NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2.

- Linear Equations in Two Variables Class 9 Ex 4.1
- Linear Equations in Two Variables Class 9 Ex 4.2
- Linear Equations in Two Variables Class 9 Ex 4.3
- Linear Equations in Two Variables Class 9 Ex 4.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Linear Equations in Two Variables |

Exercise |
Ex 4.2 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

**Ex 4.2 Class 9 Maths 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 ,

**Solution:**

**(iii)** We know that, a linear equation in two variables has infinitely many solutions. Here, given equation is a linear equation in two variables x and y, so it has infinitely many solutions.

**Ex 4.2 Class 9 Maths Question 2.**

**Write four solutions for each of the following equations :**

(i) 2x + y = 7,

(ii) πx + y = 9

(iii) x = 4y Sol.

**Solution:**

**(i)** Given equation, 2x + y=7 …(i)

Now, putting x = 0, in eq. (i), we get 2(0) + y=7

⇒ 0 + y = 7 => y=7

So, x = 0 and y = 7 is a solution of given equation.

On putting x = 1 in eq (i), we get

2(1) + y =7 ⇒ y=7-2 ⇒ y=5

So, x = 1 and y = 5 is a solution of given equation

On putting x = 2, in eq. (i) we get, 2(2) + y = 7 ⇒ y= 7 – 4 ⇒ y= 3

So, x = 2 and y = 3 is a solution of given equation.

On putting x = 3 in eq(i), we get 2(3) + y =7

⇒ y= 7-6 ⇒ y = 1

So, x = 3 and y = 1 is a solution of given equation.

Hence, four out of the infinitely many solutions of the given equation are (0, 7), (1, 5) (2, 3) and (3,1).

**(ii)** Given equation πx + y =9 – – – (i)

On putting x = 0 in eq. (i) we get, π(0) + y=9

⇒ y = 9 – 0 =>y = 9

So, x = 0 and y = 9 is a solution of given equation.

On putting x = 1 in eq. (i), we get π(1) + y= 9

y-9-π

So, x = 1, and y = (9 -π) is a solution of given equation.

On putting x = 2 in eq (i), we get π(2) + y=9

y = 9 – 2π

So x = 2 and y = (9 – 2π) is a solution of given equation.

On putting x = -1 in eq (i), we get π(-9) + y=9

y = 9 + π

So x = -1 and y = (9 + π) is a solution of given equation.

Hence, four out of the infinitely many solutions of the given equation are

(0, 9), {1,(9 – π)}, {2,(9 – 2π)}, {-1, (9 + π)}

**(iii)** Given equation x = 4y …(i)

On putting x = 0, in eq. (i), we get 4y=0 ⇒ y=0

So, x – 0 and y = 0 is a solution of given equation.

On putting x = 1 in eq. (i), we get 4y=1 ⇒ y= \(\frac { 1 }{ 4 } \)

So, x -1 and y = \(\frac { 1 }{ 4 } \) is a solution of given equation.

On putting x – 4, in eq (i), we get

4y= 4

⇒ y= – \(\frac { 4 }{ 4 } \) =>y = 1

So, x = 4 and y = 1 is a solution of given equation, on putting x = – 4 in eq. (i), we get

4y= -4 ⇒ y= \(\frac { -4 }{ 4 } \) = -1

So, x = – 4 and y = -1 is a solution of given equation.

Hence, four out of the infinitely many solutions of given equation are

(0, 0), (1, \(\frac { 1 }{ 4 } \)), (4,1) and (-4, -1).

**Ex 4.2 Class 9 Maths 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) (\( \sqrt { 2 }\) ,4\( \sqrt { 2 }\) )

(v) (1, 1) .

**Solution:**

**(i)** Given equation is x – 2y = 4.

(0, 2) means x = 0 and y = 2

On putting x = 0 and y = 2 in L.H.S., get

L.H.S. = x – 2y = 0 – 2(2) = – 4

But R.H.S. = 4

L.H.S ≠ R.H.S

Hence (0, 2) is not a solution of x – 2y – 4.

**(ii)** Given equation is x – 2y = 4 (2, 0) means x – 2 and y = 0

On putting x = 2 and y = 0 in L.H.S., we get

L.H.S. = x – 2y = 2 – 2(0) = 2-0 = 2

But – R.H.S. = 4

∴ L.H.S = R.H.S.

Hence (2, 0) is not a solution of x – 2y – 4.

**(iii)** Given equation is x – 2y = 4 (4, 0) means x = 4 and y = 0.

On putting x = 4 and y = 0 in L.H.S., we get

L.H.S. = x – 2y = 4 – 2(0) =4-0 = 4

But R.H.S. = 4

∴ L.H.S. = R.H.S.

Hence, (4, 0) is a solution of x – 2y = 4.

**(iv)** Given equation is x – 2y = 4

(\( \sqrt { 2 }\), 4\( \sqrt { 2 }\)) means x = \( \sqrt { 2 }\) and y = 4\( \sqrt { 2 }\)

On putting x = \( \sqrt { 2 }\) and y = 4\( \sqrt { 2 }\) in L.H.S., we get

L.H.S. = x – 2y = \( \sqrt { 2 }\) – 2(4\( \sqrt { 2 }\)) = \( \sqrt { 2 }\)-8\( \sqrt { 2 }\)

= \( \sqrt { 2 }\)(1 – 8) = – 7\( \sqrt { 2 }\)

But R.H.S. = 4

∴ L.H.S ≠ R.H.S.

Hence (\( \sqrt { 2 }\), 4\( \sqrt { 2 }\)) is not a solution of x – 2y = 4

**(v)** Given equation is x – 2y = 4

(1, 1) means x = 1 and y = 1

On putting x = 1 and y = 1 in L.H.S. we get

L.H.S. = x – 2y = 1 – 2(1) = 1 – 2 = -1 But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

Hence, (1, 1) is not a solution of x – 2y = 4.

**Ex 4.2 Class 9 Maths Question 4.**

Find the value by k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

**Solution:**

If x = 2, y = 1 is a solution by the equation 2x + 3y = k, then these value will satisfy the equation.

∴ 2 x 2+3 x 1 = k ⇒ k=4+3=7

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