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Mathematics

For equation 1x+1x−5=310\dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10}; one value of x is :

  1. −53-\dfrac{5}{3}

  2. 10

  3. -10

  4. 5

Quadratic Equations

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Answer

Given,

⇒1x+1x−5=310⇒x−5+xx(x−5)=310⇒2x−5x2−5x=310⇒10(2x−5)=3(x2−5x)⇒20x−50=3x2−15x⇒3x2−15x−20x+50=0⇒3x2−35x+50=0⇒3x2−30x−5x+50=0⇒3x(x−10)−5(x−10)=0⇒(3x−5)(x−10)=0⇒3x−5=0 or x−10=0⇒3x=5 or x=10⇒x=53 or x=10.\Rightarrow \dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{x - 5 + x}{x(x - 5)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x - 5}{x^2 - 5x} = \dfrac{3}{10} \\[1em] \Rightarrow 10(2x - 5) = 3(x^2 - 5x) \\[1em] \Rightarrow 20x - 50 = 3x^2 - 15x \\[1em] \Rightarrow 3x^2 - 15x - 20x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 35x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 30x - 5x + 50 = 0 \\[1em] \Rightarrow 3x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (3x - 5)(x - 10) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow 3x = 5 \text{ or } x = 10 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = 10.

Hence, Option 2 is the correct option.

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