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Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.2 | Set 3

  • Last Updated : 16 May, 2021

Question 23. Solve the following system of linear equations graphically and shade the region between the two lines and the x-axis.

(i) 2x + 3y = 12 and x – y = 1

Solution: 

Given that, 2x + 3y = 12 and x – y = 1

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Now, 2x + 3y = 12



x = (12-3y)/2

When y = 2, we get x = 3

When y = 4, we get x = 0

So, the following table giving points on the line 2x + 3y = 12

x03
y42

Now, x – y = 1

x = y + 1

When y = 0, we get x = 1

When y = 1, we get x = 2

So, the following table giving points on the line x – y = 1

x12
y01

So, the graph of the equations 2x + 3y = 12 and x – y = 1:

From the graph we conclude that the two lines intersect at P (3, 2)

Hence, x = 3 and y = 2 is the solution of the given system of equations.

(ii) 3x + 2y – 4 = 0 and 2x – 3y – 7 = 0

Solution: 

Given that, 3x + 2y – 4 = 0 and 2x – 3y – 7 = 0

Now, 3x + 2y – 4 = 0

x = (4 – 2y)/3

When y = 5, we get x = – 2



When y = 8, we get x = – 4

So, the following table giving points on the line 3x + 2y – 4 = 0

x-2-4
y58

We have,

2x – 3y – 7 = 0

x = (3y + 7)/2

When y = 1, we get x = 5

When y = -1, we get x = 2

So, the following table giving points on the line 2x – 3y – 7 = 0

x52
y1-1

So, the graph of the equations 3x + 2y – 4 = 0 and 2x – 3y – 7 = 0:



From the graph we conclude that the two lines intersect at P(2,-1)

Hence, x = 2 and y = -1 is the solution of the given system of equations.

(iii) 3x + 2y – 11 = 0 and 2x – 3y + 10 = 0

Solution: 

Given that, 3x + 2y – 11 = 0 and 2x – 3y + 10 = 0

Now, 3x + 2y – 11 = 0

x = (11 – 2y)/3

When y = 1, we get x = – 3

When y = 4, we get x = 1

So, the following table giving points on the line 3x + 2y – 11 = 0 

x31
y14

Now, 

2x – 3y + 10 = 0

x = (3y-10)/2

When y = 0, we get x = – 5

When y = 2, we get x = – 2

So, the following table giving points on the line 2x – 3y + 10 = 0

x-5-2
y02

So, the graph of the equations 3x + 2y – 11 = 0 and 2x – 3y + 10 = 0:

From the graph we conclude that the two lines intersect at P(1, 4)

Hence, x = 1 and y = 4 is the solution of the given system of equations.

Question 24. Draw the graphs of the following equations on the same graph paper:

2x + 3y = 12 and x – y = 1

Solution: 



Given that, 2x + 3y = 12 and x – y = 1

Now, 2x + 3y = 12

x = (12 – 3y)/2

When y = 0, we get x = 6

When y = 2, we get x = 3

So, the following table giving points on the line 2x + 3y = 12:

x63
y02

Now,

x – y = 1

x = 1 + y

When y = 0, we get x = 1

When y = -1, we get x = 0

So, the following table giving points on the line x – y = 1:

x10
y0-1

So, the graph of the equations 2x + 3y = 12 and x – y = 1:

From the graph we conclude that the two lines intersect at A (3, 2)

Also, we observe that the lines meet y-axis B (0, – 1) and C (0, 4)

Hence, the vertices of the required triangle are A (3, 2), B (0,-1), and C (0, 4).

Question 25. Draw the graphs of x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and the x-axis.

Solution: 

Given that, x – y + 1 = 0 and 3x + 2y – 12 = 0

Now, x – y + 1 = 0



x = y – 1

When y = 3, we get x = 2

When y = -1, we get x = -2

So, the following table giving points on the line x – y + 1 = 0

x2-2
y3-1

We have,

3x + 2y – 12 = 0

x = (12 – 2y)/3

When y = 6, we get x = 0

When y = 3, we get x = 2

So, the following table giving points on the line 3x + 2y – 12 = 0

x02
y63

So, the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0:

From the graph we conclude that the two lines intersect at A(2, 3)

Also, we observe that the lines meet x-axis B(-1, 0) and C(4, 0)

So, x = 2 and y = 3 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. 

