1. ## 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?

2. 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.

3. 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?

4. 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 $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

5. 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