# solving rational equations

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• Jul 6th 2008, 11:27 AM
Hypertension
solving rational equations
I thought I got the whole concept but when I came across these problems, my answer came no where near that of the correct answer.

1) $\displaystyle \frac{x}{x-3} + x = \frac{7x-18}{x-3}$

2) $\displaystyle \frac3{x} + \frac{x}{x+2} = - \frac{2}{x+2}$

I need to solve for "x". Help?
• Jul 6th 2008, 11:32 AM
galactus
Quote:

1) $\displaystyle \frac{x}{x-3} + x = \frac{7x-18}{x-3}$
The idea is to shed the denominators to get rid of the fractions. Do this by multiplying through by the LCD. In this case, x-3.

Multiply through by x-3

$\displaystyle (x-3)\cdot\frac{x}{x-3}+(x-3)x=(x-3)\cdot\frac{7x-18}{x-3}$

Then we get $\displaystyle x+x(x-3)=7x-18$

Now, it's easy to solve, huh?. Remember to check your solutions by plugging them back in. This one will give two solutions, but only one may be good.
• Jul 6th 2008, 11:44 AM
Hypertension
Thanks.

That's about how far I get and then I start running into trouble.
After simplifying this:
$\displaystyle x+x(x-3)=7x-18$

I get this:
$\displaystyle x+x^2-3x=7x-18$

= $\displaystyle x^2-2x=7x-18$

= $\displaystyle x^2-9x=-18$

Am I doing it wrong? :(
• Jul 6th 2008, 11:52 AM
Simplicity
Quote:

Originally Posted by Hypertension
Thanks.

That's about how far I get and then I start running into trouble.
After simplifying this:
$\displaystyle x+x(x-3)=7x-18$

I get this:
$\displaystyle x+x^2-3x=7x-18$

= $\displaystyle x^2-2x=7x-18$

= $\displaystyle x^2-9x=-18$

Am I doing it wrong? :(

No, that is correct but the method is incomplete. You must continue.

Notice, you have $\displaystyle x^2-9x=-18$ which is $\displaystyle x^2-9x+18=0$.

You can solve by factorising into 2 brackets and then equating each bracket to 0. Hence finding two values for $\displaystyle x$. If one of the solution give the denominator to equal 0 then reject this solution.

EDIT: If are struggling to factorise, and when the quadratic equation is in the form $\displaystyle ax^2 + bx + c =0$ then you can use the quadratic formula which is $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
• Jul 6th 2008, 12:25 PM
Hypertension
Ok, I've got it now.

x=6 or x=3

But in this case x would equal 6 because when you plug in the 3 for x you get 3 equaling zero as the answer. I just didn't expect to have have to factor, since I haven't had a problem involving that yet.

And for problem two, I got
x=-3

Thank you guys