Chapter – 3 : Graph | Chapter Solution Class 9

Let us work out – 3.2

Question 1

I plot and write the following points on graph paper Position (on the axis or in the quadrant)

(i) (3,0), (ii) (0,8), (iii) (-5,0), (iv) (0,-6), (v) (6,4), (vi) (-7,4) (vii) (9,-9), (viii) (-4,-5)

Solution:

(i) On the positive x-axis

(ii) on the positive y-axis

(iii) on the Negative x-axis

(iv) on the negative y-axis

(v) 1st. Quadrant

(vi) 2nd. Quadrant

(vii) 4th. Quadrant

(viii) 3rd. Quadrant

Question 2

On the graph paper, let us draw XOX’ and YOY’ on the two perpendicular axes and plot any 5 points that lie in the third quadrant 29.

Solution:

The 5 points found in the third quadrant are (-3,-2), (-5,-4), (-10,-7) and (-12, -9)

Question 3

Let’s express the following statements in simultaneous linear equations:

1. The total price of 3 copies of exercises and 2 pens is Rs 55 and the total price of 4 copies of exercises and 3 pens is Rs 75.
2. The sum of the two numbers is 80 and triple the difference of those two numbers is 20 more than the larger number.
3. When 2 is added to the numerator and denominator of a fraction, its value will be 7/9 and when 3 is subtracted from both the numerator and denominator, its value will be 1/2
4. The digit in the tens places twice the digit in the one’s place of a two-digit number. If the digits switch places, the resulting number will be 27 less than the original number.

Solution:

(i) Let the price of exercise copies be ‘x’

and  the price of  Pens be ‘y’

According to the problem,

3x + 2y = 55

4x + 3y = 75

(ii) Let the 1st number be x and the 2nd number be y

According to the problem,

x + y = 80

3(x – y) – x = 20

(iii) Let the numerator be x and denominator be y. according to the problem,

\frac{x + 2}{y + 2} = \frac{7}{9}, \frac{x - 3}{y - 3} = \frac{1}{2}

(iv) Let the digit in the unit place be x and the digit in the ten’s place be y.

The number is x + 10y

according to the problem,

x = 2y, (10x + y) − (x + 10y) = 27

Question 4

Let us express the following statements into linear equations in two variables and draw the graph of the equation.

1. At present, the age of Sujata’s father is more than the age of Sujata by 26 years.
2. The sum of the two numbers is 15.
3. If the numerator and denominator of a fraction are increased by 2, the fraction will be 7/9
4. The perimeter of our rectangular yard is 80m.
5. Of the two numbers, 5 times of larger number equals 8 times the smaller one.

Solution:

(i) Let us suppose Sujata’s father’s age is ‘x’ years and Sujata’s age is ‘y’ years.

According to the problem,

x = 26+ y

If y = -10,  x = 26 + (-10)= 26-10 = 16

y = -11, x = 26 + (-11) = 26-11 = 15

y = -12, x = 26 + (-12) = 26 – 12 = 14

x	16	15	14
y	-10	-11	-12

(ii) Let the 1^st number be ‘x’

and Let the 2^nd number be ‘y’

According to the conditions,

x + y = 15, or, x = 15 – y

If y = 0, x = 15 – 0 = 15

y = 1, x = 15 – 1 = 14

y = 2, x = 15 – 2 = 13

x	15	14	13
y	0	1	2

(iii) Let the numerator be ‘x’ and denominator be ‘y’.

\frac{x+2}{y+2}=\frac{7}{9}

or, 9x + 18 = 7y + 14

or, 9x =  y + 14 – 18

or, 9x =  7y – 4

or, x =\frac{7y-4}{9}

If y=7, x=\frac{7 \times 7-4}{9}=\frac{45}{9}=5

y=16, x=\frac{7 \times 16-4}{9}=\frac{112-4}{9} =\frac{108}{9}=12

y=25, x=\frac{7 \times 25-4}{9}=\frac{175-4}{9}

\frac{171}{9}=19

x	5	12	19
y	7	16	25

(iv) Let the length of a rectangle be ‘x’

and the breadth  of a rectangle be ‘y’

∴ Given, Perimeter = 80 or, 2(x + y) = 80

or, x + y = 40

or, x = 40 – y

If y = 20, x = 40 – 20 = 20

y = 25, x = 40 – 25 = 15

y = 30, x = 40 – 30 = 10

x	20	15	10
y	20	25	30

(v) Let the 1st number be x and 2nd number be ‘y’.

