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Exercise - 4.2

Linear Equations in Two Variables

**Question-1 :-** Which one of the following options is true, and why? y = 3x + 5 has

(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. As for every value of x, there will be a value of y satisfying the above equation and vice-versa. Hence, the correct answer is (iii).

**Question-2 :-** Write four solutions for each of the following equations:

(i) 2x + y = 7

(ii) πx + y = 9

(iii) x = 4y

(i) 2x + y = 7 For x = 0, 2(0) + y = 7 y = 7 Therefore, (0, 7) is a solution of this equation. For x = 1, 2(1) + y = 7 y = 5 Therefore, (1, 5) is a solution of this equation. For x = −1, 2(−1) + y = 7 y = 9 Therefore, (−1, 9) is a solution of this equation. For x = 2, 2(2) + y = 7 y = 3 Therefore, (2, 3) is a solution of this equation.

(ii) πx + y = 9 For x = 0, π(0) 0 + y = 9 y = 9 Therefore, (0, 9) is a solution of this equation. For x = 1, π(1) + y = 9 y = 9 − π Therefore, (1, 9 − π) is a solution of this equation. For x = 2, π(2) + y = 9 y = 9 − 2π Therefore, (2, 9 −2π) is a solution of this equation. For x = −1, π(−1) + y = 9 y = 9 + π (−1, 9 + π) is a solution of this equation.

(iii) x = 4y For x = 0, 0 = 4y y = 0 Therefore, (0, 0) is a solution of this equation. For y = 1, x = 4(1) = 4 Therefore, (4, 1) is a solution of this equation. For y = −1, x = 4(−1) x = −4 Therefore, (−4, −1) is a solution of this equation. For x = 2, 2 = 4y y = 2/4 y = 1/2 Therefore, (2, 1/2) is a solution of this equation.

**Question-3 :-** Check which of the following are solutions of the equation x – 2y = 4 and which are not:

(i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) (√2, 4√2) (v) (1, 1)

(i) (0, 2) Putting x = 0 and y = 2 in the L.H.S of the given equation, x − 2y = 0 − 2 x 2 = − 4 ≠ 4 L.H.S ≠ R.H.S Therefore, (0, 2) is not a solution of this equation.

(ii) (2, 0) Putting x = 2 and y = 0 in the L.H.S of the given equation, x − 2y = 2 − 2 × 0 = 2 ≠ 4 L.H.S ≠ R.H.S Therefore, (2, 0) is not a solution of this equation.

(iii) (4, 0) Putting x = 4 and y = 0 in the L.H.S of the given equation, x − 2y = 4 − 2(0) = 4 = R.H.S L.H.S. = R.H.S. Therefore, (4, 0) is a solution of this equation.

(iv) (√2, 4√2) Putting x = √2 and y = 4√2 in the L.H.S. of the given equation, x - 2y = √2 - 2(4√2) = √2 - 8√2 = -7√2 ≠ 4 L.H.S. ≠ R.H.S. Therefore, (√2, 4√2) is not a solution of given equation.

(v) (1, 1) is not a solution of this equation. Putting x = 1 and y = 1 in the L.H.S of the given equation, x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4 L.H.S ≠ R.H.S Therefore, (1, 1) is not a solution of this equation.

**Question-4 :-** Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Putting x = 2 and y = 1 in the given equation, 2x + 3y = k 2(2) + 3(1) = k 4 + 3 = k k = 7 Therefore, the value of k is 7.

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