How do you solve #7/(x+1)=6/(x-5)# and check for extraneous solutions?

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Answer 1 · 2016-08-26T21:19:47

#x= 41#

Before we even start with solving, we can immediately exclude two values of x as solutions.

The denominator of a fraction may not be equal to 0 so any value of x which causes this may not be a solution.

If #x+1 = 0 rArr x = -1 " " and " " if x-5 = 0 rArr x = 5#

Therefore #x !=-1 and x!=5#

We have an equation with one term on each side.
Cross-multiply.

#7/(x+1)=6/(x-5)#

#7(x-5) = 6(x+1)#

#7x-35 = 6x+6#

#x= 41#