – 18 = 2y + 18

or, 3y – 2 y = 18 + 18

Putting the value of in equation (i),

x = \frac{y(y + 9)}{3}

or, x = \frac{\cancel{36}^{12}(36 + 9)}{\cancel{3}}

or, x = 12 × 45 = 540

\therefore \quad Distance between house and my maternal uncle house = 540km and speed of the car = 36km.

15. Mohit will write such a two digit number that 4 times, of the sum total of the two digits will be 3 more of the number and if the digits arè reversed, the number will be increased by 18. Let us calculate the number which Mohit and would write.

Solution:

Let the digit in the unit place be y and in the ten’s place bex So the number be 10x + y

According to the 1st condition of the problem;

4(x + y) + 3 = 10x + y

or, 4x + 4 y + 3 – 10x – y = 0

or, – 6x + 3 y = – 3

or, – 2x + y = – 1 – (i)

According to the 2nd condition of the problem,

(10 y + x) – (10x + y) = 18

or, 10 y + x – 10x – y = 18

or, – 9x + 9 y = 18

or, – x + y = 2 – (ii)

\therefore x = 3

Now putting the value ofx in equation (i),

– 2x + y = – 1

or, – 2 × 3 + y = – 1

or, – 6 + y = – 1

or, y = – 1 + 6

10x + y

= 10 × 3 + 5

= 30 + 5

= 35 (Ans.)

16. I shall write a two digit number, the sum of two – digits of which is 14 and if 29 is subtracted from the number, the two digits will be equal. Let us form the simultaneous equations and by solving them let us see what will be the two – digit number.

Solution:

Let the digit in the unit place be y

and Let the digit in the ten’s place bex

So the number be 10x + y

According to the 1st condition of the problem,

x + y = 14 – (i)

If 29 is substracted from the number,

10x + y – 29 = 10x + y – 30 + 1

= 10(x – 3) + (y + 1)

According to the 2nd condition of the problem,

x – 3 = y + 1

or, x – y = 1 + 3

or, x – y = 4 – (ii)

\therefore \quad x+\cancel{y}=600 - (i) \\ \underline {x-\cancel{y}=1200 - (ii)} \\

or, 2x = 18

or, x = \frac{\cancel{18}^{9}}{2} \\

x + y = 14

or, 9 + y = 14

or, y = 14 – 9

10x + y

= 10 × 9 + 5

= 90 + 5

= 95 (Ans.)

17. Rahamatchacha covers 30 miles in 6 hours in downstream and returns the same distance in 10 hours in upstream by his boat. Let us calculate and write the speed to Rahamatchacha’s boat in still water and the speed of the stream too.

Solution:

Let the speed of the boat in still water = x miles/hr.

and the speed of the current = y miles/hr

\therefore speed of the boat against the currect = (x -y) miles/hr. and speed of the boat along the current = (x +y) miles/hr.

According to the 1st condition of the problem,

6(x +y) = 30

or, x +y = 5 – (i)

According to the 2nd condition of the problem,

10(x -y) = 30

or, x -y = 3 – (ii)

\therefore \quad x + \cancel{y}= 5 - (i) \\ \underline {x - \cancel{y} = 3 - (ii)} \\

or, 2x = 8

or, x = \frac{\cancel8^{4}}{\cancel2}

Now putting the value ofx in equation (i),

x + y = 5

or, 4 + y = 5

or, y = 5 – 4

\thereforey = 1

\therefore \quad Speed of the boat = 4 mils/hr

and speed of the current = 1 miles/hr. (Ans.)

18. Leaving Howrah station after 1 hour a train is late by 1 hour for spcial reason and then running with \frac{3}{5} th of the its initial speed it reaches its destination after 3 hours. If the special reason would be 50km. away from its first place, then the train would reach its destination 1 hour 20 minutes before its previous time. Let us calculate the distance that the train had covered and the original speed of the train.

Solution :

Let the total distance =x km

and the speed of the train =y km / hr

\therefore \quad Time taken by the train to reach the distance = \frac{x}{y} hrs.

