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Solve : 3(3x−12x+3)−2(2x+33x−1)=53\Big(\dfrac{3x - 1}{2x + 3}\Big) - 2\Big(\dfrac{2x + 3}{3x - 1}\Big) = 5

Quadratic Equations

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Answer

Solving,

⇒3(3x−12x+3)−2(2x+33x−1)=5⇒3(3x−1)2−2(2x+3)2(2x+3)(3x−1)=5⇒3(3x−1)2−2(2x+3)2=5(2x+3)(3x−1)⇒3(9x2−6x+1)−2(4x2+12x+9)=5(6x2−2x+9x−3)⇒27x2−18x+3−8x2−24x−18=5(6x2+7x−3)⇒19x2−42x−15=30x2+35x−15⇒30x2−19x2+35x+42x−15+15=0⇒11x2+77x=0⇒11x(x+7)=0⇒11x=0 or x+7=0⇒x=0 or x=−7.\Rightarrow 3\Big(\dfrac{3x - 1}{2x + 3}\Big) - 2\Big(\dfrac{2x + 3}{3x - 1}\Big) = 5 \\[1em] \Rightarrow \dfrac{3(3x - 1)^2 - 2(2x + 3)^2}{(2x + 3)(3x - 1)} = 5 \\[1em] \Rightarrow 3(3x - 1)^2 - 2(2x + 3)^2 = 5(2x + 3)(3x - 1) \\[1em] \Rightarrow 3(9x^2 - 6x + 1) - 2(4x^2 + 12x + 9) = 5(6x^2 - 2x + 9x - 3) \\[1em] \Rightarrow 27x^2 - 18x + 3 - 8x^2 - 24x - 18 = 5(6x^2 + 7x - 3) \\[1em] \Rightarrow 19x^2 - 42x - 15 = 30x^2 + 35x - 15 \\[1em] \Rightarrow 30x^2 - 19x^2 + 35x + 42x - 15 + 15 = 0 \\[1em] \Rightarrow 11x^2 + 77x = 0 \\[1em] \Rightarrow 11x(x + 7) = 0 \\[1em] \Rightarrow 11x = 0 \text{ or } x + 7 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -7.

Hence, x = 0 or x = -7.

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