By question, 5x = 8y

x=\frac{8 y}{5}

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

f y=5, x=\frac{8 \times 5}{5}=8

If y=10, x=\frac{8 \times 10}{5}=16

Question 5

Let us draw the graph of the following equation:

(i) x = 5

(ii) y + 2 = 0

(iii) x = 3-4y

(iv) 3x – 7y = 21

(v) 5x – 3y = 8

(vi) 2x + 3y = 11

(vii) x/3 + y/4 = 0

(viii) 6x – 7y = 12

(ix) x + y 10 = 0

(x) y = 5x-3

(xi) y = 0

Solution:

(i) x = 5

Here the value of x is 5 from any real value of y. Talking about the integral values of y, we get a table as follows.

On graph paper, the two perpendicular axes XOX’ and YOY’ are drawn and the points (5,2), (5,4) and (5,6) are plotted and joined them.

x	5	5	5
y	2	4	6

(ii) y + 2 = 0

y = – 2

Here the value of y is -2 for any real value of x. Talking about the integral values of x, we get a table as follows.

On graph paper, the two perpendicular axes XOX’ and YOY’ are drawn and the points (2,-2), (4,-2) and (6, -2) are plotted and joined them.

x	2	4	6
y	-2	-2	-2

(iii) x = 3 – 4y

If y = 0, x = 34 × 0 = 30 = 3

y = 1, x = 3 – 4 × 1 = 3-4= – 1

y= 2, x = 3 – 4 × 23 – 8 =- 5

On a graph paper by drawing the two perpendicular axes XOX’ and YOY’ and along both sides axes taking the side of the smallest squares as 1 unit the points (3,0), (−1,1) & (−5,2) are plotted. After joining these points we get a straight-line AB.

x	3	-1	-5
y	0	1	2

(iv) 3x – 7y = 21, or, 3x = 21 + 7y

∴ x = \frac{21+7y}{3}

If y=0, x=\frac{21+7 \times 0}{3}=\frac{21+0}{3}=\frac{21}{3}=7

If y=3, \quad x=\frac{21+7 \times 3}{3}=\frac{21+21}{3}=\frac{42}{3}=14

If y=6, x=\frac{21+7.6}{3}=\frac{21+42}{3}=\frac{63}{3}=21

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (7,0), (14,3) and (21,6) are plotted and joined them. Thus the straight line AB is obtained.

x	7	14	21
y	0	3	6

(v) 5x – 3y = 8, or, 5x = 8 + 3y

\therefore x=\frac{8+3 y}{5}

If y=-1, x=\frac{8+3(-1)}{5}=\frac{8-3}{5}=\frac{5}{5}=1

If y=4, \quad x=\frac{8+3 \times 4}{5}=\frac{8+12}{5}=\frac{20}{5}=4

If y=9, \quad x=\frac{8+3 \times 9}{5}=\frac{8+27}{5}=\frac{35}{5}=7

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (-1,1), (4,4) and (7,9) are plotted and joined them. Thus the straight line AB is obtained.

x	1	4	7
y	-1	4	9

(vi) 2x + 3y = 11, or, 2x = 11 – 3y.

If y=1, \quad x=\frac{11-3 \times 1}{2}=\frac{11-3}{2}=\frac{8}{2}=4

If y=3, \quad x=\frac{11-3 \times 3}{2}=\frac{11-9}{2}=\frac{2}{2}=1

If y=5, \quad x=\frac{11-3 \times 5}{2}=\frac{11-15}{2}=\frac{-4}{2}=-2

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (4,1), (1,3) and (-2,5) are plotted and joined them. Thus the straight line AB is obtained.

x	4	1	-2
y	1	3	5

(vii) \frac{x}{3}+\frac{y}{4}=0

or, \frac{4 x+3 y}{12}=0

or, 4 x+3 y=0

or, 4 x=-3 y

\therefore \quad x=\frac{-3 y}{4}

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

If y=8, \quad x=\frac{-3 \times 8}{4}=-6

If y=12, \quad x=\frac{-3 \times 12}{4}=-9

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (-3,4),(-6,8) and (-9,12) are plotted and joined them. Thus the straight line AB is obtained.

x	-3	-6	-9
y	4	8	12

(viii) 6x – 7y = 12

or, 6x = 12 + 7y

or, x = \frac{12+7y}{6}

If y=0, x=\frac{12+7 \times 0}{6}=\frac{12+0}{6}=\frac{12}{6}=2

If y=6, x=\frac{12+7 \times 6}{6}=\frac{12+42}{6}=\frac{54}{6}=9

If y=-6, x=\frac{12+7 \times(-6)}{6}=\frac{12-42}{6}=\frac{-30}{6}=-5

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (2,0), (9,6) and (-5,-6) are plotted and joined them. Thus the straight line AB is obtained.

