How do you solve #1/(x^2-5x)=(x+7)/x-1# and check for extraneous solutions?

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Answer 1 · 2017-01-17T02:23:38

#{36/7}#

Recall that #x^2 - 5x# can be written as #x(x - 5)#.

#1/(x(x - 5)) = (x + 7)/x - 1#

Put on a common denominator:

#1/(x(x -5)) = ((x + 7)(x - 5))/(x(x - 5)) - (x(x -5))/(x(x - 5))#

#1 = x^2 +7x - 5x - 35 - (x^2 - 5x)#

#1 = x^2 + 7x - 5x - 35 - x^2 + 5x#

#36 = 7x#

#x = 36/7#

This is not extraneous, as #x != 0, 5#.

Hopefully this helps!