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Wewaii [24]
3 years ago
13

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

Mathematics
2 answers:
Murljashka [212]3 years ago
8 0
Please see the pic, I'd solved in it.

Andrews [41]3 years ago
3 0
\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}
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