x	2	9	-5
y	0	6	-6

(ix) x + y – 10 = 0,

or, x = 10 – y

If y = 2, x = 10 – 2 = 8

y = 4, x = 10 – 4 = 6

y = 6, x = 10 – 6 = 4

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (8,2), (6,4) and (4,6) are plotted and joined them. Thus the straight line AB is obtained.

x	8	6	4
y	2	4	6

(x) y = 5x – 3

or,  5x – 3 = y

or,  5x = y + 3

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

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

If y=7, \quad x=\frac{7+3}{5}=\frac{10}{5}=2

If y=12, \quad x=\frac{12+3}{5}=\frac{15}{5}=3

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (1,2), (2,7) and (3,12) are plotted and joined them. Thus the straight line AB is obtained.

x	1	2	3
y	2	7	12

(xi) y = 0

On the graph paper, the two perpendicular axes XOX’ & YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (2,0), (4,0) and (6,0) are plotted and joined them. Thus the straight line AB is obtained.

x	2	4	6
y	0	0	0

 

Let us express the following statements into simultaneous linear equations and solve them graphically.

Question 6 (i)

At present Rajat’ maternal uncle is 16 years older than Rajat, 8 years later, his maternal uncle’s age will be 2 times his age. Let us calculate the present age of Rajat and that of his maternal uncle graphically.

Solution:

Let the present age of Rajat be x and that of his maternal uncle be y.

according to 1st condition of the question,

x – y = 16………. (i)

or, x = 16 + y

If y = 0, x = 16 + 0 = 16

y = 1, x = 16 + 1 = 17

y = 2, x = 16 + 2 = 18

x	16	17	18
y	0	1	2

According to 2nd condition of the question,

x + 8 = 2(y + 8), or,    x + 8 = 2y + 16

or, x = 2y + 16 – 8.

∴  x = 2y + 8…….(ii)

If y = 0, x = 0 + 8 = 8

y = 1, x = 2 × 1 + 8 = 2 + 8 = 10

y = 2, x = 2 × 2 +8 = 4 + 8 = 12

x	8	10	12
y	0	1	2

It is observed that the two straight lines. AB & CD intersect at point P. The coordinates of Pis (24,8)

∴  x = 24, y = 8

satisfy both the equation (i) & (ii)

∴ The present age of Rajat = 24 years and his maternal uncle’s age = 8 years (Ans.)

Question 6 (ii)

The sum of two numbers is 15 and the difference is 3. Let us write the numbers by solving them graphically.

Solution:

Let the two numbers be x and y.

According to 1st condition of the problem,

x + y = 15…….(i)

x = 15 – y

If y = 0, x = 150 = 15

y = 1, x = 15 – 1 = 14

y = 2, x = 15 – 2 = 13

x	15	14	13
y	0	1	2

According to 2nd condition of the problem,

x – y = 3 …….. (ii)

x = 3 + y

If y = 0, x = 3+0 = 3

y = 1, x = 3 + 1 = 4

y = 2, x = 3 + 2 = 5

x	3	4	5
y	0	1	2

It is observed that the two straight lines AB and CD intersect at point P. The co-ordinate of Pis (9,6) satisfy both equation (i) & (ii)

∴ 1 st number = 9. 2nd number 6 Ans.

Question 6 (iii)

If 3 is subtracted from the numerator and 2 is added to the denominator, the fraction will be 1/3 and if 4 is subtracted from the numerator and 2 is subtracted from the denominator, the fraction will be 1/2, Let 2. We construct the equation of the statement and write the fraction by solving graphically.

Solution:

Let the numerator be x and the denominator be y. According to 1st condition of the problem.

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

3 x-9=y+2

or, 3 x=y+2+9

or, 3 x=y+11

\therefore \quad x=\frac{y+11}{3}............(i)

If y=1, \quad x=\frac{1+11}{3}=\frac{12}{3}=4

y=4, \quad x=\frac{4+11}{3}=\frac{15}{3}=5

y=7, \quad x=\frac{7+11}{3}=\frac{18}{3}=6

x	4	5	6
y	1	4	7

According to the 2nd condition of the problem.

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

or, 2x-8=y-2

or, 2x=y-2+8, or, 2x=y+6

\therefore \quad x=\frac{y+6}{2}......(ii)

If y=0, \quad x=\frac{0+6}{2}=\frac{6}{2}=3

y=2, \quad x=\frac{2+6}{2}=\frac{8}{2}=4

y=4, \quad x=\frac{4+6}{2}=\frac{10}{2}=5

It is observed that the two straight lines A B and C D intersect at point P. The coordinate of  P is (5,4)

∴ x = 5, y = 4

Satisfy both equation (i) & (ii) The fraction = \frac{5}{4} (Ans)

x	3	4	5
y	0	2	4

Question 6 (iv)

The perimeter of Rohit’s rectangular garden is 60 m. If the length of the garden is increased by 2 m and the breadth is decreased by 2 m, the area of the garden is decreased by 24 sq.m. Let us write the length and breadth of the garden by solving graphically.