\therefore \quad Speed of the train =y km / hr.

\therefore \quad The distance travelled in 1 hr =y km.

\therefore \quad Remaining distance = (x -y) km.

After accident the speed of the train = \frac{3}{5} ofy = \frac{3y}{5}km.

\therefore Time taken to travel (x -y) km \frac{x -y}{\frac{3y}{5}}= hr. = \frac{5(x -y)}{3y} hr

\therefore Total time taken by the train = 1 + 1 + \frac{5(x -y)}{3y} hr. = 2 + \frac{5x - 5y}{3y} h r = \frac{6y + 5x - 5y}{3y} = \frac{5x +y}{3y}

According to the 1st condition of the problem,

\qquad \frac{5x +y}{3y} - \frac{x}{y} = 3 \\

\text { or, } \frac{5x +y - 3x}{3y} = 3 \\

\text { or, } \frac{2x +y}{3y} = 3 \\

\text { or, } 2x +y = 9y \\

\text { or, } 2x = 9y -y \\

\text { or, } 2x = 8y \\

\text { or, }x = 4y \\

\therefore \quadx – 4y = 0 – (i)

If the place be 50Km. away from its first place, the train will

\therefore Time taken by the train to travely = 50 Km = \frac{\mathrm{y} + 50}{\mathrm{y}} hr.

Now, remaining distance =x – (y + 50) =x -y – 50 Km.

\therefore Time taken by the train to travel this distance

= \frac{x -y - 50}{\frac{3y}{5}} hr

= \frac{5(x -y - 50)}{3y}hr \therefore Total time taken by the train to reach the destination

= \frac{y + 50}{y} + 1 + \frac{5(x - y - 50)}{3y}

= \frac{y + 50}{y} + \frac{5(x -y - 50)}{3y} + 1

= \frac{3y + 150 + 5x - 5y - 250 + 3y}{3y}

= \frac{5x +y - 100}{3y}

According to the 2nd condition of the problem,

\frac{5x +y}{3y} - \frac{5x +y - 100}{3y} = \frac{4}{3}

\left[\because 1\right. hour 20 minits \left. = \frac{80}{60} hr = \frac{4}{3} hr\right]

or, \frac{\cancel{5x} + \cancel{y} - \cancel{5x} - \cancel{y} + 100}{\cancel{3y}} = \frac{4}{\cancel{3}}

or, \frac{\cancel{100}}{\mathrm{y}} = \frac{\cancel{4}}{1}

Putting the value ofy in equation (i),

x – 4y = 0

or,x – 4 × 25 = 0

or,x – 100 = 0

\therefore \quad Total distance travelled by the train = 100 Km.

Speed of the train = 25 Km./hr. (Ans.)

19. Mousumi divides a two digit number with the sum of its digits and gets quotient as 6 and remainder as 6 . But if she divides the number interchanging the digits with the sum of its digits she will get quotient as 4 and remainder as 9. Let us determine the number that Mousumi has taken by forming simultaneous equations.

Solution :

Let the digit in the unit place bey

and the digit in the ten’s place bex

So the number is 10x +y

According to the 1st condition of the problem,

10x +y = 6(x + y) + 6

or, 10x + y = 6x + 6y + 6

or, 10x + y – 6x – 6y = 6

\therefore \quad 4x – 5y = 6 – (i)

According to the 2nd condition of the problem,

10y + x = 4(x +y) + 9

or, 10y + x = 4x + 4y + 9

or, 10y + x – 4x – 4y = 9

or, – 3x + 6y = 9

\therefore \quad – x + 2y = 3 – (ii)

\therefore \quad 4x – 5y = 6 – (i)

-x + 2y = 3 – (ii)

Multiplying equation (i) by 2 and equation (ii)

8x - 10y = 12 \\ - 5x + 10y = 15 \\or, \underline {3x = 27}

Now putting the value ofx in equation (i);

– x + 2y = 3

or, – 9 + 2y = 3

or, 2y = 12

or, y = \frac{\cancel{12}^6}{2}

\therefore \quad The required number is :

10x +y

= 10 × 9 + 6

= 90 + 6 = 96 (Ans.)

