How do you subtract 3/(7b)-8/(3b^2)?

2 Answers
Oct 28, 2016

frac{9b-56}{21b^2}

Explanation:

3/(7b)-8/(3b^2)

First find the common denominator.

The least common multiple of 3 and 7 is 21.

The least common multiple of b and b^2 is b^2.

So, the common denominator is 21b^2.

Multiply the first term by (3b)/(3b) and the second term by 7/7.

(3b)/(3b) * 3/(7b) - 7/7 * 8/(3b^2)

(9b)/(21b^2) - 56/(21b^2)

frac{9b-56}{21b^2}

Oct 28, 2016

3/(7b)-8/(3b^2)

=3/(7b)*1-8/(3b^2)*1

=3/(7b)*(3b)/(3b)-8/(3b^2)*7/7

=(9b)/(21b^2)-56/(21b^2)

=(9b-56)/(21b^2)

The fraction above cannot be simplified, so this is the fraction you'd be looking to get if you were to subtract the second fraction (right) from the first fraction (left).