# Question #f621d

Jan 20, 2018

If $p + q = 4$, then $y$ can take any value. Otherwise, $y = 0$.

#### Explanation:

It's a bit weird for the problem to ask to "solve for $y$" because if we tried to do so, we'd get "$y = 0$, provided $p + q \ne 4$" without a way to express $y$ in terms of $p , q$...anyway,

we have $p y + q y = - 4 y + 8 y$. Let's factor out $y$ from both sides to make things simpler:

$\left(p + q\right) y = 4 y$. Then we can subtract $4 y$ from both sides, and factor out $y$ again from the left:

$\left(p + q - 4\right) y = 0$.

This is true either when $p + q = 4$, in which case $y$ can be anything, or when $y = 0$.