Exercise 6.2

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

(i) 3x + 4y = 7

Solution:

To find solutions, we can substitute arbitrary values for one variable and solve for the other.

Solution 1: Let x = 0. Then, 4y = => y = 7422.

So, ( , 7447 ) is a solution.

Solution 2: Let y = 0. Then, 3x = 7 => x = 7337.

So, ( 73`3/7, ) is a solution.

Solution 3: Let x = 1. Then, + 4y = 7 => 4y = => y = .

So, ( , ) is a solution.

(ii) y = 6x

Solution:

Solution 1: Let x = 0. Then, y = . So, ( , ) is a solution.

Solution 2: Let x = 1. Then, y = . So, ( , ) is a solution.

Solution 3: Let x = -1. Then, y = . So, ( , ) is a solution.

(iii) 2x − y = 7

Solution:

Solution 1: Let x = 0. Then, -y = => y = . So, ( , ) is a solution.

Solution 2: Let y = 0. Then, 2x = 7 => x = 7227. So, ( 7227, ) is a solution.

Solution 3: Let x = 1. Then, - y = 7 => y = . So, ( , ) is a solution.

(iv) 13x − 12y = 25

Solution:

Solution 1: Let x = 0. Then, -12y = 25 => y = −25122512. So, (0, -25/12) is a solution.

Solution 2: Let y = 0. Then, 13x = 25 => x = 251325. So, ( 2513−2513, ) is a solution.

Solution 3: Let x = 1. Then, - 12y = 25 => -12y = => y = . So, ( , ) is a solution.

(v) 10x + 11y = 21

Solution:

Solution 1: Let x = 0. Then, 11y = 21 => y = 21111121. So, ( , 21111121 ) is a solution.

Solution 2: Let y = 0. Then, 10x = 21 => x = 21101021. So, ( 2110−2110, ) is a solution.

Solution 3: Let x = 1. Then, + 11y = 21 => 11y = => y = . So, ( , ) is a solution.

(vi) x + y = 0

Solution:

Solution 1: Let x = 0. Then, y = . So, ( , ) is a solution.

Solution 2: Let x = 1. Then, y = . So, (, ) is a solution.

Solution 3: Let x = -1. Then, y = . So, (, ) is a solution.

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

(i) 8x − y = 34

Solution:

For (0, a): 8() - = 34 => a = -3434

For (b, 0): 8b - 0 = 34 => b = 34834 = 174172

Therefore, ( , -3434 ) and ( 174−174, 0) are the solutions.

(ii) 3x = 7y − 21

Solution:

For (0, a): 3() = 7a7y - 21 => 7a = => a =

For (b, 0): = 7() - => 3b = -2121 => b =

Therefore, (0, ) and (, 0) are the solutions.

(iii) 5x − 2y + 3 = 0

Solution:

For (0, a): 5() - + 3 = 0 => 2a = => a = 3223

For (b, 0): - 2() + 3 = 0 => 5b = -33 => b = −3535

Therefore, (0, 3223) and (−355/3, 0) are the solutions.

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

(i) (0, 2)

Solution:

Given x − 5y = 10

=> 2() - 5() = ≠= 10.

(0, 2) a solution of the equation. NoYes

(ii) (0, –2)

Solution:

=> 2() - 5() =≠ 10.

(0, –2) solution of the equation. YesNo

(iii) (5, 0)

Solution:

=> 2() - 5() =≠ 10.

(5, 0) solution of the equation. YesNo

(iv) (2√3, -√3)

Solution:

=> 2(2√3√3) - 5(-√3√3) = 4√3√3 + 5√34√3 = √3 ≠= 10.

(2√3, -√3) solution of the equation. NoYes

(v) (1/2, 2)

Solution:

=> 2(122) - 5(212) = 1 - = -99 ≠= 10.

(1/2, 2) solution of the equation. NoYes

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:

Substituting x = 2 and y = 1 into the equation, we get:

=> 2() + 3() = k

=> + = k

=> k =

The resultant equation is 2x + 3y = 7

Solution 1: Let x = 0. Then, 3y = 7 => y = 7337. So, (, 7337) is a solution.

Solution 2: Let y = 0. Then, 2x = 7 => x = 7227. So, (7/2, 0) is a solution.

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:

Substituting x = 2 - α and y = 2 + α into the equation, we get:

=> 3(2-α2+α) - 2(2+α2-α) + 6 = 0

- 3α-3α - - 2α + 6 = 0

86 - 5αα = 0

5α = 8

α = 8558

The resultant equation remains 3x - 2y + 6 = 0.

Solution 1: Let x = 0. Then, -2y + 6 = 0

=> 2y = 6 => y = .

So, (, ) is a solution.

Solution 2: Let y = 0.

Then, 3x + 6 = 0 => x = .

So, (, ) is a solution.

Solution 3: Let x = 1.

Then, 3 - 2y + 6 = 0 => 2y = => y = 9229.

So, (, 9229) is a solution.

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

Solution:

Substituting x = 1 and y = 1 into the equation, we get:

=> 3() + a() = 6

3 + a = 6

a =

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

Solution:

(i) Equation: 4x - 5y = 10

Solution 1: Let x = 0. Then, -5y = 10 => y = .

So, (, ) is a solution

Solution 2: Let y = 0. Then, 4x = 10 => x = 5225.

So, (5225, ) is a solution.

Solution 3: Let x = 5. Then, - 5y = 10 => -5y = -1010 => y = .

So, (, ) is a solution.

(ii) Equation: x + 2y = 6

Solution 1: Let x = 0. Then, 2y = 6 => y = .

So, (, ) is a solution.

Solution 2: Let y = 0. Then, x = .

So, (, ) is a solution.

Solution 3: Let x = 2. Then, + 2y = 6 => 2y = => y = .

So, (, ) is a solution.

(iii) Equation: -x + y = 1

Solution 1: Let x = 0. Then, y = .

So, (, ) is a solution.

Solution 2: Let y = 0. Then, -x = => x = .

So, (, ) is a solution.

Solution 3: Let x = 1. Then, + y = 1 => y = .

So, (, ) is a solution.

(iv) Equation: 2x + 3y = 9

Solution 1: Let x = 0. Then, 3y = => y = .

So, (, ) is a solution.

Solution 2: Let y = 0. Then, 2x = 9 => x = 9229.

So, (9229, ) is a solution.

Solution 3: Let x = 3. Then, + 3y = 9 => 3y = => y = .

So, (, ) is a solution.

(v) Equation: x - y = 4

Solution 1: Let x = 0. Then, -y = => y = .

So, (, ) is a solution.

Solution 2: Let y = 0. Then, x = .

So, (, ) is a solution.

Solution 3: Let x = 2. Then, - y = 4 => -y = => y = .

So, (, ) is a solution.