Solution:

Let the length of a rectangular garden be x and the breadth of a rectangular be y.

According to 1st condition of the problem,

Perimeter 60,    or, 2 (x + y) = 60

or, x + y=30……..(i)

∴ x = 30 – y

If y = 10, x = 30-10 = 20

y= 15, x = 30 – 15 = 15

y = 20, x = 30 – 20 = 10

It is observed that the two straight lines AB & CD intersect at point P. The co-ordinates of Pis (20, 10)

∴ x = 20, y = 10, satisfy both the equation (i) & (ii)

According to 2nd condition of the problem,

(x+2) (y-2) = xy – 24,   or, xy – 2x + 2y – 4 = xy – 24

or, -2x + 2y – 4 = -24,    or, 2y – 4 + 24 = 2x,

or, 2y + 20 = 2x,    2(y + 10) = 2x

∴ x = 10 + y……..(ii)

If y = 5, x = 10 + 5 = 15

y = 3, x = 10 + 3 = 13

y = 0, x = 10 + 3 = 10

∴ Length of the rectangular garden = 20m

∴ Breadth of the rectangular garden = 10m (Ans)

x	15	13	10
y	5	3	0

Question 6 (v)

A boat covers 64 km. in 16 hrs while travelling downstream and it covers 24 km. in 8 hrs. while travelling upstream. Let us write the speed of the boat in still water and the speed of the stream by solving graphically.

Solution:

Let the speed of the boat be x

and the speed of the steam be y

According to 1st condition of the problem,

16(x + y) = 64 ……… (i)

or, (x + y) = 4    ∴  x = 4 – y

If y = 0, x = 4 – 0 = 4

y = 1, x = 4 – 1 = 3

y = 2, x = 4 – 2 = 2

x	4	3	2
y	0	1	2

According to 2nd condition of the problem,

8(x – y) 24………..(ii)

or, (x – y) = 3,    or, x=3+ y

If y = 0, x = 3 + 0 = 3

y = 1, x = 3 + 1 = 4

y = 2, x = 3 + 2 = 5

x	3	4	5
y	0	1	2

It is observed that the two straight lines AB &  CD intersect at point P. The coordinates of P is (\frac{7}{2}, \frac{1}{2})

∴ x = 3.5

& y = 1.5

Satisfy both equations (i) & (ii).

∴  Speed of the boat = 3.5 km/h.

”       ”      ”      ”      ”  = 0.5 km/h.

Let us draw the graph of the following simultaneous linear equation and determine the coordinates of the point of intersection.

Question 7 (i)

x = 0 and  2x + 3y = 15

Solution:

x = 0………..(i)

For any value of y, the value of x will be zero.

Hence

x	0	0	0
y	0	1	2

and, 2x + 3y = 15

or, 2x = 15 – 3y

\therefore \quad x=\frac{15-3 y}{2}

If y=1, \quad x=\frac{15-3 \times 1}{2}=\frac{15-3}{2}=\frac{12}{2}=6

If y=3, \quad x=\frac{15-3 \times 3}{2}=\frac{15-9}{2}=\frac{6}{2}=3

If y=5, \quad x=\frac{15-3 \times 5}{2}=\frac{15-15}{2}=\frac{0}{2}=0

On the graph paper, the two perpendicular XOX and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (0,2),(0,4) and (0,6) are plotted and joined them. Thus the straight line AB is obtained and the points (6,1), (3,3) and (0,5) are plotted and joined them. Thus the straight-line CD is obtained.

x	6	3	0
y	1	3	5

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P is (0,5)

∴ x = 0, y = 5

satisfying both the equation (i) & (ii)

Question 7 (ii)

y = 5 and 2x + 3y= 11

Solution:

y = 5……….(i)

∴ For any value of x, the value of y will be 5 .

Hence

x	3	5	7
y	5	5	5

and,

\quad 2 x+3 y=11..........(ii)

or, 2 x=11-3 y

\therefore x=\frac{11-3 y}{2}

If y=1, x=\frac{11-3 \times 1}{2}=\frac{11-3}{2}=\frac{8}{2}=4

If y=3, x=\frac{11-3 \times 3}{2}=\frac{11-9}{2}=\frac{2}{2}=1

If y=3, \quad x=\frac{11-3 \times 5}{2}=\frac{11-15}{2}=\frac{-4}{2}=-2

x	4	1	-2
y	1	3	5

On the graph paper, the two perpendicular XOX’ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (3, 5), (5, 5) and (7, 5) are plotted and joined them. Thus the straight line AB is obtained and the points (4, 1), (1, 3) and (-2, 5) are plotted and joined them. Thus the straight-line CD is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P are (-2, 5)

