NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Linear Equations in Two Variables |

Exercise |
Ex 8.2 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2

**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)** A linear equation in two variables has infinitely many solutions.

**Question 2.**

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

**(i) 2x + y = 7**

**(ii) πx + y = 9**

**(iii) x = 4y**

**Solution:**

**(i)** 2x + y = 7

By inspection, x = 2 and y = 3 is a solution because for x = 2, y = 3,

2x + y = 2 x 2 + 3 = 4 + 3 = 7

Now, let us choose x = 0 with this value of x, the given equation reduces to y = 7.

So, x = 0, y = 7 is also a solution of 2x + y = 7. Similarly, taking y = 0,

the given equation reduces to 2x = 7 which has the unique solution x = \(\frac { 7 }{ 2 }\) .

So, x = \(\frac { 7 }{ 2 }\) , y = 0 is a solution of 2x + y = 7.

Finally, let us take x = 1

The given equation now reduces to 2 + y = 7 hose solution is given by y = 5.

Therefore, (1, 5) is also a solution of the given equation.

So, four of the infinitely many solutions of the given equation are (2, 3), (0, 7), ( \(\frac { 7 }{ 2 }\) , 0) and (1,5).

**(ii)** πx + y = 9

Now, let us choose x = 0 with this value of x,

the given equation reduces to y = 9 which has a unique solution y = 9.

So, x = 0, y = 9 is also a solution of πx + y = 9

Similarly, taking y = 0, the given equation reduces to x = \(\frac { 9 }{ \pi }\) So, x = \(\frac { 9 }{ \pi }\) ,y = 0 is a solution of πx + y = 9as well.

Finally, let us take x = 7 the given equation now reduces to \(\frac { 22 }{ 7 }\) . 7 + y = 9

whose solution is given by y = -13.

Therefore; (7,-13) is also a solution of the given equation.

So, four of the infinitely many solutions of the given equation are

**(iii)** x = 4y ⇒ x – 4y = 0

By inspection, x = 0, y = 0 is a solution because for x = y = 0, 0 – 4 x 0 = 0 – 0 = 0, it satisfies.

Now, let us choose x = 4 with this value of x,

the given equation reduces to y = 1 which has a unique solution y = 1.

So, x = 4, y = 1 is also a solution , of x – 4y = 0. Similarly, taking y = \(\frac { 1 }{ 2 }\) , the given equation reduces to x = 2.

So , x = 2, y = \(\frac { 1 }{ 2 }\) is a solution x – 4y = 0 as well.

Finally, let us take x = 1, the given equation now reduces to 1 – 4y = 0

whose solution is given by y = \(\frac { 1 }{ 4 }\). Therefore, (1,1/4) is also a solution of the given equation. So, four

of the infinitely many solutions of the given equation are

**Question 3.**

**Check which of the following are solution 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)**

**Solution:**

**(i)** Take x – 2y and put x = 0, y = 2,

we get 0 – 2 x 2 = 0 – 4 = -4 ≠ 4

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

**
(ii)** Take x – 2y and put x = 2, y = 0,

we get 2 – 2 x 0 = 2 – 0 = 2 ≠ 4

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

**Take x – 2y and put x = 4, y = 0;**

(iii)

(iii)

we get 4 – 2 x 0 = 4 – 0 = 4

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

**Take x – 2y and put x = √2, y = 4√2, we get**

(iv)

(iv)

√2 – 2 x 4√2 = √2 – 8√2 =-7√2 ≠ 4

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

**Take x – 2y and put x = 1, y = 1,**

(v)

(v)

we get 1 – 2 x 1 = 1 – 2 = -1 ≠ 4

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

**Question 4.**

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

**Solution:**

Take 2x + 3y = k

Put x = 2, y = 1 then we get, 2 x 2 + 3 x 1 = k

⇒ 4 + 3 = k

⇒ k = 7

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