# Solve the equation: 5/(y - 4) = 3y/(y + 2) –

Solve the equation:
$$\displaystyle\frac{{5}}{{{y}-{4}}}={3}\frac{{y}}{{{y}+{2}}}–\frac{{{2}{y}^{{2}}–{14}{y}}}{{{y}^{{2}}–{2}{y}–{8}}}$$

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$$\displaystyle\frac{{5}}{{{y}-{4}}}={3}\frac{{y}}{{{y}+{2}}}–\frac{{{2}{y}^{{2}}–{14}{y}}}{{{y}^{{2}}–{2}{y}–{8}}}$$
$$\displaystyle\frac{{5}}{{{y}-{4}}}={3}\frac{{y}}{{{y}+{2}}}–\frac{{{2}{y}^{{2}}–{14}{y}}}{{{\left({y}–{4}\right)}{\left({y}–{2}\right)}}}$$
$$\displaystyle\frac{{{5}{\left({y}+{2}\right)}}}{{{\left({y}-{4}\right)}{\left({y}+{2}\right)}}}=\frac{{{3}{y}{\left({y}–{4}\right)}}}{{{\left({y}+{2}\right)}{\left({y}-{4}\right)}}}–\frac{{{2}{y}^{{2}}–{14}{y}}}{{{\left({y}–{4}\right)}{\left({y}–{2}\right)}}}$$
$$\displaystyle{5}{y}+{10}={3}{y}^{{2}}–{12}{y}–{2}{y}^{{2}}+{14}{y}$$
$$\displaystyle-{y}^{{2}}+{3}{y}+{10}={0}$$
$$\displaystyle{y}^{{2}}–{3}{y}–{10}={0}$$
$$\displaystyle{\left({y}–{5}\right)}{\left({y}+{2}\right)}={0}$$
$$\displaystyle{y}={5}{\quad\text{or}\quad}{y}=-{2}$$