So, 

AD = y-coordinate point A (2, 3)

AD = 3 and BC = 4- ( – 1) = 4 + 1 = 5

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 5 × 3 = 15/2 = 7.5 sq. units

Question 26. Solve graphically the system of linear equation:

4x – 3y + 4 = 0 and 4x + 3y – 20 = 0

Find the area bounded by these lines and x-axis.

Solution: 

Given that, 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0

Now, 4x – 3y + 4 = 0

x = (3y – 4)/4

When y = 0, we get x = -1

When y = 4, we get x = 2

So, the following table giving points on the line 4x – 3y + 4 = 0

x2-1
y40

Now, 

4x + 3y – 20 = 0



x = (20 – 3y)/4

When y = 0, we get x = 5

When y = 4, we get x = 2

So, the following table giving points on the line 4x + 3y – 20 = 0

x52
y04

So, the graph of the equations 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0:

From the graph we conclude that the two lines intersect at A (2, 4)

Also, we observe that the lines meet x-axis B (-1, 0) and C(5, 0)

So, x = 2 and y = 4 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. 

So, we get

AD = y-coordinate point A (2, 4)

AD = 3 and BC = 5 – (-1) = 4 + 1 = 6

Hence, the area of the shaded region =1/2 × base × altitude

1/2 × 6 × 4 = 12 sq. units

Question 27. Solve the following system of linear equations graphically:

3x + y – 11 = 0 and x – y – 1 = 0

Shade the region bounded by these lines and the y-axis. Also, find the area of the region bounded by these lines and the y-axis.

Solution: 

Given that, 3x + y – 11 = 0 and x – y – 1 = 0

Now, 3x + y – 11 = 0

y = 11 – 3x

When x = 0, we get y = 11

When x = 3, we get y = 2

So, the following table giving points on the line 3x + y – 11 = 0

x03
y112

We have

x – y – 1 = 0

y = x – 1

When x = 0, we get y = -1

When x = 3, we get y = 2

So, the following table giving points on the line x – y – 1 = 0

x03
y-12

So, the graph of the equations 3x + y – 11 = 0 and x – y – 1 = 0:

From the graph we conclude that, the two lines intersect at A (3, 2)

We also observe that the lines meet y-axis B(0, 11) and C (0, – 1)

So, x = 3 and y = 2 is the solution of the given system of equations.

AD is drawn perpendicular A on x-axis. Clearly we have,

AD = y-coordinate point A(2, 4)

AD = 3 and BC = 11- (- 1) = 11 + 1 = 12

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 12 × 3 = 18 sq. units

Question 28. Draw the graph of the following equation:

2x – 3y + 6 = 0,

2x + 3y – 18 = 0,

y – 2 = 0

Find the vertices of the triangle so obtained. Also, find the area of the triangle.

Solution: 

Given that, 2x – 3y + 6 = 0, 2x + 3y – 18 = 0, and y – 2 = 0



Now, 2x – 3y + 6 = 0

x = (3y – 6)/2

When y = 0, we get x = -3

When y = 2, we get x = 0

So, the following table giving points on the line  2x – 3y + 6 = 0

x-30
y02

Now,

2x + 3y – 18 = 0

x = 18-3y / 2

When y = 2, we get x = 6

When y = 6, we get x = 0

So, the following table giving points on the line 2x + 3y – 18 = 0

x60
y26

Now, 

y – 2 = 0

y = -2

So, the graph of the equations 2x – 3y + 6 = 0, 2x + 3y – 18 = 0, and y – 2 = 0:

From the graph of three equations, we conclude that the three lines taken in pairs 

intersect each other at points A(3, 4), B(0, 2), and C(6, 2)

Hence, the vertices of the required triangle are (3, 4), (0, 2) and (6, 2)

From the graph, we have

AD = 4 – 2 = 2

BC = 6 – 0 = 6

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 6 × 2 = 6 sq. units

Question 29. Solve the following system of equations graphically:

2x – 3y + 6 = 0 and 2x + 3y – 18 = 0

Also, find the area of the region bounded by these lines and y-axis.

