Which one of the following pairs of linear equations has/ have infinite many solutions?

(A) $9x + 3y+12= 0$, and $18x + 6y+ 24 = 0$
(B) $2x-3y= 13$, and $7x - 2y = 20$
(C) $x + y = 5, 2x + 2y = 10$
(D) $3x + y- 3 = 0$ and $2x +\frac{2}{3}y= 2$

Choose the correct answer from the options given below:

(C), (D) and (A) only

The correct answer is Option (3) → (C), (D) and (A) only

To check for infinitely many solutions between pairs of linear equations:

Two linear equations:

$a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$

have infinitely many solutions if and only if:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

(A) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$

Ratios: $\frac{9}{18} = \frac{3}{6} = \frac{12}{24} = \frac{1}{2}$

All ratios equal ⇒ Infinitely many solutions

(B) $2x - 3y = 13$ and $7x - 2y = 20$

Rewrite both in general form: $2x - 3y - 13 = 0$, $7x - 2y - 20 = 0$

Ratios: $\frac{2}{7} \ne \frac{-3}{-2}$ ⇒ Not equal ⇒ No infinite solutions

(C) $x + y = 5$ and $2x + 2y = 10$

Rewrite second as $2(x + y) = 10$ ⇒ equivalent

Ratios: $\frac{1}{2} = \frac{1}{2} = \frac{5}{10}$ ⇒ Infinitely many solutions

(D) $3x + y - 3 = 0$, $2x + \frac{3}{2}y - 2 = 0$

Ratios: $\frac{3}{2} = 1.5$, $\frac{1}{\frac{2}{3}} = \frac{3}{2} $

All ratios equal ⇒ Infinitely many solutions