Skip to content
Related Articles

Related Articles

Improve Article

Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.2 | Set 2

  • Last Updated : 16 May, 2021

Question 14. Solve the following of equation graphically

X – 2y + 11 = 0

3x + 6y + 33 = 0

Solution: 

Given that, x – 2y + 11 = 0 and 3x + 6y + 33 = 0

Now, x – 2y + 11 = 0

x = 2y -11

When y = 5, we get x = -1



When y = 4, we get x = -3

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

x-1-3
y5-4

Now, 3x – 6y + 33 = 0

x = (6y – 33)/3

When y = 6, we get x = -1

When y = 5, we get x = -1

So, the following table giving points on the line 3x – 6y + 33 = 0

x1-1
y65

So, the graph of the equations x – 2y + 11 = 0 and 3x + 6y + 33 = 0:



From the graph we conclude that the graphs of the two equations are coincident

Hence, the system of equations has infinitely many solutions.

Question 15. Solve the following of equation graphically

3x – 5y = 20

6x – 10y = – 40

Solution: 

Given that, 3x – 5y = 20 and 6x – 10y = – 40

Now, 3x – 5y = 20

x = (5y + 20)/3

When y = -1, we get x = 5

When y = – 4, we get x = 0

So, the following table giving points on the line 3x – 5y = 20

x50
y-1-4

Now, 6x – 10y = – 40

x = (6y – 40)/6

When y = 4, we get x = 0

When y = 1, we get x = – 5

So, the following table giving points on the line 6x – 10y = – 40

x0-5
y41

So, the graph of the equations 3x – 5y = 20 and 6x – 10y = – 40:

From the graph we conclude that there is no common point between these two lines.

Hence, given systems of equations is in consistent.

Question 16. Solve the following of equation graphically

x – 2y = 6

3x – 6y = 0

Solution: 



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

Now, x – 2y = 6

x = 6 + 2y

When y = 0, we get x = 6

When y = -2, we get x = 2

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

x62
y0-2

Now, 3x – 6y = 0

= x = 2y

When y = 0, we get x = 0

When y = 1, we get x = 2

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

x02
y01

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

From the graph we conclude that both the lines are parallel. So, the two lines have no common point.

Hence, the given system of equations is in-consistent.

Question 17. Solve the following of equation graphically

2y – x = 9

6y – 3x = 21

Solution: 

Given that, 2y – x = 9 and 6y – 3x = 21

Now, 2y – x = 9

x = -9 + 2y

When y = 3, we get x = -3



When y = 4, we get x = -1

So, the following table giving points on the line 2y – x = 9

x-3-1
y34

Now, 6y – 3x = 21

x = 2y – 7

When y = 2, we get x = -3

When y = 3, we get x = -1

So, the following table giving points on the line 6y – 3x = 21

x-3-1
y23

Graph of the given equations 2y – x = 9 and 6y – 3x = 21:

From the graph we conclude that both the lines are parallel. So, the two lines have no common point.

Hence, the given system of equations is in-consistent.

Question 18. Solve the following of equation graphically

3x – 4y – 1 = 0,

2x – 8/3y + 5 = 0

Solution: 

Given that, 3x – 4y – 1 = 0 and 2x – 8/3y + 5 = 0

Now,  

3x – 4y -1 = 0

x = (4y + 1)/3

When y = 2, we get x = 3

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

So, the following table giving points on the line 3x – 4y – 1 = 0

x-13
y-12

Now,

2x – 8/3y + 5 = 0

x = (8y – 15)/6

When y = 0, we get x = – 2.5

When y = 3, we get x = 1.5

So, the following table giving points on the line 2x – 8/3y + 5 = 0

x-2.51.5
y03

Graph of the given equations 3x – 4y – 1 = 0 and 2x – 8/3y + 5 = 0:

From the graph we conclude that both the lines are parallel. So, the two lines have no common point.

Hence, the given system of equations is inconsistent.

