How do you solve #-2/(x-1)=(x-8)/(x+1)#?

1 Answer
Grace
Jun 20, 2017

The answer is #x = 2, 5#.

Explanation:

We see that the equation is a proportion, so we can cross-multiply to start off. Cross-multiplying, the equation becomes:

#-2(x+1) = (x-1)(x-8)#

We can then distribute the -2 on the left side, while multiplying out the right-hand side using FOIL (or any method you learned).

#-2x - 2 = x^2 - 9x + 8#

Adding #2x# and #2# to both sides gives us

#0 = x^2 - 7x + 10#

Now, we can factor the equation to get the zeroes of the equation or the values for #x#. We see that #-5*-2 = 10# (the third term in the expansion) and #-5+(-2)=-7# (the second term in the expansion), so those should be the second terms in each binomial. This is just one method of factoring; there are many other ways you could use. After factoring, we get:

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

To get the zeroes, we see that if #x-5 = 0#, then #x=5#, and if #x-2 = 0#, then #x=2#.

Therefore, #x = 2, 5#.

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