20. When faridabibi put oranges in some boxes, she observed that if she would put 20 oranges more in each box, the 3 boxes would be less required. But if she would put 5 oranges less in each box, 1 more box would be required. Forming simulaneous equations, let us calculate how many boxes and oranges did faridabadi have?

Solution :

Let x be no. of oranges andy be the no. of oranges in one box.

According to the 1st condition of the problem,

\quad \frac{x}{y + 20} - \frac{x}{y} = - 3 \\

\text { or, } \frac{x}{y + 20} = \frac{x}{y} - 3 \\

\text { or, } \frac{x}{y + 20} = \frac{x - 3y}{y} \\

\text { or, }(x - 3y)(y + 20) =xy \\

\text { or, }xy + 20x - 3y^{2} - 60y =xy \\

\text { or, } 20x = 3y^{2} + 60y \\

\text { or, } 20x = 3y(y + 20) \\

\therefore \quad x = \frac{3y(y + 20)}{20} – (i)

According to the 2nd condition of the problem,

\frac{x}{y - 5} = \frac{x}{y} + 1 \\

\text { or, } \frac{x}{y - 5} = \frac{x +y}{y} \\

\text { or, }(x +y)(y - 5) =xy \\

\text { or, }xy - 5x +y^{2} - 5y =xy \\

\text { or, } 5x =y^{2} - 5y \\

\text { or, }x = \frac{y(y - 5)}{5} – (ii)

Compairing equations (i) and (ii)

\frac{3\cancel y + 20)}{\cancel{20}^{4}} = \frac{\cancel y(y - 5)}{\cancel5}[\because y \neq 0]

or, 4(y – 5) = 3(y + 20)

or, 4y – 20 = 3y + 60

or, 4y – 3y = 60 + 20

\therefore y = 80

Now putting the value of y in equation (ii),

x = \frac{y(y - 5)}{5} = \frac{^{16}{\cancel{80}}(80 - 5)}{\cancel5} = 16 × 75 = 75

\therefore \quad Number of oranges = 1200

and number of boxes = \frac{^{15}1\cancel{200}}{\cancel{80}} = 15 (Ans)

## Short answer type questions :

(i) If x = 3 + and y = \frac{2 t}{3} – 1, then for what value of t, x = 3 y

Solution : Given,

x = 3 t, y = \frac{2 t}{3} – 1

Since x = 3y, x = 3t

or, 3y = 3 t

or, y = t

\therefore \quad y = \frac{2 t}{3} – 1 implies

or, t = \frac{2 t}{3} – 1

or, t = \frac{2 t - 3}{3}

or, 3t = 2t – 3

or, 3t – 2t = – 3

\therefore \quad t = – 3 (Ans)

(ii) For what value of k, the two equa tions 2x + 5y = 8 and 2x – Ky = 3 will have no solutions.

Solution : the given equations are

2x + 5y = 8

2x – ky = 3

they have no solution if,

\frac{\cancel2}{\cancel2} = \frac{5}{ - k}

or, -k = 5

\therefore \quad k = – 5 (Ans)

(iii) If x, y are real numbers and (x – 5)2 + ( x – y)2 = 0 then wha t are the values of x and y ?

Solution: Given

(x – 5)2 + ( x – y)2 = 0

The sum of two positive quantities are zero.

Then each quantity will be zero.

\therefore \quad x – 5 = 0, x – y = 0

\therefore \quad x = 5, or, y = x

\therefore \quad x = y = 5 (Ans)

(iv) If x2 + y2 – 2x + 4y = – 5, then wha t are the value of x and y ?

Solution : Given

x2 + y2 – 2x + 4y = – 5

or, x2 + y2 – 2x + 4y + 5 = 0

or, x2 – 2x + y2 + 4y + 4 + 1 = 0

or, (x)2 – 2.x.1 + (1)2 + ( y)2 + 2.y.2 + (2)2 = 0

or, (x – 1)2 + ( y + 2)2 = 0

the sum of two positive quan ti ties are zero, then each quantity will be zero.