∴  x = -2, y = 5, satisfying both the equation (i) & (ii)

Question 7 (iii)

x + y = 12 and x – y = 2 x = 12-y

Solution:

x + y = 2………(i)     ∴ x = 12 – y

If y = 2, x = 12 – 2 = 10

y = 4, x = 12 – 4 = 8

y = 6, x = 2 + 6 = 8

x	10	8	6
y	2	4	6

and x – y = 2…… (ii)

∴ x = 2 + y

If y = 2, x = 2 + 2 = 4

y = 4, x = 2 + 4 = 6

y = 6, x = 2 + 6 = 8

x	4	6	8
y	2	4	6

On the graph paper, the two perpendicular XOX’ and YOY’

are drawn and taking the side of the smallest square as 1 unit, the points (10, 2), (8, 4) and (6, 6) are plotted and joined them. Thus the straight line AB is obtained and the points (4, 2), (6, 4) and (8, 8) are plotted and joined them. Thus the straight-line CD is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P is (7, 5) x = 7, y = 5, satisfy both the equation (i) & (ii)

(iv) 3x – 5y =  16 and 2x – 9y = 5

3x – 5y = 16…………(i)

or,  3x = 16 + 5y

\therefore \quad x =\frac{16+5 y}{3}

If y=2, \quad x=\frac{16+5(-2)}{3}=\frac{16-10}{3}=\frac{6}{3}=2

y=1, \quad x=\frac{16+5 \times 1}{3}=\frac{16+5}{3}=\frac{21}{3}=7

y=4, \quad x=\frac{16+5 \times 4}{3}=\frac{16+20}{3}=\frac{36}{3}=12

and 2x – 9y = 5……..(ii)

or, 2x = 5 + 9y

\therefore \quad x=\frac{5+9 y}{2}

If y=1, x=\frac{5+9 \times 1}{2}=\frac{5+9}{2}=\frac{14}{2}=7

If y=3, \quad x=\frac{5+9 \times 3}{2}=\frac{5+27}{2}=\frac{32}{2}=16

If y=-1, x=\frac{5+9(-1)}{2}=\frac{5-9}{2}=\frac{-4}{2}=-2

It is observed that the two straight lines A B & CD intersect at the point P. The co-ordinates of P is (7, 1)

∴ x = 7, y = 1 , satisfy both the equation (i) & (ii)

Let us solve the following equations graphically.

Question 8 (i)

4x – y = 3 ;  2x + 3y = 5

Solution:

(i) 4x – y = 3 ;  2x + 3y = 5

or, 4x = 3 – y……..(i)

\therefore \quad x=\frac{3+y}{4}

If y=1, \quad x=\frac{3+1}{4}=\frac{4}{4}=1

If y=5, \quad x=\frac{3+5}{4}=\frac{8}{4}=2

If y=9, \quad x=\frac{3+9}{4}=\frac{12}{4}=3

x	1	2	3
y	1	5	9

2x + 3y = 5…………(ii)

or, 2x = 5 – 3y

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

If y=1, x=\frac{5-3 \times 3}{2}=\frac{5-3}{2}=\frac{2}{2}=1

If y=3, x=\frac{5-3 \times 3}{2}=\frac{5-9}{2}=\frac{-4}{2}=-2

If y=5, \quad x=\frac{5-3 \times 5}{2}=\frac{5-15}{2}=\frac{-10}{2}=-5

x	1	-2	-5
y	1	3	5

On the graph paper, the two perpendicular XOX’ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (1, 1), (2, 5) and (3, 9) are plotted and joined them. Thus the straight line AB is obtained and the points (1, 1), (-2, 3) and (-5, 5) are plotted and joined them. Thus the straight line CD. is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The co-ordinates of P is (1, 1)

∴ x = 1, y = 1, satisfy both the equation (i) & (ii)

Question 8 (ii)

3x – y = 5 ; 4x + 3y = 11

Solution:

(i) 3x – y = 5

or, 3x = 5 + y

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

If y=1, x=\frac{5+1}{3}=\frac{6}{3}=2

If y=4, x=\frac{5+4}{3}=\frac{9}{3}=3

If y=7, \quad x=\frac{5+7}{3}=\frac{12}{3}=4

x	2	3	4
y	1	4	7

4x + 3y = 11……..(ii)

or, 4x = 11 – 3y

\therefore x=\frac{11-3 y}{4}

Ify=1, x=\frac{11-3 \times 1}{4}=\frac{11-3}{4}=\frac{8}{4}=2

If y=5, x=\frac{11-3 \times 5}{4}=\frac{11-15}{4}=\frac{-4}{4}=-1

If y=9, \quad x=\frac{11-3 \times 9}{4}=\frac{11-27}{4}=\frac{-16}{4}=-4

x	2	-1	-4
y	1	5	9

On the graph paper, the two perpendicular XOX’ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (2, 1), (3, 4) and (4, 7) are plotted and joined them. Thus the straight line AB is obtained and the points (2, 1), (-1, 5) and (-4, 9) are plotted and joined them. Thus the straight line CD is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P are (2, 1)

