How do you solve #\frac { u - 2} { u - 5} + 1= \frac { u - 5} { u - 3}#?

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Answer 1 · 2017-12-11T07:23:01

#u = 4" "# or #" "u = -1#

Given:

#(u-2)/(u-5)+1 = (u-5)/(u-3)#

Multiply both sides by #(u-5)(u-3)# to get:

#(u-2)(u-3)+(u-5)(u-3) = (u-5)^2#

Multiply out to get:

#(u^2-5u+6)+(u^2-8u+15) = u^2-10u+25#

Simplify the left hand side:

#2u^2-13u+21 = u^2-10u+25#

Subtract the right hand side from the left and transpose to get:

#0 = u^2-3u-4 = (u-4)(u+1)#

So:

#u = 4" "# or #" "u = -1#

Note that these are both valid solutions of the original equation, since they do not cause any denominator to be #0#.