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Maharashtra State BoardSSC (English Medium) 9th Standard

Solve the following equation. 𝑥2+12⁢𝑥−20/3⁢𝑥−5 =𝑥2+8⁢𝑥+12/2⁢𝑥+3 - Algebra

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Question

Solve the following equation.

`(x^2 + 12x - 20)/(3x - 5) = (x^2 + 8x + 12)/(2x + 3)`

Sum
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Solution

`(x^2 + 12x - 20)/(3x - 5) = (x^2 + 8x + 12)/( 2x + 3 )`

Multiplying both sides by `1/4`, we get

`(x^2 + 12x - 20)/(12x - 20) = (x^2 + 8x + 12)/(8x + 12)`

Using dividendo, we get

`[(x^2 + 12x - 20) - (12x - 20)]/(12x - 20) = [(x^2 + 8x + 12) - (8x + 12)]/ (8x + 12)`

⇒ `[x^2 + 12x - 20 - 12x + 20 ]/(12x - 20) = [x^2 + 8x + 12 -  8x - 12]/(8x + 12)`

⇒ `x^2/(12x - 20) = x^2/(8x + 12)`

This equation is true for x = 0.

Therefore, x = 0 is a solution of the given equation.

If x ≠ 0, then x2 ≠ 0.

Dividing both sides by x2, we get

`1/(12x - 20)= 1/(8x + 12)`

⇒ 12x − 20 = 8x + 12

⇒ 12x − 8x = 20 + 12

⇒ 4x = 32

⇒ x = 8

Thus, the solutions of the given equation are x = 0 and x = 8.

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Application of Properties of Equal Ratios
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Chapter 4: Ratio and Proportion - Practice Set 4.3 [Page 70]

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Balbharati Mathematics 1 [English] Standard 9 Maharashtra State Board
Chapter 4 Ratio and Proportion
Practice Set 4.3 | Q (4) (i) | Page 70
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