Ex-3.2 Pair Of Linear Equations In Two Variables (Part - 1), Class 10, Math RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Solve the following system of equations graphically:

Q1. x + y = 3; 2x + 5y = 12

Sol: 2x + 5y = 12

We have,

x + y = 3

When y = 0 we have x = 3

When x = 0 we have y = 3

Thus we have the following table giving points on the line x + y = 3

Now, 2 + 5y = 12

When x = 1, we have

Thus we have the following table giving points on the line 2x + 5y = 12

Graph of the equation x + y = 3 and 2x + 5y = 12 is

Clearly two lines intersect at a point P (1,2)

Hence x = 1 and y = 2

Q2: x- 2y = 5, 2x + 3y = 10

Sol: We have, x- 2y = 5 and 2x + 3y = 10

Now, x- 2y = 5

= x = 5 + 2y

When y = 0 then, x = 5

When y = -2 then, x = 1

Thus, we have the following table giving points on the line x-2y = 5

Now ,  2x + 3y = 10 = x =  

When y = 0, then x = 5

When y = 2 , then x = 2Thus, we have the following table giving points on the line 2x + 3y = 10

Graph of the equation x-2y = 5 and 2x + 3y = 10

Clearly, two lines intersect at a point P (5,0)

Hence x = 5 and y = 0

Q3: 3x + y + 1 = 0, 2x-3y + 8 = 0

Sol: We have, 3x + y + 1 = 0 and 2x-3y + 8 = 0

Now 3x + y + 1 = 0

= y = -1-3x

When x = 0 then, x = -1

When y = -1 then, x = 2

Thus, we have the following table giving points on the line x-2y = 5

Now,    2x-3y + 8 = 0

When y = 0, then x = -4

When y = 2 , then x = 1Thus, we have the following table giving points on the line 2x + 3y = 10

Graph of the equation 3x + y + 1 = 0 and 2x-3y + 8 = 0

Clearly two lines intersect at a point P (-1, 2)

Hence x = -1 and y = 2

Q4: 2x + y-3 = 0, 2x-3y-7 = 0

Sol: We have, 2x + y-3 = 0 and 2x-3y-7 = 0

Now 2x + y-3 = 0

= y = 3-2x

When x = 0 then, x = 3

When x = 1 then, x = 1

Thus, we have the following table giving points on the line 2x + y-3 = 0

Now,    2x-3y-7 = 0

When x = 0, then y = 1

When x = 2, then y = -1Thus, we have the following table giving points on the line 2x + 3y = 10

Graph of the equation 2x + y-3 = 0 and 2x-3y-7 = 0

Clearly two lines intersect at a point P (2,-1)

Hence x = 2 and y = -1

Q5. x + y = 6

x-y = 2

Sol. We have,  x + y = 6 and x-y = 2

Now x + y = 6

= y = 6-x

When x = 2 then, y = 4

When x = 3 then, y = 3

Thus, we have the following table giving points on the line x + y = 6

Now ,    x-y = 2

= y = x-2

When x = 0, then y = -2

When x = 2 , then y = 0Thus, we have the following table giving points on the line 2x + 3y = 10

Graph of the equation x + y = 6 and x-y = 2

Clearly two lines intersect at a point P (4,2)

Hence x = 4 and y = 2

Q6. x-2y = 6

3x-6y = 0

Sol. We have,   x-2y = 6 and 3x-6y = 0

Now x-2y = 6

= x = 6 + 2y

When y = -2 then, x = 2

When y = -3 then, x = 0

Thus, we have the following table giving points on the line x-2y = 6

Now ,    3x-6y = 0

= x = 2y

When y = 0, then y = 0

When y = -1 , then x = 2

Thus, we have the following table giving points on the line 3x-6y = 0

Graph of the equation x-2y = 6 and 3x-6y = 0

Clearly two lines are parallel to each other. So, the two lines have no common point.

Hence the given system has no solutions.

Q7. x + y = 4

2x-3y = 3

Sol. We have,   x + y = 4 and 2x-3y = 3

Now x + y = 4

= x = 4-y

When y = 0 then, x = 4

When y = 2 then, x = 2

Thus, we have the following table giving points on the line x + y = 4

Now ,    2x-3y = 3

When y = 1, then x = 3

When y = -1 , then x = 0

Thus, we have the following table giving points on the line 2x-3y = 3

Graph of the equation x + y = 4 and 2x-3y = 3

Clearly two lines intersect at a point P(3,1)

Hence x = 3 and y = 1

Q8. 2x + 3y = 4

 x -y + 3 = 0

Sol. So we have, 2x + 3y = 4 and x-y + 3 = 0

Now 2x + 3y = 4

When y = 0 then, x = 2

When y = 2 then, x = -1

Thus, we have the following table giving points on the line x + y = 4

Now ,    x-y + 3 = 0

= x = y-3

When y = 3, then x = 0

When y = 4, then x = 1

Thus, we have the following table giving points on the line x-y + 3 = 0

Graph of the equation 2x + 3y = 4 and x-y + 3 = 0

Clearly two lines intersect at (-1, 2)

