Solve and Perform the indicated operation.

Question

1. (6a^2/b) / (3a^4/5b^2)
2. [(x^2 - 2x - 35) / (x^2 - 4x - 21)] / [(x^2 + 9x + 20) / (x^2 - x - 12)]

Answer 1

Hi there! For the first question, you're dividing a fraction by a fraction. Remember, the rule for dividing fractions is you take the first fraction and multiply it by the reciprocal of the second fraction. In other words, flip the second fraction. If you do that, you get

(6a^2/b) x (5b^2/3a^4)

Now, multiply your numerators and your denominators separately:

30a^2b^2

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3a^4b

30 divided by 3 is 10:

10a^2b^2

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a^4b

If you remember your exponent rules (subtract exponents in a division problem!), you'll know you can now cancel some of those a's and b's. For a, 2-4 is -2, which just means that a^2 stays in the denominator. For b, 2-1 is 1, so b stays in the numerator.

So, 10b/a^2

For the second question, you need to factor your quadratics AND use the flip-the-second-fraction rule. So:

[(x^2 - 2x - 35) / (x^2 - 4x - 21)] X [(x^2 - x - 12)/(x^2 + 9x + 20)]

If you factor your quadratics:

(x-7)(x+5) X (x-4)(x+3)

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(x-7)(x+3) (x+5)(x+4)

You can see that (x-7) cancels in the first fraction, as do (x+5) and (x+3), which leaves:

x-4

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x+4