Question

Find two solutions for each of the following equations:
$\begin{array}{c}4x+3y=12\\ 2x+5y=0\\ 3y+4=0\end{array}$

Answer 1

1) $4x+3y=12$

for $y=4$

$4x+12=12$ $x=0$

for $y=0$

$4x+0=12$ $x=3$

$(0,4)$ & $(3,0)$ are $2$ solution

2) $2x+5y=0$

for $y=-2$

$2x-10=0$ $x=5$

for $y=-4$

$2x-20=0$ $x=10$

$(5,-2)$ and $(10,-4)$ are $2$ solutions

3) $3y+4=0$

$y=\frac{-4}{3}$ is only solution