Question

okay i need help again. i;m supposed to solve this equation, i came up with two equations being equal to each other
(x+2)/(x^2-16) + (x-3)/(x^2-3x-4) = (2x-3)/(x^2+5x+4)

Answer 1

you made mistake somwhere, because it has one solution

Answer 2

so what was my mistake?

Answer 3

show your work and we will shot you mistake :P

Answer 4

:) okay, i factored the denominators and got this
(x+2)/(x+4)(x-4) + (x-3)/(x+1)(x-4) = (2x+3)/(x+1)(x+4)
multiplying everything by the LCD which is (x+4)(x-4)(x+1) i got this
(x+2)(x+1) + (x-3)(x+4) = (2x-3)(x-4)
x^2+3x+2+x^2+x-12=2x^2-11x+12
2x^2+4x-10=2x^2-11x+12
that's what i have

Answer 5

\[\frac{x+2}{x^2-16} + \frac{x-3}{x^2-3x-4} = \frac{2x-3}{x^2+5x+4}\]
\[\frac{x+2}{(x+4)(x-4)} + \frac{x-3}{(x+1)(x-4)} = \frac{2x-3}{(x+1)(x+4)}\]
\[(x+2)(x+1)+(x-3)(x+4)=(2x-3)(x-4)\]
\[x^2+3x+2+x^2+x-12=2x^2-11x+12\]
\[2x^2+4x-10=2x^2-11x+12\]
yeah that's correct
but they are not equal, subtract both sides by \(2x^2-11x+12\)

Answer 6

it was my 1000th reply btw :P

Answer 7

so x=22/15?

Answer 8

no
\[1\frac{7}{15}\]
:P

Answer 9

:) lol okay, thank you