SOLUTION: u+5/u-7 + 1= u-1/u-5

Question 280347: u+5/u-7 + 1= u-1/u-5
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
(In the future, please post fractions in the form:
{numerator)/(denominator)
Tutors will be more likely to respond if the problems are posted clearly. As it is, I am guessing that your equation is:

If this is correct then it should be posted as:
(u+5)/(u-7) + 1= (u-1)/(u-5)

Most people would agree that equations without fractions are easier to work with than the ones that have fractions. So whenI am faced with equations like this, my first goal is to get rid of the fractions ASAP. The quickest way to do this is to multiply both sides of the equation by the Lowest Common Denominator (LCD) of all the denominators (on both sides). The LCD for the fractions of this equation is:
(u-7)(u-5)
so this is what we will multiply by:

On the left side we need to use the Distributive Property to multiply:

Now we can cancel:

leaving:

And the fractions are gone. To solve this we start by simplifying:

Since this is a quadratic equation we want one side of the equation to be zero. So we'll subtract the entire right side from both sides:

Now we factor (or use the Quadratic Formula). This factors easily:
(u-1)(u-3) = 0
From the Zero Product PRoperty we know that this product is zero only if one fo the factors is zero. So:
u-1 = 0 or u-3 = 0
Solving these we get:
u = 1 or u = 3

When we eliminated the fractions we multiplied both sides by the LCD, (u-5)(u-7). Depending on the value of u, (u-5)(u-7) could be zero. Whenever you multiply both sides of an equation by some expresssion that might be zero, it is important (not just a good idea) to check your answer.

Checking u = 1:

Check!

Checking u = 3:

Check!

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