Chapter 4 Linear Equations in Two Variables Ex-4.1 |
Chapter 4 Linear Equations in Two Variables Ex-4.3 |
Chapter 4 Linear Equations in Two Variables Ex-4.4 |

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).

Write four solutions for each of the following equations:

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

**Answer
2** :

(i) 2*x* + *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) π*x* + *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* = 4*y*

For *x* = 0,

0 = 4*y*

⇒ *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 = 4*y*

Therefore, is a solution of this equation.

Check which of the following are solutions of the equation *x* −2*y *= 4 and which are not:

(i) (0, 2 (ii) (2, 0) (iii) (4, 0)

**Answer
3** :

(i) (0, 2)

Putting *x* = 0 and *y* = 2 in the L.H.Sof the given equation,

*x *− 2*y* = 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 *− 2*y* = 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 *− 2*y *= 4 −2(0)

= 4 = R.H.S

Therefore, (4, 0) is a solution of this equation.

Putting and in the L.H.S of the given equation,

L.H.S *≠ *R.H.S

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 *− 2*y* = 1 −2(1) = 1 − 2 = − 1 ≠ 4

L.H.S *≠ *R.H.S

Therefore, (1, 1) is not a solution of this equation.

**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.

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