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(B) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:$1$. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$,the

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$(1-b), (2-c), (3-a)$

Vedclass · Answer

(B) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:$1$. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$,the system has no solution (empty set).$2$. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the system has infinitely many solutions (infinite set).$3$. If $\frac{a_1}{a_2} 
eq \frac{b_1}{b_2}$,the system has a unique solution (singleton set).Analysis:$1. x+2y=3, 2x+4y=5$: Here $\frac{1}{2} = \frac{2}{4} 
eq \frac{3}{5}$. This corresponds to $b$ (empty set).$2. 2x+3y=6, x+\frac{3}{2}y=3$: Here $\frac{2}{1} = \frac{3}{1.5} = \frac{6}{3} = 2$. This corresponds to $c$ (infinite set).$3. 3x-y=0, x-y+6=0$: Here $\frac{3}{1} 
eq \frac{-1}{-1}$. This corresponds to $a$ (singleton set).Thus,the correct match is $(1-b), (2-c), (3-a)$.

Part $I$	Part $II$
$1. x+2y=3, 2x+4y=5$	$a. \text{Solution set is a singleton set.}$
$2. 2x+3y=6, x+\frac{3}{2}y=3$	$b. \text{Solution set is an empty set.}$
$3. 3x-y=0, x-y+6=0$	$c. \text{Solution set is an infinite set.}$
	$d. \text{Solution set contains two members.}$

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