How do you solve $\frac{7}{x + 1} = \frac{6}{x - 5}$ $7/(x+1)=6/(x-5)$ and check for extraneous solutions?

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Answer 1 · 2016-08-26T21:19:47

$x = 41$ $x= 41$

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.

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

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

$7x-35 = 6x+6$

$x= 41$

How do you solve 7/(x+1)=6/(x-5)7x+1=6x−57/(x+1)=6/(x-5) and check for extraneous solutions?