∴ x = 2, y = 1, satisfy both the equation (i) & (ii)

Question 8 (iii)

3x – 2y = 1 ; 2x – y = 3

Solution:

(i) 3x – 2y……(i)

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

If y=1, x=\frac{1+2 \times 1}{3} =\frac{1+2}{3}=\frac{3}{3}=1

If y=4, x =\frac{1+2 \times 4}{3} =\frac{1+8}{3}=\frac{9}{3}=3

If y=7, x=\frac{1+2 \times 7}{3} =\frac{1+14}{3}=\frac{15}{3}=5

x	1	3	5
y	1	4	7

2x – y = 3…….(ii)

or, 2x = 3 – y

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

If y=1,  x=\frac{3+1}{2}=\frac{4}{2}=2

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

If y=5, x=\frac{3+5}{2}=\frac{8}{2}=4

x	2	3	4
y	1	3	5

On the graph paper, the two perpendicular XOX’ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (1, 1), (3, 4) and (5, 7) are plotted and joined them.

Thus the straight line AB is obtained and the points (2, 1), (3. 3) and (4, 5) are plotted and joined them. Thus the straight-line CD is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P is (4, 5)

∴ x = 4, y = 5, satisfy both the equation (i) & (ii)

Question 8 (iv)

2x + 3y = 12; 2x = 3y

Solution:

(i) 2x + 3y = 12

or, 2x = 12 – 3y

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

If y=0, \quad x=\frac{12-3 \times 0}{2}=\frac{12-0}{2}=\frac{12}{2}=6\\

y=2, \quad x=\frac{12-3 \times 2}{2}=\frac{12-6}{2}=\frac{6}{2}=3\\

y=4, \quad x=\frac{12-3 x 4}{2}=\frac{12-12}{2}=\frac{0}{2}=0\\

x	6	3	0
y	0	2	4

2x = 3y =12……..(ii)

or, 2x = 12 + 3y

\therefore \quad x=\frac{12+3 y}{2} \\

If y=0, \quad x=\frac{12+3 \times 0}{2}=\frac{12+0}{2}=\frac{12}{2}=6 \\

y=2, \quad x=\frac{12+3 x 2}{2}=\frac{12+6}{2}=\frac{18}{2}=9 \\

y=4, \quad x=\frac{12+3 x 4}{2}=\frac{12+12}{2}=\frac{24}{2}=12

x	6	9	12
y	0	2	4

On the graph paper, the two perpendicular XOX’ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (6, 0), (3, 2) and (0, 4) are plotted and joined them. Thus the straight line AB is obtained and the points.(6, 0), (9, 2) and (12, 4) are plotted and joined them. Thus the straight-line CD is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P is (6, 0)

∴ x = 6, y = 0, satisfy both the equation (i) & (ii)

Question 8 (v)

5x − 2y = 1; 3x + 5y = 13

Solution:

5x − 2y = 1

or, 5x = 1 + 2y…….(i)

\therefore \quad x=\frac{1+2 y}{5}\\

If y=2, \quad x=\frac{1+2 \times 2}{5}=\frac{1+4}{5}=\frac{5}{5}=1 \\

y=7, \quad x=\frac{1+2 \times 7}{5}=\frac{1+14}{5}=\frac{15}{5}=3\\

y=12, \quad x=\frac{1+2 \times 12}{5}=\frac{1+24}{5}=\frac{25}{5}=5\\

x	1	3	5
y	2	7	12

3x + 5y = 13……(ii)

or, 3x = 13 – 5y

\therefore \quad x=\frac{13-5 \mathrm{y}}{3}

If y=2, x=\frac{13-5 \times 2}{3}=\frac{13-10}{3}=\frac{3}{3}=1 \\

y=5, \quad x=\frac{13-5 \times 5}{3}=\frac{13-25}{3}=\frac{-12}{3}=-4 \\

y=8, \quad x=\frac{13-5 \times 8}{3}=\frac{13-40}{3}=\frac{-27}{3}=-9 \\

x	1	-4	-9
y	2	5	8

On the graph paper, the two perpendicular XOX′ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (1, 2), (3, 7) and (5, 12) are plotted and joined them. Thus the straight line AB is obtained and the points (1, 2),(−4, 5) and (−9, 8) are plotted and joined them. Thus the straight-line CD is obtained.

