Test 3 Section 4 #30

This one is yucky. Let’s attack it by substituting. First, we’re told that b=c-\frac{1}{2}, so let’s reflect that (I’m going to use the decimal version of \frac{1}{2} for simplicity):

    \begin{align*}3x+(c-0.5)&=5x-7\\3y+c&=5y-7\end{align*}

Now, let’s solve both equations for c:

    \begin{align*}3x+(c-0.5)&=5x-7\\c-0.5&=2x-7\\c&=2x-6.5\\\\3y+c&=5y-7\\c&=2y-7\end{align*}

Because we now have two different equations with c isolated, we can substitute again here to get an equation that only contains xs and ys.

    \begin{align*}2x-6.5&=2y-7\\2x&=2y-0.5\\x&=y-0.25\end{align*}

That translates to “x is y minus \frac{1}{4},” so the answer is A.