Question

Can you solve (4x-3)/(x+2)+(15)/(x-3)

Answer 1

Please see the pic, I'd solved in it.

Answer 2

$\frac{4x-3}{x+2}$ + $\frac{15}{x-3}$ = 0
1st step: We need to have a common denominator.
So we do this:
$\frac{(4x-3)*(x-3)}{(x+2)*(x-3)}$ + $\frac{(15)*(x+2)}{(x-3)*(x+2)}$ = 0
2nd step: Distribute them
$\frac{4x^{2}-3x-12x+9}{(x-3)(x+2)}$ + $\frac{15x+30}{(x+2)(x-3)}$ = 0
3rd step: Do the rest of the equation!
We can put it on one fraction, its fine.
$\frac{4x^2-15x+9+15x+30}{(x+2)(x-3)}$ = 0
$\frac{4x^2+39}{(x+2)(x-3)}$ = 0 (-15 and 15 cancels out)
4th step: take the denominator to the other side
4x²+39 = 0 (anything × 0 = 0)
5th step: Take 39 to the other side
4x² = -39
6th step: Divide by 4 on both sides
$\frac{4x^2}{4}$ = $-\frac{39}{4}$
(4 and 4 cancels out)
x² = $-\frac{39}{4}$
7th step: Take the square root to keep x alone
x = $-\sqrt\frac{39}{4}$
It can also be written as:
x = $-\frac{ \sqrt{39} }{ \sqrt{4} }$
Final answer:
x = $-\frac{ \sqrt{39} }{2}$