Question 570645
Solve:
{{{(2x)/(x+3)+5/x -4 = 18/(x^2+3x)}}} Simplify. The LCD is{{{x^2+3x}}}
{{{((2x^2+5x+15-4x^2-highlight(12x)))/(x^2+3x) = 18/(x^2+3x)}}} so we have:
{{{-2x^2-7x+15 = 18}}} Subtract 18 from both sides.
{{{-2x^2-7x-3 = 0}}} Multiply through by -1 to get:
{{{2x^2+7x+3 = 0}}} Factor.
{{{(2x+1)(x+3) = 0}}} Apply the zero product rule:
{{{2x+1 = 0}}} or {{{x+3 = 0}}} so...
{{{x = -1/2}}} or {{{x = -3}}}
Notice:  The left side becomes...
{{{(2x(x)+5(x+3)-4(x(x+3)))/x(x+3)}}} which, when simplified becomes:
{{{(2x^2+5x+15-4x^2-12x)/x(x+3)}}}={{{(-2x^2-7x+15)/(x^2+3x)}}}