## College Algebra (6th Edition)

a. Restriction values $: -4$ b. Solution set $:\{-3\}$
$\frac{3}{x+4} - 7 = \frac{-4}{x+4}$ a. $x =-4$ makes the denominator zero. So, $x \ne -4$ b. $\frac{3}{x+4} - 7 = \frac{-4}{x+4} ; x \ne -4;$ $\frac{3-7(x+4)}{x+4} = \frac{-4}{x+4} ; x \ne -4;$ $\frac{3-7x-28}{x+4} = \frac{-4}{x+4} ; x \ne -4;$ $\frac{-7x-25}{x+4} = \frac{-4}{x+4} ; x \ne -4;$ Multiply both sides by $(x+4)$ to clear fraction. $(x+4)(\frac{-7x-25}{x+4} )=(x+4)( \frac{-4}{x+4}) ; x \ne -4;$ $-7x-25 = -4$ $-7x = -4+25$ $-7x =21$ Divide both sides by $-7$ $x=-3$