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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता ९ वी

Solve the following equation. 10x2+15x+635x2-25x+12=2x+3x-5 - Algebra

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प्रश्न

Solve the following equation.

`[10x^2 + 15x + 63]/[5x^2 - 25x + 12] = (2x + 3)/( x -5)`

बेरीज

उत्तर

`[10x^2 + 15x + 63]/[5x^2 - 25x + 12] = (2x + 3)/( x -5)`

By Alternando,

⇒ `[10x^2 + 15x + 63]/(2x + 3) =[5x^2 - 25x + 12]/( x -5)`

Multiplying both sides by `1/(5x)`, we get

`[10x^2 + 15x + 63]/(10x^2 + 15x) =[5x^2 - 25x + 12]/( 5x^2 -25x)`

Using dividendo, we get

`[(10x^2 + 15x + 63) - (10x^2 + 15x)]/(10x^2 + 15x) =[(5x^2 - 25x + 12) - ( 5x^2 -25x)]/( 5x^2 -25x)`

⇒ `(63)/[10x^2 + 15x] = 12/[5x^2 - 25x]`

⇒ `(63)/{5x (2x + 3)} = 12/{5x(x - 5)}`

⇒ `(63)/(2x + 3) = 12/(x - 5)`

⇒ `63(x - 5) = 12(2x + 3)`

⇒ `63x - 315 = 24x + 36`

⇒ `63x - 24x = 315 + 36`

⇒ `39x = 351`

⇒ x = `351/39`

⇒ x = 9

Thus, the solution of the given equation is x = 9.

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Application of Properties of Equal Ratios
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Ratio and Proportion - Practice Set 4.3 [पृष्ठ ७०]

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बालभारती Algebra (Mathematics 1) [English] 9 Standard Maharashtra State Board
पाठ 4 Ratio and Proportion
Practice Set 4.3 | Q (4) (ii) | पृष्ठ ७०
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