How do you solve u(u - 5) + 8u = u(u + 2) - 4 ?

May 27, 2017

See a solution process below:

Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{u} \left(u - 5\right) + 8 u = \textcolor{b l u e}{u} \left(u + 2\right) - 4$

$\left(\textcolor{red}{u} \times u\right) - \left(\textcolor{red}{u} \times 5\right) + 8 u = \left(\textcolor{b l u e}{u} \times u\right) + \left(\textcolor{b l u e}{u} \times 2\right) - 4$

${u}^{2} - 5 u + 8 u = {u}^{2} + 2 u - 4$

${u}^{2} + \left(- 5 + 8\right) u = {u}^{2} + 2 u - 4$

${u}^{2} + 3 u = {u}^{2} + 2 u - 4$

Next, subtract $\textcolor{red}{{u}^{2}}$ from each side of the equation to eliminate this term while keeping the equation balanced:

$- \textcolor{red}{{u}^{2}} + {u}^{2} + 3 u = - \textcolor{red}{{u}^{2}} + {u}^{2} + 2 u - 4$

$0 + 3 u = 0 + 2 u - 4$

$3 u = 2 u - 4$

Now, subtract $\textcolor{red}{2 u}$ from each side of the equation to solve for $u$ while keeping the equation balanced:

$- \textcolor{red}{2 u} + 3 u = - \textcolor{red}{2 u} + 2 u - 4$

$\left(- \textcolor{red}{2} + 3\right) u = 0 - 4$

$1 u = - 4$

$u = - 4$