alka8q7

2021-11-17

For Exercise, solve the equation.
$\frac{48}{{m}^{2}-4m}+3=\frac{12}{m-4}$

Annie Midgett

Step 1
We factor the denominators
$\frac{48}{{m}^{2}-4m}+3=\frac{12}{m-4}$
$\frac{48}{m\left(m-4\right)}+3=\frac{12}{m-4}$
LCD is $m\left(m-4\right)$
Then we multiply both sides by LCD so that we get rid of the fractions
$\frac{48}{m\left(m-4\right)}+3=\frac{12}{m-4}$
$m\left(m-4\right)\left[\frac{48}{m\left(m-4\right)}+3\right]=m\left(m-4\right)\frac{12}{m-4}$
$48+3m\left(m-4\right)=12m$
Step 2
Then distribute 3m, combine the like terms and then make right side equal to 0.
Then factor
$48+3m\left(m-4\right)=12m$
$48+3{m}^{2}-12m=12m$
$3{m}^{2}-24m+48=0$
$3\left({m}^{2}-8m+16\right)=0$
$3\left(m-4\right)\left(m-4\right)=0$
$3{\left(m-4\right)}^{2}=0$
Set each factor equal to 0 and solve for m
$m-4=0$
$m=4$
For $m=4$ we get denominator equal to 0 in the original equation. So $m=4$ can not be a solution