How to solve $\frac{2x+1}{2x-3}+\frac{7x\:}{9-4x^2}=1+\frac{x-4}{2x+3}$ for $x$?

Question

Can somebody explain me this one!

$\frac{2x+1}{2x-3}+\frac{7x\:}{9-4x^2}=1+\frac{x-4}{2x+3}$

My book says the answer is $x_1 = 0$; $x_2 = 6$.

I tried to solve it and got stuck somewhere in:

$4x^2+x+3/(2x-3)(3x-1) = 0$

I can't find my mistake.

Answer 1

Note that $$9 - 4x^2 = -(4x^2 - 9) = -(2x+3)(2x-3)$$

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$$\frac{2x+1}{2x-3}+\frac{7x\:}{9-4x^2}=1+\frac{x-4}{2x+3}\frac{2x+1}{2x-3}-\frac{7x\:}{(2x+3)(2x-3)}=1+\frac{x-4}{2x+3}$$

Now, multiply both sides of the equation by $(2x - 3)(2x+ 3)$:

$$(2x+1)(2x+3) - 7x = (2x+3)(2x-3) +(x-4)(2x-3)$$

Simplify by expanding as needed, and forming a quadratic equation (all terms on left, with $0$ on the right).

$$\begin{align} &\quad 4x^2 +8x+3 - 7x -(4x^2 -9) - (2x^2 - 11x +12) = 0\\ & \iff -2x^2 +12x = 0 \\ &\iff -2x(x -6) = 0 \\ &\implies x = 0\, \text{ or }\,x = 6\end{align}$$