Solve for x: (i) (x - 1)/(x - 2) + (x - 3)/(x - 4) = (3)1/3; x ≠ 2, 4

Question

Solve for x:

(i) \(\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3};x\neq2,4\) 

(ii) \(\frac{1}{x}-\frac{1}{x-2}=3,x\neq0,2\)

(iii) \(x+\frac{1}{x}=3,x\neq0\)

(iv) \(\frac{16}{x}-1=\frac{15}{x+1},x\neq0,-1\)

Answer 1

(i) \(\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3};x\neq2,4\)

LCM of the denominator is (x - 2)(x - 4)Now further solve it,

⇒ 3(x– 1)(x – 4) + 3(x – 2)(x – 3) = 10(x – 2)(x – 4)

⇒ 3(x² - 4x - x + 4) + 3(x² - 3x - 2x + 6) =10(x² - 4x - 2x + 8)

⇒ 3(x² - 5x + 4) + 3(x² - 6x+6) =10(x² - 6x + 8)

⇒ 3x² + 12 – 15x + 3x² + 18 – 18x = 10x² – 60x + 80

⇒ 6x² - 30x + 30 = 10x² – 60x + 80

⇒ 6x² - 10x² - 30x + 60x + 30 - 80 = 0

⇒ - 4x² + 30x - 50 = 0

⇒ 4x² – 30x + 50 = 0

⇒ 2x² – 15x + 25 = 0

Factorize it by splitting the middle term.

⇒ 2x² – 10x – 5x + 25 = 0

⇒ 2x(x – 5) – 5(x – 5) = 0

⇒ (2x – 5)(x – 5) = 0⇒2x – 5 = 0

\(x=\frac{5}{2}\)

⇒ x – 5 = 0 x = 5

Thus, \(x=5,\frac{5}{2}\)

(ii) \(\frac{1}{x}-\frac{1}{x-2}=3,x\neq0,2\)

LCM of the denominator is x (x - 2)

⇒ x – 2 – x = 3x (x - 2)

⇒ x – 2 – x = 3x² – 6x

⇒ 3x² – 6x + 2 = 0

Using \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Here a = 3, b = -6, c = 2

(iii) \(x+\frac{1}{x}=3,x\neq0\)

LCM of denominators is x.

\(\Rightarrow \frac{x^2+1}{x}=3\)

⇒ x² + 1 = 3x

⇒ x² – 3x + 1 = 0

Using \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

(iv) \(\frac{16}{x}-1=\frac{15}{x+1},x\neq0,-1\)

Use cross multiplication to get, 

(16 – x)(x + 1) = 15x

⇒ 16x – x² + 16 – x = 15x

⇒ 16x –15x - x² + 16 – x = 0

⇒ - x² + 16 = 0

⇒ x² = 16

⇒ x = ±4