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

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

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Answer 1 · 2017-07-22T01:58:08

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

Explanation:

$\frac{6}{x + 3} = \frac{9 x}{x^{2} - 9}$ $6/(x+3)=(9x)/(x^2-9)$

Simplify by making the common denominator between the two fractions.

$\frac{6 (x - 3)}{x^{2} - 9} = \frac{9 x}{x^{2} - 9}$ $(6(x-3))/(x^2-9)=(9x)/(x^2-9)$

You can now get rid of the denominators.

$6 (x - 3) = 9 x$ $6(x-3)=9x$

Solve for x.

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

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

1 Answer

Explanation:

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How do you solve 6x+3=9xx2−9\frac { 6} { x + 3} = \frac { 9x } { x ^ { 2} - 9}?

1 Answer

Explanation:

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