Algebra question

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July 31st 2009, 04:10 PM
marie7

Algebra question

Solve for x:

(x^2 - 4x + 4x - 16) - (x^2 - 2x - 3x + 6) = 0
July 31st 2009, 04:18 PM
Chris L T521

Quote:

Originally Posted by marie7

Solve for x:

(x^2 - 4x + 4x - 16) - (x^2 - 2x - 3x + 6) = 0

Have you tried to simply that?

Note that its the same as $x^2-4x+4x-16-x^2+2x+3x-6=0$ .

Can you try to simplify the left side first and then solve for x?
July 31st 2009, 04:21 PM
marie7

So it would be -5x -10 = 0 right? Then where would I go from there? Because the question I'm doing are quadratic equations and then there's this one which is random and doesn't have the format to do a quadratic calculation on it.
July 31st 2009, 04:23 PM
marie7

-5x = 10
x = -2 is the answer I've gotten.

The original equation is [(x + 4) / (x +2)] - [(x - 3) / (x - 4)] = 0 and I simplified it to the equation I mentioned in my first post. -2 doesn't fit into the original equation. Maybe I've done something wrong.
July 31st 2009, 04:27 PM
Chris L T521

Quote:

Originally Posted by marie7

So it would be -5x -10 = 0 right? Then where would I go from there? Because the question I'm doing are quadratic equations and then there's this one which is random and doesn't have the format to do a quadratic calculation on it.

I get $5x-22=0$

From there, solve it like you would solve a linear equation!

Don't let this mess with you. Sometimes when you deal with this kind of stuff, you may not get a quadratic equation to solve!
July 31st 2009, 04:29 PM
marie7

I thought -16 + 6 is -10? not -22
July 31st 2009, 04:32 PM
Chris L T521

Quote:

Originally Posted by marie7

I thought -16 + 6 is -10? not -22

In the original expression, it was $x^2+4x-4x-16-(x^2-2x-3x+6)=0\implies x^2-16-x^2+5x-6=0$ $\implies 5x-16-6=0\implies 5x-22=0$

So it should be $x=-\frac{22}{5}$
July 31st 2009, 04:34 PM
marie7

I'll copy and paste this just in case you didn't see it in my above post:

"The original equation is [(x + 4) / (x +2)] - [(x - 3) / (x - 4)] = 0 and I simplified it to the equation I mentioned in my first post. your answer doesn't fit into the original equation. Maybe I've done something wrong in my calculations."

http://www.mathhelpforum.com/math-he...c/progress.gif

Note that I've just edited the first (x - 4) to (x +4) in the equation above. It was a typo, so sorry!
July 31st 2009, 04:42 PM
Chris L T521

Quote:

Originally Posted by marie7

The original equation is [(x - 4) / (x +2)] - [(x - 3) / (x - 4)] = 0 and I simplified it to the equation I mentioned in my first post. -2 doesn't fit into the original equation. Maybe I've done something wrong.

Note that I've just edited the first (x - 4) to (x +4) in the equation above. It was a typo, so sorry!

If that's the case then, you should get

$(x-4)(x+4)-(x+2)(x-3)=0\implies x^2-16-(x^2-x-6)=0$ $\implies x^2-16-x^2+x+6=0\implies x-10=0\implies x=10$
July 31st 2009, 04:45 PM
marie7

Thank you very much for your help! *gives you cookies*