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Solve :

2(2x−1x+3)−3(x+32x−1)=52\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5

x ≠ -3, 12\dfrac{1}{2}

Quadratic Equations

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Answer

Substituting, 2x−1x+3\dfrac{2x - 1}{x + 3} = a in above equation

⇒2a−31a=5⇒2a−3a=5⇒2a2−3a=5⇒2a2−3=5a⇒2a2−5a−3=0⇒2a2−6a+a−3=0⇒2a(a−3)+1(a−3)=0⇒(2a+1)(a−3)=0⇒2a+1=0 or a−3=0⇒a=−12 or a=3.\Rightarrow 2a - 3\dfrac{1}{a} = 5 \\[1em] \Rightarrow 2a - \dfrac{3}{a} = 5 \\[1em] \Rightarrow \dfrac{2a^2 - 3}{a} = 5 \\[1em] \Rightarrow 2a^2 - 3 = 5a \\[1em] \Rightarrow 2a^2 - 5a - 3 = 0 \\[1em] \Rightarrow 2a^2 - 6a + a - 3 = 0 \\[1em] \Rightarrow 2a(a - 3) + 1(a - 3) = 0 \\[1em] \Rightarrow (2a + 1)(a - 3) = 0 \\[1em] \Rightarrow 2a + 1 = 0 \text{ or } a - 3 = 0 \\[1em] \Rightarrow a = -\dfrac{1}{2} \text{ or } a = 3.

Considering, a = 3

⇒2x−1x+3=3⇒2x−1=3(x+3)⇒2x−1=3x+9⇒3x−2x=−1−9⇒x=−10.\Rightarrow \dfrac{2x - 1}{x + 3} = 3 \\[1em] \Rightarrow 2x - 1 = 3(x + 3) \\[1em] \Rightarrow 2x - 1 = 3x + 9 \\[1em] \Rightarrow 3x - 2x = -1 - 9 \\[1em] \Rightarrow x = -10.

Considering, a = -12\dfrac{1}{2}

⇒2x−1x+3=−12⇒2(2x−1)=−1(x+3)⇒4x−2=−x−3⇒4x+x=−3+2⇒5x=−1⇒x=−15.\Rightarrow \dfrac{2x - 1}{x + 3} = -\dfrac{1}{2} \\[1em] \Rightarrow 2(2x - 1) = -1(x + 3) \\[1em] \Rightarrow 4x - 2 = -x - 3 \\[1em] \Rightarrow 4x + x = -3 + 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\dfrac{1}{5}.

Hence, x = −15,−10.-\dfrac{1}{5}, -10.

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