Question 19. Determine graphically the vertices of the triangle, the equations of whose sides are given below,

(i) 2y – x = 8, 5y – x = 14 and y – 2x = 1

(ii) y = x, y = 0 and 3x + 3y = 10

Solution: 



(i) 2y – x = 8

5y – x = 14

y – 2x = 1

Now, 2y – x = 8

x = 2y – 8

When y = 2, we get x = – 4

When y = 4, we get x = 0

So, the following table giving points on the line 2y – x = 8

x-40
y24

Now, 5y – x = 14

x = 5y – 14

When y = 2, we get x = 1

When y = 3, we get x = 1

So, the following table giving points on the line 5y – x = 14

x-41
y23

We have,

y – 2x = 1

x = (y-1)/2

When y = – 1, we get x = 1

When y = 3, we get x = 1

So, the following table giving points on the line y – 2x = 1

x-11
y13

So, the graph of the equation 2y – x = 8, 5y – x = 14 and y – 2x = 1:

From the graph of the lines represent by the given equation, we conclude that the lines taken in pairs intersect at points A (- 4, 2) B(1, 3) and C(2, 5)

Hence, the vertices of the triangle are A(- 4, 2) B(1, 3) and C(2, 5)

(ii) The given systems of equations are:

y = x

y = 0

3x + 3y = 10

We have, y = x

When x = 1, we get y = 1

When x = -2, we get y = – 2

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

x1-2
y3/73/4

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

From the graph of the lines represent by the given equation, we conclude that the lines taken in pairs intersect at points

A(0, 0) B(10/3, 0) and C(5/3, 5/3)

Hence, the vertices of the triangle are A(0, 0), B(10/3, 0) and C(5/3, 5/3).

Question 20. Determine graphically whether the system of equations x – 2y = 2, 4x – 2y = 5 is consistent or in-consistent

Solution: 

Given that, x – 2y = 2 and 4x – 2y = 5

Now, x – 2y = 2

x = 2 + 2y



When y = 0 then, x = 2

When y = -1 then, x = 0

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

x20
y0-1

Now, 4x – 2y = 5

x = (5+2y)/4

When y = 0, then x = 5/4

When y = 1, then x = 7/4

So, the following table giving points on the line 4x – 2y = 5

x5/47/4
y01

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

From the graph we conclude that the two lines intersect at (1, 0)

Hence, the system of equations is consistent.

Question 21. Determine by drawing graphs, whether the following system of linear equation has a unique solution or not:

(i) 2x – 3y = 6 and x + y = 1

(ii) 2y = 4x – 6 and 2x = y + 3

Solution: 

(i) Given that, 2x – 3y = 6 and x + y = 1

Now, 2x – 3y = 6

x = (6 + 3y)/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

x30
y0-2

Now, x + y = 1

x = 1 – y

When y = 0, we get x = 1

When y = 1, we get n x = 0

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

x01
y10

So, the graph of the given equations 2x – 3y = 6 and x + y = 1:

(ii) Given that, 2y = 4x – 6 and 2x = y + 3

x = (6 + 2y)/4

Now, 2y = 4x – 6

When y = -1, we get x = 1

When y = 5, we get x = 4

So, the following table giving points on the line 2y = 4x – 6

x14
y-15

Now, 2x = y + 3

x = (y + 3)/2

When y = 1, we get x = 2

When y = 3, we get x = 3

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

x23
y13

So, the graph of the given equations 2y = 4x – 6 and 2x = y + 3:

From the graph we conclude that the graphs of the two equations are consistent.



Hence, the system of equations has infinitely many solutions.

Question 22. Solve graphically each of the following system of linear equations. Also, find the coordinates of the points where the lines meet the axis of y.

