# Thread: Algebraic equation with fractions

1. ## Algebraic equation with fractions

Hi, I need help with this equation!

$\frac {9} {x-10} - \frac {210}{x^2 -100} = 1$

I have tried to cross multiply and factor the question and end up with

$x(-9x^2-9x+1200) = 0$

If that is in the ballpark can you help me factor that or am I completely wrong?

The book asks for

Solutions are x1= and x2= . Important Note: x1 denotes the smaller solution.

2. Sorry but this is unreadable. Please use brackets where they're needed or learn some basic LaTeX. There is an entire subforum here dedicated to learning to use LaTeX.

3. Huzzah, I have fixed my original post. Sorry for the inconvenience.

4. Originally Posted by tofuwrice
Hi, I need help with this equation!

$\frac {9} {x-10} - \frac {210}{x^2 -100} = 1$

I have tried to cross multiply and factor the question and end up with

$x(-9x^2-9x+1200) = 0$

If that is in the ballpark can you help me factor that or am I completely wrong?

The book asks for

Solutions are x1= and x2= . Important Note: x1 denotes the smaller solution.
Cross multiplying is a method that is "in the ballpark", but the easier way is to get a common denominator.

Note that $x^2 - 100 = (x - 10)(x + 10)$.

So $\frac{9}{x - 10} - \frac{210}{x^2 - 100} = 1$

$\frac{9(x + 10)}{(x - 10)(x + 10)} - \frac{210}{(x - 10)(x + 10)} = 1$

$\frac{9(x + 10) - 210}{(x - 10)(x + 10)} = 1$

$\frac{9x + 90 - 210}{x^2 - 100} = 1$

$\frac{9x - 120}{x^2 - 100} = 1$

$9x - 120 = 1(x^2 - 100)$

$9x - 120 = x^2 - 100$

$0 = x^2 - 9x + 20$

$0 = (x - 4)(x - 5)$

$x - 4 = 0$ or $x - 5 = 0$

$x = 4$ or $x = 5$.

5. omg, i am literally smacking my head. I can't believe i didn't see this simple solution. Thanks a lot!