# Solve for x:(i) (ii) (iii) (iv)

(i)

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 – 5 = 0

x=5

Thus,

(ii)

LCM of the denominator is x (x-2)

⇒

⇒ x – 2 – x = 3x(x-2)

⇒ x – 2 – x = 3x2 – 6x

⇒ 3x2 – 6x + 2 = 0

Using

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

(iii)

LCM of denominators is x.

⇒ x2 + 1 = 3x

⇒ x2 – 3x + 1 = 0

Using

(iv)

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

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