(i) 2x – 5y + 4 = 0 and 2x + y – 8 = 0

Solution: 

Given that, 2x – 5y + 4 = 0 and 2x + y – 8 = 0

Now, 2x – 5y + 4 = 0

x = (5y – 4)/2

When y = 2, we get x = 3

When y = 4, we get x = 8

So, the following table giving points on the line 2x – 5y + 4 = 0

x38
y24

Now, 2x + y – 8 = 0

x = (-y + 8)/2

When y = 4, we get x = 2

When y = 2, we get x = 3

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

x38
y24

So, the graph of the given equations 2x – 5y + 4 = 0 and 2x + y – 8 = 0

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

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

We also observe that the lines represented by 2x – 5y + 4 = 0 and 2x + y – 8 = 0 meet y-axis at A(0, 4/5) and B(0, 8) respectively.

(ii) 3x + 2y = 12 and 5x – 2y = 4

Solution: 

Given that, 3x + 2y = 12 and 5x – 2y = 4

Now, 3x + 2y = 12

x = (12 – 2y)/3

When y = 3, we get x = 2

When y = -3, we get x = 6

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

x26
y3-3

Now, 5x – 2y = 4

x = (4 + 2y)/5

When y = 3, we get x = 2

When y = -7, we get x = -2

So, the following table giving points on the line 5x – 2y = 4

x2-2
y3-7

So, the graph of the given equations 3x + 2y = 12 and 5x – 2y = 4:

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

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

We also observe that the lines represented by 3x + 2y = 12 and 5x – 2y = 4 meet y-axis at A(0, 6) and B(0,-2) respectively.

(iii) 2x + y – 11 = 0 and x – y – 1 = 0

Solution: 

Given that, 2x + y – 11 = 0 and x – y – 1 = 0

Now, 2x + y = 11

y = 11 – 2x

When y = 4, we get x = 3



When y = – 5, we get x = 1

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

x45
y31

Now, x – y = 1

y = x – 1

When x = 2, we get y = 1

When y = 3, we get y = 2

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

x23
y12

So, the graph of the given equations 2x + y – 11 = 0 and x – y – 1 = 0:

From the graph we conclude that the two intersect at P (4, 3)

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

We also observe that the lines represented by 2x + y = 11 and x – y = 1 meet y-axis at A(0, 11) and B(0,-1) respectively.

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

Solution: 

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

Now, 2x – y – 4 = 0

X = 7 – 2y

When y = 1, we get x = 5

When y = -2, we get x = 3

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

x53
y12

Now, 2x – y – 4 = 0

y = 2x – 4

When x = 2, we get y = 0

When y = 0, we get y = -4

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

x20
y0-4

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

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

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

We also observe that the lines meet y-axis at A(0, 3.5) and B(0,- 4) respectively.

(v) 3x + y – 5 = 0 and 2x – y – 5 = 0

Solution: 

Given that, 3x + y – 5 = 0 and 2x – y – 5 = 0

Now, 3x + y – 5 = 0

y = 5 – 3x

When x = 1, we get y = 2

When x = 2, we get y = -1

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

x12
y2-1

Now,  2x – y – 5 = 0

y = 2x – 5

When x = 0, we get y = -5

When x = 2, we get y = -1

So, the following table giving points on the line 2x – y – 5 = 0

x02
y-5-1

So, the graph of the given equations 3x + y – 5 = 0 and 2x – y – 5 = 0:

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

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

We also observe that the lines meet y-axis at A(0, 5) and B(0, – 5) respectively.

(vi) 2x – y – 5 = 0 and x – y – 3 = 0

Solution: 

Given that,

2x – y – 5 = 0        ……(i)

x – y – 3 = 0          ……(ii)



The two points satisfying (i) can be listed in a table as,

x13
y-31

The two points satisfying (ii) can be listed in a table as,

x15
y-2-2

Now, the graph of equations (i) and (ii) can be drawn as,

Graph of the equation (i) and (ii) It is seen that the solution of the given system of equation is given by x = 2, y = -1.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points (0, – 3) and (0, – 5) respectively.

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




My Personal Notes arrow_drop_up
Recommended Articles
Page :