Solution: 

Given that, 2x – 3y + 6 = 0 and 2x + 3y – 18 = 0

Now, 2x – 3y + 6 = 0

y = (2x + 6)/3

When x = 0, we get y = 2

When x = -3, we get y = 0

So, the following table giving points on the line 2x – 3y + 6 = 0 

x0-3
y26

Now, 

2x + 3y – 18 = 0

x = (18 – 3y)/2

When y = 2, we get x = 6

When y = 6, we get x = 0

So, the following table giving points on the line 2x + 3y – 18 = 0

x60
y26

So, the graph of the equations 2x – 3y + 6 = 0 and 2x + 3y – 18 = 0:

From the graph we conclude that, the two lines intersect at A (3, 4).

Hence x = 3 and y = 4 is the solution of the given system of equations.

Also, from the graph, we have

AD = x-coordinate point A(3, 4) = 3

BC = 6 – 2 = 4

Hence, the area of the shaded region = 1/2 × base × altitude

1/2 × 4 × 3 = 6 sq. units

Question 30. Solve the following system of linear equation graphically;

4x – 5y – 20 = 0 and 3x + 5y – 15 = 0

Determine the vertices of triangle formed by the lines representing the above equation and the y-axis.

Solution: 

Given that, 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0

Now, 4x – 5y – 20 = 0

x = (5y + 20)/4



When y = 0, we get x = 5

When y = – 4, we get x = 0

So, the following table giving points on the line 4x – 5y – 20 = 0 

x50
y0-4

We have

3x + 5y – 15 = 0

x = (5y + 20)/5

When y = 0, we get x = 5

When y = – 4, we get x = 0

So, the following table giving points on the line 3x + 5y – 15 = 0

x50
y03

So, the graph of the equations 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0:

From the graph we conclude that the two lines intersect at A(5, 0). 

Hence, x – 5, y – 0 is the solution of the given system of equations.

So, the lines meet the y-axis at B(0, -4) and C(0, 3) respectively.

Hence, the vertices of the triangle are (5, 0), (0,- 4), and (0, 3)

Question 31. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle forms by these lines and y-axis. Calculate the area of the triangle forms.

Solution: 

Given that, 5x – y = 5 and 3x – y = 3

Now, 

5x – y = 5

y = 5x – 5

So, the following table giving points on the line 5x – y = 5

x012
y-505

Now, 3x – y = 3

y = 3x – 3

So, the following table giving points on the line 3x – y = 3

x012
y-303

So, the graph of the equations 5x – y = 5 and 3x – y = 3:

It can be observed that the required triangle is ABC

Hence, the coordinates of its vertices A (1, 0) B(0, – 3) and C(0, -5)

Now, AB = 3.2, BC = 5.1

s = (a + b + c) /2 = (3.2 + 2 + 5.1)/2 = 5.2

Area of triangle ABC = √s(s – a)(s – b)(s – c)

= √5.2(5.2 – 3.2)(5.2 – 2)(5.2 – 5.1)

= √3.328

= 1.8 sq.unit

Question 32. Form the pair of linear equation in the following problems, and find their solution graphically:

(i) 10 students of class X took part in the mathematics quiz. If the number of girls is 4 more than the number of boys. Find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost Rs.50, whereas 7 pencils and 5 pens together cost Rs.46.

Find the cost of one pencil and one pen.

(iii) Champa went to a sale to purchase some pants and skirts. When her friends asked her how many of each kind she had bought, she answered, “the number of skirts is two less than twice the number of pants purchase”. Also, ” the number of skirts is four less than four times the number of pants purchased.” Help her friends to find how many pants and skirts Champa bought.

Solution: 

(i) Let the number of girls and boys in the class be x and y respectively.



According to the question, 

x + y = 10 and x – y = 4 are the given equations

Now, x + y = 10

x = 10 – y

So, the following table giving points on the line x + y = 10

x456
y654

Now, x – y = 4

x = 4 + y

So, the following table giving points on the line x – y = 4

x543
y10-1

So, the graph of the equations x + y = 10 and x – y = 4:

From the graph we conclude that the two lines intersect each other at point (7, 3).

So, x = 7 and y = 3

Hence, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of one pencil and one pen Rs. x and Rs. y respectively.

According to the question, we get,

5x + 7y = 50

7x + 5y = 50

Now, 5x + 7y = 50

x = (50 – 7y)/5

So, the following table giving points on the line 5x + 7y = 50

x310-4
y5010

Now, 7x + 5y = 46,

x = (46 – 5y)/7

So, the following table giving points on the line 7x + 5y = 46,

x83-2
y-2512

So, the graph of the equations 5x + 7y = 50 and 7x + 5y = 50:

From the graph we conclude that the two lines intersect each other at the point (3, 5)

So, x = 3 and y = 5

Hence, the cost of the pencil and the pen are 3 and 5 respectively.