It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P is (1, 2)

∴ x = 1, y = 2, satisfy both the equation (i) & (ii)

Question 9

Lets us determine the solution of the given equation graphically:

3x + 2y = 12 = 9x – 2y

Solution:

3x + 2y = 12……(i)

or, 3x = 12 – 2y

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

If y=0, \quad x=\frac{12-2 \times 0}{3}=\frac{12-0}{3}=\frac{12}{3}=4 \\

y=3, \quad x=\frac{12-2 \times 3}{3}=\frac{12-6}{3}=\frac{6}{3}=2 \\

y=6, \quad x=\frac{12-2 \times 6}{3}=\frac{12-12}{3}=\frac{0}{3}=0 \\

x	4	2	0
y	0	3	6

9x – 2y = 12……..(ii)

or 9x = 12 + 2y

\therefore \quad x=\frac{12+2 y}{9}\\

If y=3, \quad x=\frac{12+2 \times 3}{9}=\frac{12+6}{9}=\frac{18}{9}=2 \\

y=12, \quad x=\frac{12+2 \times 12}{9}=\frac{12+24}{9}=\frac{36}{9}=4 \\

y=-6, \quad x=\frac{12+2(-6)}{9}=\frac{12-12}{9}=\frac{0}{9}=0\\

x	2	4	0
y	3	12	-6

On the graph paper, the two perpendicular XOX’ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (4, 0),(2, 3) and (0, 6) are plotted and joined them. Thus the straight line AB is obtained and the points (2, 3), (4, 12) and (0, −6) are plotted and joined them. Thus the straight-line CD is obtained. It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P is (2, 3)

∴ x = 2,y = 3, satisfy both the equation (i) & (ii)

Question 10

Let us draw the graph of the equation \frac{x}{3}+\frac{y}{4}=2 and Calculate the area of the triangle formed by the graph and axes and write the area.

Solution:

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

or, \frac{4 x+3 y}{12}=\frac{2}{1}\\

or, 4x + 3y=  24

or, 4x = 24 – 3

\therefore \quad x=\frac{24-3 y}{4} \\

x	6	3	0
y	0	4	8

If y=0, \quad x=\frac{24-3 \times 0}{4}=\frac{24-0}{4}=\frac{24}{4}=6 \\

y=4, \quad x=\frac{24-3 \times 4}{4}=\frac{24-12}{4}=\frac{12}{4}=3 \\

y=8, \quad x=\frac{24-3 \times 8}{4}=\frac{24-24}{4}=\frac{0}{4}=\\

On the graph paper, the two perpendicular XOX′ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (6,0),(3,4) and (0,8) are plotted and joined them. Thus the straight line AB is obtained.

Area =\frac{1}{2} \times \text { base } \times \text { altitude } \\

= \frac{1}{2} × 6 × 8 = 24 Sq. units

Question 11

Let us draw the graph of the three equations x = 4, y = 3 and 3x + 4y = 12 and determine the area of the triangle formed by the graph.

Solution:

x = 4

for any value of y,

the value of x will be 4

∴

x	4	4	4
y	2	4	6

y = 3

for any value of x,

the value of y will be 3

∴

x	2	4	6
y	3	3	3

x = 4y = 12 or, 3x = 12 − 4y

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

If y=0, \quad x=\frac{12-4 \times 0}{3}=\frac{12-0}{3}=\frac{12}{3}=4\\

y=3, \quad x=\frac{12-4 \times 3}{3}=\frac{12-12}{3}=\frac{0}{3}=0 \\

y=6, \quad x=\frac{12-4 \times 6}{3}=\frac{12-24}{3}=\frac{-12}{3}=-4\\

x	4	0	-4
y	0	3	6

Here we get the required \bigtriangleup PQR by the graph of three given equations.

∴ The area of \bigtriangleup PQR

=\frac{1}{2} \times base \times altitude\\

=\frac{1}{2} \times 3 \times 4\\

=6 \mathrm{Sq}. Units.\\

Question 12

Let us draw the graph of the equation y =\frac{x+2}{3}

From the graph let’s determine the value of y where x=2 and the value of x where y=3

Solution:

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

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

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

x=7, \quad y=\frac{7+2}{3}=\frac{9}{3}=3\\

x	1	-2	7
y	1	0	3

On the graph paper, the two perpendicular XOX′ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (1, 1),(−2, 0) and (7, 3) are plotted and joined them. Thus the straight line AB is obtained.

Question 13

Let us solve graphically: \frac{3 x-1}{2}=\frac{2 x+6}{3}

Solution:

Let us determine the point of intersection by drawing the graph of the equations.

y=\frac{3 x-1}{2} \text { and } y=\frac{2 x+6}{3}

The coordinates of the point of intersection will be the required solution.

