How do you solve #5-2/(x-6) = (10-2x)/(x-6)#?

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Answer 1 · 2017-05-29T06:31:13

no solution

Let's give everything the same denominator. Fortunately, two of the three components already share the same denominator #(x-6)#.

If we multiply #5# by #(x-6)/(x-6)#, all the components will be "combinable":

#(5x-30)/(x-6) - 2/(x-6) = (10-2x)/(x-6)#

Combine like-terms

#((5x-30)-(2))/(x-6) = (10-2x)/(x-6)#

Multiply by (#x-6#) on both sides

#5x-30-2 = 10-2x#

Simplify

#5x-32=10-2x#

Add 2x to both sides

#7x-32=10#

Add #32# to both sides

#7x=42#

Divide by #7# on both sides

#x=6#

Just to check our work, let's solve the equation, replacing #x# with #6#:

#5- 2/(6-6) = (10-2xx6)/(6-6)#

#5-2/0=(-2)/0#

Uh oh! We're dividing by ZERO! that's illegal, so there are no solutions.