(iii) Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :

y = 2x – 2    …..(i)

y = 4x – 2   ……(ii)

The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. 

So, they are given in the following table.

x20
y = 2x – 22-2

Hence, the graphic representation is as follows:

The two lines intersect at the point (1, 0). 

So, x = 1, y = 0 is the required solution of the pair of linear equations, 

i.e., the number of parts she purchased is 1 and she did not buy any skirt.

Question 33. Solve the following system of equations graphically:

Shade the region between the lines and y-axis

(i) 3x – 4y = 7 and 5x + 2y = 3

(ii) 4x – y = 4 and 3x + 2y = 14

Solution : 

(i) Given that, 3x – 4y = 7 and 5x + 2y = 3

Now, 3x – 4y = 7,

y = (3x – 7)/4

When x = 1, we get y = – 1

When x = – 3, we get y = – 4

So, the following table giving points on the line 3x – 4y = 7

x1-3
y-1-4

Now, 5x + 2y = 3,

y = (3 – 5x)/2

When x = 1, we get y = – 1



When x = 3, we get y = – 6

So, the following table giving points on the line 5x + 2y = 3

x13
y-1-6

So, the graph of the equations 3x – 4y = 7 and 5x + 2y = 3:

From the graph we conclude that the two lines intersect at A(1,-1)

Hence, x = 1 and y = -1 is the solution of the given system of equations.

(ii) Given that, 4x – y = 4 and 3x + 2y = 14

Now, 4x – y = 4

y = 4x – 4

When x = 0, we get y = – 4

When x = -1, we get y = – 8

So, the following table giving points on the line 4x – y = 4 

x0-1
y-4-8

Now, 3x + 2y = 14,

y = (14 – 3x)/2

When x = 0, we get y = 7

When x = 4, we get y = 1

So, the following table giving points on the line 3x + 2y = 14

x04
y71

So, the graph of the equations 4x – y = 4 and 3x + 2y = 14:

From the graph we conclude that the two lines intersect at A (2, 4)

Hence, x = 2 and y = 4 is the solution of the given system of equations.

Question 34. Represent the following pair of equations graphically and write the coordinates of points where the lines intersect the y-axis

x + 3y = 6 and 2x – 3y = 12

Solution: 

Given that, x + 3y = 6 and 2x – 3y = 12

Now, x + 3y = 6,

y = (6 – x)/3

When x = 0, we get y = 2

When x = 3, we get y = 1

So, the following table giving points on the line x + 3y = 6 

x03
y21

Now, 2x – 3y = 12,

y = (2x – 12)/3

When x = 0, we get y = – 4

When x = 6, we get y = 0

So, the following table giving points on the line 2x – 3y = 12

x06
y-40

So, the graph of the equations x + 3y = 6 and 2x – 3y = 12

From the graph we conclude that the two lines meet the y-axis at B(0, 2) and C(0, -4) respectively.

Hence, the required coordinates are (0, 2) and (0, -4)

Question 35. Given the linear equation 2x + 3y – 8 = 0, write another in two variables in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines (iii) coincident lines.

Solution: 

(i) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be intersecting. 

We must have, a1/a2 ≠ b1/b2

So the other linear equation can be 5x + 6y – 16 = 0

a1/a2 = 2/5

b1/b1 = 3/6 = 1/2

c1/c2 = -8/-16 = 1/2

(ii) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be parallel

We must have, a1/a2 = b1/b2 ≠ c1/c2

So, the other linear equation can be 6x + 9y + 24 = 0

a1/a2 = 2/6 = 1/3

b1/b2 = 3/9 = 1/3

c1/c2 = -8/-24 = 1/3 

(iii) For the two lines a1x + b1x + c1 = 0  and a2x + b2x + c2 = 0 to be coincident 

We must have, a1/a2 = b1/b2 = c1/c2

So, the other linear equation can be 6x + 9y + 24 = 0

a1/a2 = 2/8 = 1/4

b1/b2 = 3/12 = 1/4

c1/c2 = -8/-32 = 1/4




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