\therefore \quad x – 1 = 0, y + 2 = 0

\therefore \quad x = 1, \therefore y = – 2

\therefore \quad x = 1, y = – 2  (Ans)

(v) For what value of r, the two equations r x – 3y – 1 = 0 and (4 – r) x – y + 1 = 0 would have solution?

Solution : Given

r x – 3 y – 1 = 0

(4 – r) x – y + 1 = 0

they have no solution if,

\frac{r}{4 - r} = \frac{ - 3}{ - 1}

or, – 12 + 3r = – r

or, 3r + r = 12

or, 4r = 12

\therefore \quad r = 3 (Ans)

(vi) Let us write the equation a_{1} x + b_{1} y + c_{1} = 0 in the form of y = m x + c where m and c are constant.

Solution : the given equation is

a_{1} x + b_{1} y + c_{1} = 0

or, b_{1} y = - a_{1} x - c_{1}

or, y = \frac{ - a_{1}}{b_{1}} x - \frac{c_{1}}{b_{1}}

or, y = \left(\frac{ - a_{1}}{b_{1}}\right) x + \left(\frac{ - c_{1}}{b_{1}}\right)

Which is the form of y = m x + c where

m = \frac{ - a_{1}}{b_{1}}, \quad c = \frac{ - c_{1}}{b_{1}} (Ans)

(vii) For what value of K, the two equations K x – 21 y + 15 = 0 and 8 x – 7 y = 0 have only one solution?

Solution : Given,

K x – 21y + 15 = 0

8x – 7y = 0 – (ii)

By comparing the values \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \ and \ \frac{c_{1}}{c_{2}} of the equation (i) a_{1} x + b_{1} y + c_{1} = 0 and the equation (ii) a_{2} x + b_{2} y + c_{2} = 0 we get,

a_{1} = K, b_{1} = - 21 \text { and } C_{1} = 15

a_{2} = 8, b_{2} = - 7 \text { and } C_{2} = 0

\therefore \quad the two equations (i) and (ii) will have no solution

or, \frac{K}{8} = \frac{\cancel{21}^{3}}{7}

\therefore \quad the two equations (i) and (ii) will have only solution for all values of K except 24 . (Ans.)

(viii) For what value of a and b the two equations 5 x + 8 y = 7 and (a + b) x + (a – b) y = (2a + b + 1) have infinite number of solutions.

Solution : The given equations

5x + 8 y = 0 – (i)

(a + b) x + (a – b) y = (2a + b + 1)

B y comparing the values \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} and \frac{c_{1}}{c_{2}}of the equation (i) a_{1} x + b_{1} y + c_{1} = 0 and the equation (ii) a_{2} x + b_{2} y + c_{1} = 0 we get,

a_{1} = 5, b_{1} = 8 \text { and } C_{1} = - 7 \\

a_{2} = (a + b), b_{2} = (a - b) \text { and } C_{2} = - (2 a + b + 1)

The equations (i) and (ii) will have more than one solution, :

if \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

or, \frac{5}{a + b} = \frac{8}{a - b} = \frac{ + 7}{ + (2 a + b + 1)}

or, \frac{5}{a + b} = \frac{8}{a - b}  \quad Again \frac{8}{a - b} = \frac{7}{2 a + b + 1}

or, 8a + 8b = 5a – 5b \quad or, 16a + 8b + 8 = 7a – 7b

or, 8a – 5a + 8b + 5b = 0 \quad or, 16a – 7a + 8b + 7b = – 8

or, 3a + 13b = 0 \quad or, 9a + 15b = – 8 – (iv)

\therefore 3a + 13b = 0 – (iii)

9a + 15b = – 8 – (iv)

Multiplying equation (iii) by 3 and equation (iv) by 1

(By subtracting)

or, b = \frac{8}{24}

Putting the value of b in equation (iii),

3a + 13b = 0

or, 3a + 13 \times \frac{1}{3} = 0

or, 3a + \frac{13}{3} = 0

or, 3a = \frac{ - 13}{3}

\therefore \quad a = \frac{ - 13}{9}

\therefore \quad a = - \frac{13}{9}, b = \frac{1}{3} (Ans.)

