How do you evaluate #frac { x + y } { 8x - y } - \frac { 7x } { y - 8x }#?

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Answer 1 · 2017-12-04T12:25:10

See a solution process below:

First, multiply the second fraction by #-1/-1#. This is equal to #1# and therefore does not change the value of the fraction:

#(x + y)/(8x - y) - (-1/-1 xx (7x)/(y - 8x)) =>#

#(x + y)/(8x - y) - (-1 xx 7x)/(-1(y - 8x)) =>#

#(x + y)/(8x - y) - (-7x)/((-1 xx y) - (-1 xx 8x)) =>#

#(x + y)/(8x - y) - (-7x)/(-y - (-8x)) =>#

#(x + y)/(8x - y) - (-7x)/(-y + 8x) =>#

#(x + y)/(8x - y) - (-7x)/(8x - y)#

Now, with the fractions over common denominators we can subtract the numerators:

#((x + y) - (-7x))/(8x - y) =>#

#(x + y + 7x)/(8x - y) =>#

#(x + 7x + y)/(8x - y) =>#

#(1x + 7x + y)/(8x - y) =>#

#((1 + 7)x + y)/(8x - y) =>#

#(8x + y)/(8x - y)#