# Telangana SCERT Class 9 Math Solution Chapter 6 Linear Equations in Two Variables Exercise 6.2

## Telangana SCERT Solution Class IX (9) Math Chapter 6 Linear Equations in Two Variables Exercise 6.2

(1.) Find three different solutions of the each of the following equations.

i) 3x + 4y = 7 ii) y = 6x iii) 2x − y = 7 iv) 13x − 12y = 25 v) 10x + 11y = 21 vi) x + y = 0

Solution:

(1)(i) Given,

3x + 4y = 7

For, x = 0 we get                                               x,y = (0, 7/4)

3 x 0 + 4y = 7

Or, y = 7/4

For,

y = 0

3 x X + 4 x 0 = 7                                                  ∴ x,y = ( ,0)

Or, x =7/3

For x = 1,

3x 1 + 4y = 7                                                        ∴ x,y = (1, 1)

Or, 4y = 4

Or, y = 1

∴ The three different solutions are (0, 7/4), (7/3 ,0), (1, 1)

(ii) Given,

Y = 6x

For, y = 0,

6x = 0                                                                    ∴ x, y = (0, 0)

x = 0

For, x = 0

y = 6 x 0                                                                ∴ x, y = (0, 0)

= 0

For, x = 1

y = 6 x 1                                                                ∴ (1, 6) = x, y

= 6

∴ Three different Solutions are (0, 0) (0, 0), (1, 6)

(iii) given,

2x – y = 7

For, x = 0                                              x, y = (0, – 7)

2 x 0 – y = 7

Or, y = -7

For, y = o                                             x, y = ( , 0)

2x – 0 = 7

Or, x = 7/2

For x = 1                                               x, y = (1, -5)

2 x 1 – y = 7

Or, y = -5

Three different Solutions are (0, -7) (7/2, 0), (1, -5)

(iv) 13x – 12y = 25

For x = 0,

13 x 0 – 12 x Y = 25                           (x, y) = (0, -25/12)

Or, y = – 25/12

For y = 0,

13X – 12 x 0 = 25                               (x, y) = (25/13, 0)

Or, x = 25/13

For, x = 1,

13 x 1 – 12y = 25                                (x, y) = (1, 1)

Or, y = (25-13)/12

= 1

Three different solutions (0, – 25/12), (25/13 , 0), (1, 1)

(v) 10x + 11y = 21

For, x = 0

0 + 11y = 21                                         (x, y) = (0,

y = 21/11

For y = 0

10x + 0 = 21                                         (x, y) = (21/10 , 0)

Or, x = 21/10

For, x = 1

10 x 1 + 11y = 21                                x, y = (1, 1)

Or, y = (21-10)/11

= 1

Three different Solutions are (0, 21/11), (21/10 , 0), (1, 1)

(vi) x + y = 0

For x = 1,                                              x, y = (1, – 1)

y = – 1

For, y = 1,                                            x, y = (-1, 10)

x = -1

For x = 2                                               x, y = (2, -2)

y = -2

Three different solutions are (1, – 1), (-1, 10), (2, -2)

(2) If (0, a) and (b, 0) are the solutions of the following linear equations. Find ‘a’ and ‘b’.

i) 8x − y = 34 ii) 3x = 7y − 21 iii) 5x − 2y + 3 = 0

Solution:

(2) (i) Given,

8x – y = 34                                           Solution (0, a); (b, 0)

For, (0, a)

8 x 0 – a = 34

Or, a = -34

For (b, 0)

8 x b – 0 = 34

Or, b = 34/8

Or, b = 17/4

(ii) 3x = 7y – 21                                  Solution (0, a); (b, 0)

For, (0, a)

3 x 0 = 7 x a – 21

Or, a = 21/7

a = 3

For (b, 0)

3 x b = 7 x 0 – 21

Or, b = -21/3

Or, b = -7

(iii) Given

5x – 2y + 3 = 0                                    Solution (0, a); (b, 0)

For, (0, a)

5 x 0 – 2 x a + 3 = 0

Or,  – 2a = – 3

Or, a =

For (b, 0)

5 x b – 2 x 0 + 3 = 0

Or, b = -3/5

(3) Check which of the following are solutions of the equation 2x − 5y = 10

i) (0, 2) ii) (0, –2) iii) (5, 0) iv) 2√3, √3 v) (1/2, 2)

Solution:

Given,

2x – 5y = 10

(i) For (0, 2)

LHS                                                    RHS

2 x 0 – 5 x 2                                         10

= – 10

Hence it is not a solution

LHS ≠ RHS

(ii) For (0, – 2)

