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Question

Solve for x 
$$\frac{4x-3}{2x+1}-10\left ( \frac{2x+1}{4x-3} \right )=3;x\neq -\frac{1}{2},\frac{3}{4}$$


Solution

$$\dfrac{4x-3}{2x+1}-10\left(\dfrac{2x+1}{4x-3}\right)=3$$    $$x\neq \dfrac{-1}{2},\dfrac{3}{4}$$
On solving this we get
$$\dfrac{(4x-3)^{2}-10(2x+1)^{2}}{(2x+1)(4x-3)}=3$$
$$16x^{2}+9-24x-40x^{2}-10-40x=3(8x^{2}-2x-3)$$
$$16x^{2}-40x^{2}-24x^{2}-24x-40x+6x+9-10+9=0$$
$$-48x^{2}-58x+8=0$$
$$48x^{2}+58x-8=0$$
$$\Rightarrow 48x^{2}+58x-8=0$$ On dividing by $$2$$
$$24x^{2}+29x-4=0$$
$$x=\dfrac{-29\pm \sqrt{1225}}{48}=\dfrac{-29\pm 35}{48}$$
$$x=\dfrac{6}{48},\dfrac{-64}{48}$$
Hence values of x are $$\left(\dfrac{1}{8}, \dfrac{-2}{3}\right)$$.

1179609_1350745_ans_afcd3dc8d81945b3bd7f5fa05d2b087f.jpg

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