ok so i have this problem.

$\displaystyle \frac{x+4}{x-2} - \frac{x-3}{x-4}$

So i know i should multiply both sides by the denominator then factorize then subtract , then use quadratic formula right?

But i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

2. Originally Posted by el123
ok so i have this problem.

$\displaystyle \frac{x+4}{x-2} - \frac{x-3}{x-4}$

So i know i should multiply both sides by the denominator then factorize then subtract , then use quadratic formula right?

But i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

$\displaystyle \frac{x+4}{x-2} - \frac{x-3}{x-4}=0$?

To use the quadratic function, one side must be equal to 0. Did your question not include an equality? You can't find the value of x without having the function equal something (and if you did have any value instead of 0, you could bring it to the x side to get 0); Even if you simplify that down to the form $\displaystyle ax^2 + bx +c$, you don't know what that function is equal to, x could have an infinite number of possibilities. Otherwise I'd say your instructor wants you to assume it is equal to 0 and do as you are planning to.

3. yeah my bad i forgot , yes it is equal to zero.

4. Originally Posted by el123
ok so i have this problem.

$\displaystyle \frac{x+4}{x-2} - \frac{x-3}{x-4}$

So i know i should multiply both sides by the denominator then factorize then subtract , then use quadratic formula right?

But i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

Turns out to be a very similar problem.

5. Originally Posted by el123
...i keep getting mixed up , or maybe i am missing somthing. Can somebody please help me.

6. In case you haven't worked it out yet:

$\displaystyle \frac{x+4}{x-2}-\frac{x-3}{x-4}=0$

$\displaystyle (x+4)(x-4)-(x-3)(x-2)=0$

$\displaystyle (x^2-16)-(x^2-5x+6)=0$

$\displaystyle 5x-22=0$

$\displaystyle x=\frac{22}{5}$