LHS                                                   RHS

2 x 0 – 5 x (-2)                                    10

= +10

= RHS

Hence (0, -2) is a solution 2x – 5y = 10

(iii) For (5, 0)

LHS                                                      RHS

2 x 0 – 5 x 0                                         10

= 10

= RHS

Hence (5, 0) is a solution of 2x – 5y = 10

(iv) For (2√3, – √3)

LHS                                                                   RHS

2 x 2√3 – 5 x (- √3)                                            10

= 4√3 + 5√3

= √3(9)

= 9√3

≠RHS

Hence (2√3, – √3) is not a solution of 2x – 5y = 10

(v) For (1/5, 2)

LHS                                                                   RHS

2 x 1/5 – 5 x 2                                                     10

= 1 – 10

= – 9

≠RHS

Hence (1/5, 2) is not a solution of 2s – 5y = 10

(4) Find the value of k, if x = 2, y =1 is a solution of the equation 2x + 3y = k. Find two more solutions of the resultant equation.

Solution:

Given,

2x + 3y = K                                                           Solution x = 2, y = 1

2 x 2 + 3 x 1 = K

Or, K = 4 + 3

= 7

The resultant equation is 2x + 3y = 7

Now, For x = 0

2 x 0 + 3 x y = 7

Or, y = 7/3

For x = 1

2 x 1 + 3y = 7

Or, y = 5/3

(0,7/3) and (1,5/3) are two more solution of equation 2x + 3y = 7

(5) If x = 2 – α and y = 2 + α is a solution of the equation 3x – 2y + 6 = 0 find the value of ‘α ’. Find three more solutions of the resultant equation.

Solution:

Given,

3x – 2y + 6 = 0                    x = 2 – ∝, y = 2 + ∝ are two solution of the given equation

3 (2 – ∝) -2 (2 + ∝) + 6 = 0

Or, 6 – 3∝ – 4 – 2∝ + 6 = 0

Or, 8 – 5∝ = 0

Or, 5∝ = 8

Or, ∝ = 8/5

For, x = 0

3 x 0 – 2y + 6 = 0

Or, – 2y = -6

Or, y = 3

For x = 1

3 x 1 – 2y + 6 = 0

Or,-  2y = – 9

Or, y = 9/2

For y = 0

3 x X – 2 x 0 + 6 = 0

Or, 3x = – 6

Or, x = – 2

(0, 3) (1,9/2), (-2, 0) are three more solution of the above given equation.

(6) If x = 1, y = 1 is a solution of the equation 3x + ay = 6, find the value of ‘a’.

Solution:

Given, 3x + ay = 6, x = 1, y = 1 are two solution

For, (1, 1)

3 x 1 + a x 1 = 6

Or, a = 6 – 3

= 3

(7.) Write five different linear equations in two variables and find three solutions for each of them?

Solution:

(i) x + y + 1 = 0

For, x = 0

0 + y + 1 = 0

Or, y = – 1

For, x = 1

1 + y + 1 = 0

y = – 2

For y = 0

x + 0 + 1 = 0

x = – 1

Three different solutions are (0, -1) (1, -2) (-1, 0)

(ii) x + y – 1 = 0

For, x = 0

0 + y – 1 = 0

Or, y = 1

For, x = 1

1 + y – 1 = 0

Or, y = 0

For, y = 0

X + 0 – 1 = 0

Or, x = 1

Three different solutions are (0, 1) (1, 0)(1, 0)

(iii) 2x + 3y – 6 = 0

For, x = a

2 x 0 + 3y – 6 = 0

Or, y = 6/3 = 2

For y = 0

2 x X + 3 x 0 – 6 = 0

Or, 2x = 6

Or, x = 3

For x = 1

2 x 1 + 3y – 6 = 0

Or, 3y = 6 – 2

Or, y =

Three different solutions are (0, 2) (3, 0) (1,4/3)

(iv) 2x + 4y – 8 = 0

For x = 0

2 x 0 + 4y = 8

Or, y = 2

For y = 0

2 x + 4 x 0 = 8

Or, x = 4

For x = 1

2 x 1 + 4y = 8

Or, 4y = 6

Or, y = 6/4 = 3/2

Three different solutions are (0, 2) (4, 0) (1,3/2)

(v) 2x + 4y + 8 = 0

For x = 0

2 x 0 + 4y = -8

Or, y = -2

For, x = 1

2 x 1 + 4y = – 8

Or, y = (-10)/4 = (-5)/2

For, y = 0

2 x X + 0 = – 8

Or, x = -4

Three different solutions are (0, -2) (1,-5/2), (-4, 0)

Updated: September 21, 2021 — 4:37 pm

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