\(\frac{4x-3}{x+2}\) + \(\frac{15}{x-3}\) = 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 = \(-\frac{ \sqrt{39} }{2}\)
