# How do you solve |5x + 2| < 7?

$5 x + 2 < 7 \mathmr{and} 5 x + 2 > - 7 , 5 x < 5 \to x < 1 \mathmr{and} 5 x > - 9 \to x > - \frac{9}{5}$
$\left(- \frac{9}{5}\right) < x < 1$
Use the definition of absolute value of x to solve. That is, $| x | = x \mathmr{if} x \ge 0 , \mathmr{and} - x \mathmr{if} x < 0$. Then solve for x