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

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 1^st 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 = 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

(By adding)

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.

4x+\cancel{6y}=14 \\9x+\cancel{6y}=24 \\(-)(-) \quad(-)

(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

(By adding)

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.

22 x+33 y=352 \\-27 x+33 y=9 \\(+) \quad(-) \quad(-)

(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

(By adding)

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

(By adding)

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}

(By adding)

7x = 14

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

\therefore \quad x = 2

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

\therefore \quad y = 6

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

\therefore \quad x =\frac{6}{5}

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}

\therefore \quad y=\frac{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}

(By adding)

or, 62x = 434

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

\therefore \quad x = 7

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}}

\therefore \quad y = 9

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)

i.e. \quad x+y=5 x y-(i) \\\quad\underline {x-y=9 x y-(ii)}

(By adding)

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

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

\therefore \quad y=\frac{1}{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

\therefore \quad x=\frac{-1}{2}

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

\therefore \quad x=\frac{5}{4}

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

\therefore \quad y = 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

(By adding)

\frac{77}{a}=11

or, 11a = 77

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

\therefore \quad \mathrm{a}=7

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 \mathrm{b}=1\\

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

(By adding)

or, 2x = 8

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

\therefore \quad x=4

Now putting the value of x  in equation (iii)

x + y = 7

or, 4 + y = 7

or, y = 7 – 4

\therefore \quad y=3

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}}}

(By adding)

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

\therefore \quad y=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

∴ \quad \frac{x+a}{a}=\frac{y+b}{b}\\

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}

(By adding)

or, 28 x = 28

or, \quad x=\frac{\cancel{28}}{\cancel{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)

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

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

Solution:

\quad 2 x+\frac{3}{y}=5 — (i)

5 x-\frac{2}{y}=3 — (ii)

Taking equation (i)

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

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}}\\

∴ \quad \mathrm{y}=1\\

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)

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

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

Solution:

\quad \frac{x}{2}+\frac{y}{3}=1  — (i)

\frac{x}{3}+\frac{y}{2}=1 — (ii)

Taking equation (i)

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

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}}\\

\therefore \quad y=3\\

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} \\

\therefore \quad x=\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\\

\therefore \quad y=\frac{2}{3}\\

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 \\

\therefore \quad x=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\\

\therefore \quad x=\frac{100+y}{9}- (\text{iii})\\

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} \\

\therefore \quad y=8\\

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 \\

\therefore \quad x=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\\

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

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}}\\

\therefore \quad y=4\\

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

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

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

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

\therefore \quad x=4\\

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)\\

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

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}} \\

\therefore \quad y=3

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

\therefore \quad x=1-y \\

=1-3 \\

=-2 \\

\therefore \quad x=-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}}\\

\therefore \quad x=3 (\text{Ans.})\\

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

(\text{a}) \quad3 x-y=7\\

2 x+4 y=0\\

(\text{b}) \quad\frac{x}{2}+\frac{y}{3}=2=\frac{x}{4}+\frac{y}{2}\\

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}}\\

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

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

\therefore \quad x =\frac{7+y}{3} \\

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

=\frac{7-1}{3}

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

\therefore \quad x=2

By graphically,

Taking equation (i) Taking equation (ii)

3x - 7 = 7\\

or, 3 x=7+y\\

\therefore \quad x=\frac{7+y}{3}\\

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 \\

\therefore \quad =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\\

\therefore \quad 3 x=12-2y\\

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\\

\therefore \quad x=8-2 y\\

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}} \\

\therefore \quad y=6\\

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

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

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}\\

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

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\\

\therefore \quad 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 \\

\therefore \quad x=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}\\

\therefore \quad y=\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 \\

\therefore \quad x =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\\

\therefore \quad y=\frac{1}{5}\\

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} \\

\therefore \quad x =\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\\

\therefore \quad y=\frac{\cancel{13}^{1}}{\cancel{26}_{2}}=\frac{1}{2}

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}} \\

=2 \\\therefore \quad x=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}\\

\therefore \quad x=\frac{5 y-3}{2 y}-(\mathrm{iii})\\

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}}\\

\therefore \quad y=1\\

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 \\

\therefore \quad x =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}\\

\therefore \quad x=\frac{6-2 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\\

\therefore \quad 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{\cancel{18}^{6}}{5}}{\cancel{3}} \\

\therefore \quad x =\frac{6}{5}\\

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}\\

\therefore \quad x=\frac{3 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 \\

\therefore \quad y =8\\

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

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

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

=6 \\

\therefore \quad x=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}\\

\therefore \quad x=\frac{2 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}}\\

\therefore \quad y=10\\

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 \\

\therefore \quad x=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 \\

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

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}}\\

\therefore \quad y=5\\

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 \\

\therefore \quad x =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}}\\

\therefore \quad y=9\\

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 \\

\therefore \quad x=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),