alka8q7

2021-11-17

For Exercise, solve the equation.

$\frac{48}{{m}^{2}-4m}+3=\frac{12}{m-4}$

Annie Midgett

Beginner2021-11-18Added 7 answers

Step 1

We factor the denominators

$\frac{48}{{m}^{2}-4m}+3=\frac{12}{m-4}$

$\frac{48}{m(m-4)}+3=\frac{12}{m-4}$

LCD is$m(m-4)$

Then we multiply both sides by LCD so that we get rid of the fractions

$\frac{48}{m(m-4)}+3=\frac{12}{m-4}$

$m(m-4)[\frac{48}{m(m-4)}+3]=m(m-4)\frac{12}{m-4}$

$48+3m(m-4)=12m$

Step 2

Then distribute 3m, combine the like terms and then make right side equal to 0.

Then factor

$48+3m(m-4)=12m$

$48+3{m}^{2}-12m=12m$

$3{m}^{2}-24m+48=0$

$3({m}^{2}-8m+16)=0$

$3(m-4)(m-4)=0$

$3{(m-4)}^{2}=0$

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

Answer: No solution

We factor the denominators

LCD is

Then we multiply both sides by LCD so that we get rid of the fractions

Step 2

Then distribute 3m, combine the like terms and then make right side equal to 0.

Then factor

Set each factor equal to 0 and solve for m

For

Answer: No solution