How do you solve $\frac { 6} { x + 3} = \frac { 9x } { x ^ { 2} - 9}$ ?

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Answer 1 · 2017-07-22T01:58:08

Use the formula of multiplying polynomials:
$(x+y)(x-y)=x^2-y^2$
and use Least Common Denominator (LCD) to isolate and solve for x.

$6/(x+3)=(9x)/(x^2-9)$

Simplify by making the common denominator between the two fractions.

$(6(x-3))/(x^2-9)=(9x)/(x^2-9)$

You can now get rid of the denominators.

$6(x-3)=9x$

Solve for x.

$-3x=18$
$x=-6$

How do you solve \frac { 6} { x + 3} = \frac { 9x } { x ^ { 2} - 9}\frac { 6} { x + 3} = \frac { 9x } { x ^ { 2} - 9}?