How to solve for $x$ in $2(x-5) + 4 (x-3) = -30$ In $2(x-5) + 4 (x-3) = -30$... I'm very confused as to how to solve for $x$, the correct response is $-2$, but I keep getting $4/3$.
 A: $$2(x-5)+4(x-3)=-30\implies 2x-10+4x-12=-30\\ \implies 6x=-30+10+12=(-8)\\ \implies x=\frac{-8}{6}=\frac{-4}{3}$$
The printed solution in whatever book you're using is wrong since the answer isn't $(-2)$.
You can verify that by value plugging.
A: Assuming you transcribed the problem correctly, your answer is correct.
\begin{align*}
2(x - 5) + 4(x - 3) & = -30\\
x - 5 + 2(x - 3) & = -15 && \text{divide both sides of the equation by $2$}\\
x - 5 + 2x - 6 & = -15 && \text{distribute}\\
3x - 11 & = -15 && \text{combine like terms}\\
3x & = -4 && \text{add $11$ to each side of the equation}\\
x & = -\frac{4}{3} && \text{divide each side of the equation by $3$}
\end{align*} 
Check:  We can verify our answer by direct substitution.  If $x = -\dfrac{4}{3}$, then
\begin{align*}
2(x - 5) + 4(x - 3) & = 2\left(-\frac{4}{3} - 5\right) + 4\left(-\frac{4}{3} - 3\right)\\
& = 2\left(-\frac{4}{3} - \frac{15}{3}\right) + 4\left(-\frac{4}{3} - \frac{9}{3}\right)\\
& = 2\left(-\frac{19}{3}\right) + 4\left(-\frac{13}{3}\right)\\
& = -\frac{38}{3} - \frac{52}{3}\\
& = -\frac{90}{3}\\
& = -30
\end{align*}
If you substitute $-2$ for $x$ in the expression $2(x - 5) + 4(x - 3)$, you should obtain $-34$.
