Question

If the system of linear equations $\{\begin{array}{c}3x-9y=-6\\ \frac{1}{2}x-\frac{3}{2}y=c\end{array}$ has infinitely many solutions, then the value of constant $c$ is

• A. $-6$
• B. $-3$
• C. $-2$
• D. $-1$

Answer 1

The correct option is D $-1$

Given linear equations are $3x-9y=-6,\frac{1}{2}x-\frac{3}{2}y=c$

If two linear equations have infinity many solutions, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

Hence $\frac{3}{\frac{1}{2}}=\frac{9}{\frac{3}{2}}=\frac{-6}{c}$

$\therefore \frac{-6}{c}=6$

$\therefore c=-1$