+0

# Help me

0
397
2

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

Oct 12, 2017

#2
+771
+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.

$$x=-\frac{1}{6}$$

.
Oct 12, 2017

#1
+50
+1

x=-1/6

Oct 12, 2017
#2
+771
+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.

$$x=-\frac{1}{6}$$