$$2x-5\{7-(x-6)+3x\}-28=39$$
$x = -\frac{33}{2} = -16\frac{1}{2} = -16.5$
$$2x-5\left(7-x-\left(-6\right)+3x\right)-28=39$$
$$2x-5\left(7-x+6+3x\right)-28=39$$
$$2x-5\left(13-x+3x\right)-28=39$$
$$2x-5\left(13+2x\right)-28=39$$
$$2x-65-10x-28=39$$
$$-8x-65-28=39$$
$$-8x-93=39$$
$$-8x=39+93$$
$$-8x=132$$
$$x=\frac{132}{-8}$$
$$x=-\frac{33}{2}$$
Show Solution
Hide Solution
