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Mathematics

Solve the following equation using the formula :

x−1x−2+x−3x−4=313\dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3}

Quadratic Equations

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Answer

Given,

⇒x−1x−2+x−3x−4=313⇒(x−1)(x−4)+(x−3)(x−2)(x−2)(x−4)=103⇒x2−4x−x+4+x2−2x−3x+6x2−4x−2x+8=103⇒2x2−10x+10x2−6x+8=103⇒3(2x2−10x+10)=10(x2−6x+8)⇒6x2−30x+30=10x2−60x+80⇒10x2−6x2−60x+30x+80−30=0⇒4x2−30x+50=0⇒2(2x2−15x+25)=0⇒2x2−15x+25=0⇒2x2−10x−5x+25=0⇒2x(x−5)−5(x−5)=0⇒(2x−5)(x−5)=0⇒2x−5=0 or x−5=0⇒x=52 or x=5.\Rightarrow \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 - 4x - x + 4 + x^2 - 2x - 3x + 6}{x^2 - 4x - 2x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow 3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8) \\[1em] \Rightarrow 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \\[1em] \Rightarrow 10x^2 - 6x^2 - 60x + 30x + 80 - 30 = 0 \\[1em] \Rightarrow 4x^2 - 30x + 50 = 0 \\[1em] \Rightarrow 2(2x^2 - 15x + 25) = 0 \\[1em] \Rightarrow 2x^2 - 15x + 25 = 0 \\[1em] \Rightarrow 2x^2 - 10x - 5x + 25 = 0 \\[1em] \Rightarrow 2x(x - 5) - 5(x - 5) = 0 \\[1em] \Rightarrow (2x - 5)(x - 5) = 0 \\[1em] \Rightarrow 2x - 5 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow x = \dfrac{5}{2} \text{ or } x = 5.

Hence, x = 5, 52\dfrac{5}{2}.

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