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

Q9. 2x-3y + 13 = 0

3x-2y + 12 = 0

Sol. So we have 2 x-3y + 13 = 0 and 3 x-2y + 12 = 0

Now,  2x-3y + 13 = 0

When y = 1 then, x = -5

When y = 3 then, x = -2

Thus, we have the following table giving points on the line 2x-3y + 13 = 0

Now,    3x-2y + 12 = 0

When y = 0, then x = -14

When y = 3, then x = -2

Thus, we have the following table giving points on the line x-y + 3 = 0

Graph of the equation 2x-3y + 14 = 0 and 3x-2y + 12 = 0

Clearly two lines intersect at (-2, 3)

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

Q10. 2x + 3y + 5 = 0

3x + 2y-12 = 0

Sol. So we have 2x + 3y + 5 = 0 and 3x + 2y- 12 = 0

Now,  2x + 3y + 5 = 0

When y = 1 then, x = -4

When y = -1 then, x = -1

Thus, we have the following table giving points on the line 2x + 3y + 5 = 0

Now,    3x + 2y-12 = 0

When y = 0, then x = 4

When y = 3, then x = 6

Thus, we have the following table giving points on the line 3x-2y-12 = 0

Graph of the equation 2x + 3y + 5 = 0 and 3x + 2y-12 = 0

Clearly two lines intersect at (2, 3)

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

Q11. 2x + 3y = 6

4x + 6y = 12

Sol. So we have 2x + 3y = 6 and 4x + 6y = 12

Now,  2x + 3y = 5

When y = 0 then, x = 3

When y = 2 then, x = 0

Thus, we have the following table giving points on the line 2x + 3y = 6

Now,    4x + 6y = 12

When y = 0, then x = 3

When y = 2, then x = 0

Thus, we have the following table giving points on the line 4x + 6y = 12

Graph of the equation 2x + 3y = 6 and 4x + 6y = 12

Thus the graphs of the two equations are coincident.

Hence, the system of equations has infinitely many solutions.

Q12. x-2y = 5

3x-6y = 15

Sol. So we have x-2y = 5 and 3x-6y = 15

Now,  x-2y = 5

= x = 2y + 5

When y = -1 then, x = 3

When y = 0 then, x = 5

Thus, we have the following table giving points on the line x-2y = 5

Now,    3x-6y = 15

When y = -2, then x = 1

When y = -3, then x = -1

Thus, we have the following table giving points on the line 3x-6y = 15

Q13. 3x + y = 8

6x + 2y = 16

Sol. So we have 3x + y = 8 and 6x + 2y = 16

Now,  x-2y = 5

= y = 8-3x

When x = 2 then, y = 2

When x = 3 then, y = -1

Thus, we have the following table giving points on the line 3x + y = 8

Now,    6x + 2y = 16

When x = 1, then y = 5

When x = 3, then y = -1

Thus, we have the following table giving points on the line 6x + 2y = 16

Graph of the given equation

Thus, the graphs of the two equations are coincident

Hence, the system of equations has infinitely many solutions.

Q14. X-2y + 11 = 0

3x + 6y + 33 = 0

Sol. So we have x-2y + 11 = 0 and 3x + 6y + 33 = 0

Now,  x-2y + 11 = 0

= x = 2y-11

When y = 5 then, x = -1

When y = 4 then, x = -3

Thus, we have the following table giving points on the line x-2y + 11 = 0

Now,    3x-6y + 33 = 0

When y = 6, then x = -1

When y = 5, then x = -1

Thus, we have the following table giving points on the line 3x-6y + 33 = 0

Thus, the graphs of the two equations are coincident

Hence, the system of equations has infinitely many solutions.

Q15. 3x-5y = 20

6x-10y = -40

Sol. So we have 3x-5y = 20 and 6x-10y = -40

Now,  3x-5y = 20

When y = -1 then, x = 5

When y = -4 then, x = 0

Thus, we have the following table giving points on the line 3x-5y = 20

Now,    6x-10y = -40

When y = 4, then x = 0

When y = 1, then x = -5

Thus, we have the following table giving points on the line 6x-10y = -40

Clearly, there is no common point between these two lines.

Hence, given systems of equations is in consistent.

Q16. x-2y = 6

3x-6y = 0

Sol. So we have x-2y = 6 and 3x-6y = 0

Now,  x-2y = 6

= x = 6 + 2y

When y = 0 then, x = 6

When y = -2 then, x = 2

Thus, we have the following table giving points on the line x-2y = 6

Now,    3x-6y = 0

= x = 2y

When y = 0, then x = 0

When y = 1, then x = 2

Thus, we have the following table giving points on the line 3x-6y = 0

We find the lines represented by equations x-2y = 6 and 3x-6y = 0 are parallel. So, the two lines have no common point.

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