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Solve for $x$: $$ \frac{3x+12}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}$$

Guest Oct 12, 2017

Best Answer 

 #2
avatar+648 
+2

\(\frac{3x+12}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}\)

First, you can take a 3 out of the numerator on the left.

\(\frac{3(x+4)}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}\)

Combine like factors on the top right.

\(\frac{3(x+4)}{x+4} = \frac{12x^2+8x}{2x}\)

The x+4's on the right can cancel.

\(3=\frac{12x^2+8x}{2x}\)

The left can be divided by 2x.

\(3=6x+4\)

Subtract 4 from both sides.

\(-1=6x\)

Divide both sides by 6.

\(x=-\frac{1}{6}\)

AdamTaurus  Oct 12, 2017
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2+0 Answers

 #1
avatar+50 
+1

x=-1/6

Atroshus  Oct 12, 2017
 #2
avatar+648 
+2
Best Answer

\(\frac{3x+12}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}\)

First, you can take a 3 out of the numerator on the left.

\(\frac{3(x+4)}{x+4} = \frac{4x^2 + 8x + 8x^2}{2x}\)

Combine like factors on the top right.

\(\frac{3(x+4)}{x+4} = \frac{12x^2+8x}{2x}\)

The x+4's on the right can cancel.

\(3=\frac{12x^2+8x}{2x}\)

The left can be divided by 2x.

\(3=6x+4\)

Subtract 4 from both sides.

\(-1=6x\)

Divide both sides by 6.

\(x=-\frac{1}{6}\)

AdamTaurus  Oct 12, 2017

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