How do you solve #\frac { 1} { 3y - 3} + \frac { 1} { 4y - 4} = 1#?

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Answer 1 · 2017-07-27T02:07:02

Given #1/(3y - 3) + 1/(4y - 4) = 1#

Multiply both sides by #12y-12#:

#4 + 3 = 12y-12#

#19=12y#

#y = 19/12 larr# answer.

Check:

#1/(3(19/12) - 3) + 1/(4(19/12) - 4) = 1#

#12/(3(19) - 3(12)) + 12/(4(19) - 4(12)) = 1#

#4/(19 - 12) + 3/(19 - 12) = 1#

#4/7 + 3/7 = 1#

#7/7 = 1#

This checks.