**NCERT Solutions Class 9 Mathematics**

**Exercise 4.2**

**Q.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

**Q.2.** Write four solutions for each of the following equations:

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

**Q.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) |

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

**NCERT Solutions Class 9 Mathematics**

**Exercise 4.2**

**Q.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:**

** **Any linear equation will have (iii) infinitely many solutions. For each value of x, we will get a value of y. Since there is no limit to taking the values of x, we will have infinite number of solutions.

**Q.2.** Write four solutions for each of the following equations:

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

**Solution:**

(i) Given, 2x + y = 7

⇒ y = 7 - 2x

Put x = 0 ⇒ y = 7 - 2 × 0 ⇒ y = 7

⇒ (0, 7) is the solution.

We can similarly tabulate the four solutions as follows,

x |
0 |
1 |
2 |
3 |

y |
7 |
5 |
3 |
1 |

Hence, the four solutions of the equation 2x + y = 7 could be (0, 7), (1, 5), (2, 3) and (3, 1).

(ii) πx + y = 9 ⇒ y = 9 - πx

x |
0 |
1 |
2 |
3 |

y |
9 |
9 - π |
9 - 2π |
9 - 3π |

The four solutions of the equation πx + y = 9 could be (0, 9), (1, 9-π), (2, 9-2π) and (3, 9-3π).

(iii) x = 4y

x |
0 |
4 |
8 |
12 |

y |
0 |
1 |
2 |
3 |

The four solutions of the equation πx + y = 9 may be (0, 0), (4, 1), (8, 2) and (12, 3).

**Q.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) Putting x = 0 and y = 2 in the equation x - 2y = 4, we get,

0 - 2×2 = 4 ⇒ -4 = 4

Since LHS and RHS are not equal, (0, 2) is not a solution of the given equation.

(ii) Putting x = 2 and y = 0 in the equation x - 2y = 4, we get,

2 - 2×0 = 4 ⇒ 2 = 4

Since LHS and RHS are not equal, (2, 0) is not a solution of the given equation.

(iii) Putting x = 4 and y = 0 in the equation x - 2y = 4, we get,

4 - 2×0 = 4 ⇒ 4 = 4

Since LHS and RHS are equal, (4, 0) is a solution of the given equation.

(iv) Putting x = $\sqrt{2}$ and y = 4$\sqrt{2}$ in the equation x - 2y = 4, we get,

$\sqrt{2}$ - 2×4$\sqrt{2}$ = 4

⇒ $\sqrt{2}$ - 8$\sqrt{2}$ = 4

⇒ $\sqrt{2}$(1 - 8) = 4

⇒ -7$\sqrt{2}$ = 4

Since LHS and RHS are not equal, $\sqrt{2}$, 4$\sqrt{2}$) is not a solution of the given equation.

(v) Putting x = 1 and y = 1 in the equation x - 2y = 4, we get,

1 - 2×1 = 4 ⇒ -1 = 4

Since LHS and RHS are not equal, (1, 1) is not a solution of the given equation.

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

**Solution:**

Given equation is 2x + 3y = k

Now, x = 2, y = 1 is a solution of the given equation.

Putting the value of x and y in the equation, we get

2 × 2 + 3 × 1 = k

⇒ k = 4 + 3

⇒ k = 7