How do you solve #(u-2)/(u-3) + 3= u + (u-4)/(3-u)# and find any extraneous solutions?

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Answer 1 · 2017-07-26T23:57:35

#(u-2)/(u-3)+3= u+(u-4)/(3-u)#

#(u-2)/(u-3)+3= u-(u-4)/(u-3)#

#(u-2)/(u-3)+(u-4)/(u-3)=u-3#

#(2u-6)/(u-3)=u-3#

#u-3=2*(u-3)/(u-3)#

#u-3=2#

#u=5#

1) I rewrote polynomial.

2) I left polynomials to left side and polynomials to right one.

3) I simplified polynomial due to x isn't equal to 3 condition.

4) I solved equation.