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

2 Answers
Shell
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}#

Tiago Hands
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).

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