What is (a-7)/(a^2-3a-28)*(8a-40)/(a^2+3a-40) ?

The solution to (a-7)/(a^2-3a-28)*(8a-40)/(a^2+3a-40) is 8/((a+4)(a+8))

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simplify (a-7)/(a^2-3a-28)*(8a-40)/(a^2+3a-40)

simplify

\frac{a - 7}{a ^{2} - 3 a - 28} \cdot \frac{8 a - 40}{a ^{2} + 3 a - 40}

\frac{8}{( a + 4 ) ( a + 8 )}

Solution steps

\frac{a - 7}{a ^{2} - 3 a - 28} \cdot \frac{8 a - 40}{a ^{2} + 3 a - 40}

Cancel

\frac{a - 7}{a ^{2} - 3 a - 28} : \frac{1}{a + 4}

= \frac{1}{a + 4} \cdot \frac{8 a - 40}{a ^{2} + 3 a - 40}

Cancel

\frac{8 a - 40}{a ^{2} + 3 a - 40} : \frac{8}{a + 8}

= \frac{1}{a + 4} \cdot \frac{8}{a + 8}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 8}{( a + 4 ) ( a + 8 )}

Multiply the numbers:

1 \cdot 8 = 8

= \frac{8}{( a + 4 ) ( a + 8 )}

x^{2} - 7 x y - 8 y^{2}

4 x^{2} + 2 x

6 \div 3 + 9^{2} - 4

27 \cdot 9

n - 4 \geq 13

What is (a-7)/(a^2-3a-28)*(8a-40)/(a^2+3a-40) ?
The solution to (a-7)/(a^2-3a-28)*(8a-40)/(a^2+3a-40) is 8/((a+4)(a+8))