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

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

If x=3, y=\frac{3 \times 3-1}{2}=\frac{9-1}{2}=\frac{8}{2}=4\\

If x=5, y=\frac{3 \times 5-1}{2}=\frac{15-1}{2}=\frac{14}{2}=7 \\

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

If x=0, y=\frac{2 \times 0+6}{3}=\frac{0+6}{3}=\frac{6}{3}=2\\

x=3, \quad y=\frac{2 \times 3+6}{3}=\frac{6+6}{3}=\frac{12}{3}=4\\

x=6, \quad y=\frac{2 \times 6+6}{3}=\frac{12+6}{3}=\frac{18}{3}=6\\

On the graph paper, the two perpendicular XOX′ and YOY’ are drawn and taking the side of the smallest square as 1 unit, the points (1, 1),(3,4 ) and (5, 7) are plotted and joined them. Thus the straight line AB is obtained and the points (0, 2), (3, 4) and (6, 6) are plotted and joined them. Thus the straight-line CD is obtained. It is observed that the two straight lines AB & CD intersect at point P. The coordinates of P are (1, 2)

∴ x = 1,y = 2, satisfy both the equation (i) & (ii)

Multiple Choice Questions :

Question 14 (i)

The graph of the equation is

1. Parallel to x-axis
2. Parallel to y-axis
3. not parallel to any axis.
4. Passing through origin

Solution:

(b) Parallel to the y-axis

Question 14 (ii)

The graph of the equation ay +b=0 ( a & b are constants and a≠0,b≠0 ) is

1. Parallel to x-axis
2. Parallel to y-axis
3. not parallel to any axis
4. Passing through origin

Solution:

(a) Parallel to the x-axis

Question 14 (iii)

The graph of the equation 2x + 3y = 0 is

1. Parallel to x-axis
2. Parallel to y-axis
3. pasing through origin
4. Passing through (2, 0)

Solution:

(a) passing through origin since (0, 0) satisfies the equation 2x + 3y = 0

Question 14 (iv)

The graph of the equation cx + d = 0 (c & d are constants and c≠0) will be the y-axis, when

1. d =-c
2. d = c
3. b = 2
4. b = 0

Solution:

(c) b = 2

Question 14 (v)

The graph of the equation ay + b = 0 (a & b are constants and a = 0) will be x-axis, when

1. b = a
2. d = c
3. d = 0
4. d = 1

Solution:

(d) d = 1

Short answer type questions:

Question 15 (i)

Let us write the coordinates of the point of intersection of the graph of equation 2x + 3y = 12 and the x-axis.

Solution:

The equation of the x-axis is y = 0

and other equation is 2x + 3y = 12

Putting y = 0 in 2x + 3y = 12, we get

2x + 3 × 0 = 12 or, 2x + 0 = 12

or, 2x = 12

or, x = \frac{12}{2}

∴ x = 6

The point of intersection of the graph of the equation

2x + 3y = 12 and x-axis = (6, 0)

Question 15 (ii)

Let us write the coordinates of the point of intersection of the graph of equation 2x – 3y = 12 and the y-axis.

Solution:

The equation of the y-axis is x = 0

and other equation is 2x-3y = 12

Putting x = 0 in 2x-3y = 12, we get

2 × 0 − 3y = 12 or, 0 − 3y = 12

or, −3y = 12

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

∴ y = −4 The point of intersection of the graph of equation 2x + 3y = 12 and y-axis =(0, −4)

Question 15 (iii)

Let us write the area of the triangle formed by the graph of equation 3x + 4y = 12 and coordinate axes.

Solution:

The equation of the x-axis is y = 0

and equation if the y-axis is x = 0

Putting x = 0 in 3x + 4y = 12, we get

3 × 0 + 4y = 12 or, 4y = 12

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

∴ y = 3 Putting y=0 in 3x + 4y = 12, we get

3x + 4 × 0= 12 or, 3x = 12

or, x = \frac{12}{3}

∴ x = 4

Area of the triangle AOB = 12 × base × altitude

= 12 × 4 × 3 = 6 sq. units. (Ans.)

Question 15 (iv)

(iv) Let us write the distances of the point (6, -8) from the x-axis and y-axis.

Solution:

The distance of the point (6, -8) from x-axis (y = 0)

= 0 + 8  = 8 units.

The distance of the point (6, -8) from y-axis (x = 0)

= 6 – 0

= 6 units. (Ans.)

Question 15 (v)

Let us write the angle derived from the equation x = y in the positive direction of the x-axis.

Solution:

For any values of x and y. on the graph paper, we found T’s right-angled isosceles triangle. So the straight line x = y makes 45° in the positive direction of the x-axis.