How do you write #\frac { 5b + 9y } { 6b } - \frac { 5b + 6y } { 2b } + 1# as a single fraction?

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Answer 1 · 2018-03-24T00:45:15

You may want to use #6b# as a common denominator.

Multiply the second and third term by a "disguised" #1#:

2nd term: #(5b+6y)/(2b)xx3/3=(15b+18y)/(6b)#

3rd term: #1xx(6b)/(6b)=(6b)/(6b)#

Now put it all together:

#=(5b+9y)/(6b)-(15b+18y)/(6b)+(6b)/(6b)#

#=(5b+9y-(15b+18y)+6b)/(6b)#

#=(-4b-9y)/(6b)=-(4b+9y)/(6b)#