## 22. M.C.Q.

(i) The two equation 4x + 3y = 7 and 7 x – 3y = 4 have

1. Only one solution
2. Infinite number of solutions
3. No solution
4. None of them

Solution : Given,

4x + 3y = 7 – (i)

7x – 3y = 4

we compare these equations with

a_{1} {x} + b_{1} y + c_{1} = 0 \text{ and } a_{2} x + b_{2} y + c_{2} = 0, we get

\therefore \quad a_{1} = 4, b_{1} = 3 \text{ and } c_{1} = – 7

or, a_{2} = 7, b_{2} = - 3 \text{ and } c_{2} = – 4

We see that,

\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

\therefore \quad There is only one solution

\therefore (a) is correct option. :

(ii) The two equation 3x + 6y = 15 and 6x + 12y = 30 have

1. Only one solution
2. Infinite number of solutions
3. No solution
4. None of them

Solution : The given equations are

3x + 6Y = 15 – (i)

6x + 12y = 30 – (ii)

we compare equations (i) and (ii) with

a_{1} x + b_{1} y + c_{1} = 0 \text{ and } a_{2} x + b_{2} y + c_{2} = 0, we get

\therefore \quad a_{1} = 3, b_{1} = 6 \text{ and } c_{1} = – 15.

or, a_{2} = 6, b_{2} = 12 \text{ and } c_{2} = – 30

We see that,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

\therefore \quad There is an infinite numbers of solutions

\therefore (b) is correct option.

(iii) The two equation 4x + 4y = 15 and 5x + 5y = 30 have

1. Only one solution
2. Infinite number of solutions
3. No solution
4. None of them

Solution : The given equations are

4x + 4y = 20

5x + 5y = 30

we compair equations (i) and (ii) with

a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0, we get

\therefore \quad a_{1} = 4, b_{1} = 4 and c_{1} = 20

or, \mathrm{a}_{2} = 5, \mathrm{~b}_{2} = 5 \text{ and } \mathrm{c}_{2}= 30

We see that,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

\therefore \quad There is no solutions

\therefore \quad (c) is correct option.

(iv) Which of the following equations have a solution (1, 1)

1. 2x + 3y = 9
2. 6x + 2y = 9
3. 3x + 2y = 5
4. 4x + 6y = 8

Solution : (c) is satisfy the value (1, 1)

So, (1, 1) is a solution of the equation

\therefore \quad (c) is correct option.

(v) The two equations 4 x + 3 y = 25 and 5 x – 2 y = 14 have the solutions.

1. x = 4, y = 3
2. x = 3, y = 4
3. x = 3, y = 3
4. x = 4, y = – 3

Solution : We solve the two equations 4x + 3y = 25 and 5x – 2y = 14

\therefore \quad 4x + 3y = 25 – (i)

5x – 2y = 14 – (ii)

Multiplying equation (i) by 2 and equation (ii) by 3 , we get.

8 x + \cancel{6 y} = 50 \\ \underline{15 x - \cancel{6 y} = 42} \\

or, 23x = 92

or, x = \frac{\cancel{92}^{4}}{\cancel{23}}

Putting the value of x in equation (i),

4x + 3y = 25

or, 4 × 4 + 3y = 25

or, 16 + 3y = 25

or, 3y = 25 – 16

or, 3y = 9

\therefore y = 3

y = 3

\therefore \quad (a) is correct option

(vi) The solutions of the equation x + y = 7 are

1. (1, 6), (3, -4)
2. (1, -6), (4, 3)
3. (1, ), (4,3)
4. ( -1, 6), ( -4, 3)

Solution : (c) is satisfy the the equation x + y = 7

So (c) is correct option.

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