View Full Version : Solve algebraically: 5/x+5-2/x-3 = -1/x^2+2x-15

kjones

06-01-2006, 08:23 AM

5/x+5 - 2/x-3 = -1/x^2+2x-15

If I could see it worked out I think that would really help me see the process. I know I have to multiply the denominator to remove the fraction.

Thanks. K

steve_b

06-01-2006, 08:48 AM

5/x+5 - 2/x-3 = -1/x^2+2x-15

If I could see it worked out I think that would really help me see the process. I know I have to multiply the denominator to remove the fraction.

===============

I'm going to assume that you left out some parentheses:

5/(x+5) - 2/(x-3) = -1/(x^2+2x-15)

Factor the denominator on the right side to get:

-1/[(x+5)(x-3)]

Now, if you multiply both sides of the equation by (x+5)(x-3), you will eliminate all denominators:

5(x-3) - 2(x+5) = -1

Use the Distributive Property to eliminate parentheses:

5x - 15 - 2x - 10 = -1

Gather like terms:

3x - 25 = -1

Isolate the x term to one side of the equation:

3x = 24

Divide both sides by 3:

x = 8

Check the answer in the original equation. (Your job!)

Hope that helps...

Steve

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