## Algebra 2 Common Core

$x= \frac{-6}{5}$
$|x-2| = 4x + 8$ There are two cases. Case 1: $x-2 = 4x + 8$ Case 2: $x-2= -(4x+8)$ Solving for $x$ in Case 1: Subtract $x$ from both sides of the equation. $-2 = 3x + 8$ Subtract $8$ from both sides of the equation. $-10 = 3x$ Divide both sides by $3$ to isolate $x$. $x = \frac{-10}{3}$ However, if we plug in $\frac{-10}{3}$ as $x$ into the equation, we get $|\frac{-10}{3} -2| = 4(\frac{-10}{3}) + 8$ This simplifies to $|\frac{-16}{3}| = \frac{-16}{3}$ , which is not possible, because absolute values cannot be negative. Let's solve for $x$ in Case 2: Subtract $x$ from both sides of the equation. $-2 = -5x - 8$ Add $8$ to both sides. $6 = -5x$ Divide both sides by $-5$. $x = \frac{-6}{5}$ If we plug in $\frac{-6}{5}$ for $x$, we get $| \frac{-6}{5} - 2| = 4(\frac{-6}{5}) + 8$ This simplifies to $|\frac{-16}{5}| = \frac{16}{5}$, which satisfies the equation. Thus $x$ only has one solution, $\frac{-6}{5}$