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# Can someone help??!!

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I keep getting this wrong!

Jul 8, 2017

#1
+7347
+2

$$\frac{m-4}{m^2+8m+16}-\frac{m-4}{m+4}$$

We need to get a common denominator. To do that...multiply the second fraction by $$\frac{m+4}{m+4}$$  .

$$=\frac{m-4}{m^2+8m+16}-\frac{(m-4)(m+4)}{(m+4)(m+4)}$$

Multiply the numerator and denominator out.

$$=\frac{m-4}{m^2+8m+16}-\frac{m^2-16}{m^2+8m+16}$$

Now that we have a common denominator, we can combine the fractions.

$$=\frac{m-4-(m^2-16)}{m^2+8m+16} \\~\\ =\frac{m-4-m^2+16}{m^2+8m+16} \\~\\ =\frac{\mathbf{-1}m^2+m+\mathbf{12}}{m^2+\mathbf{8}m+\mathbf{16}}$$

.
Jul 8, 2017

#1
+7347
+2

$$\frac{m-4}{m^2+8m+16}-\frac{m-4}{m+4}$$

We need to get a common denominator. To do that...multiply the second fraction by $$\frac{m+4}{m+4}$$  .

$$=\frac{m-4}{m^2+8m+16}-\frac{(m-4)(m+4)}{(m+4)(m+4)}$$

Multiply the numerator and denominator out.

$$=\frac{m-4}{m^2+8m+16}-\frac{m^2-16}{m^2+8m+16}$$

Now that we have a common denominator, we can combine the fractions.

$$=\frac{m-4-(m^2-16)}{m^2+8m+16} \\~\\ =\frac{m-4-m^2+16}{m^2+8m+16} \\~\\ =\frac{\mathbf{-1}m^2+m+\mathbf{12}}{m^2+\mathbf{8}m+\mathbf{16}}$$

hectictar Jul 8, 2017
#2
+1

Thank you

Guest Jul 8, 2017