Chapter 4 Linear Equations in Two Variables Ex 4.2 Solutions
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
Answer - 1 : -
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
Answer - 2 : -
(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) + 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
⇒
Therefore, 
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) 
(v) (1, 1)
Answer - 3 : -
(i) (0, 2)
Putting x = 0 and y = 2 in the L.H.Sof the given equation,
x − 2y = 0 −2×2 = − 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.Sof 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.Sof the given equation,
x − 2y = 4 −2(0)
= 4 = R.H.S
Therefore, (4, 0) is a solution of this equation.
(iv)Putting 
and 
in the L.H.S of the given equation, L.H.S ≠ R.H.S
Therefore, is not a solution of this equation.
(v) (1, 1)
Putting x = 1 and y = 1 in the L.H.Sof 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.
Answer - 4 : -
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.