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

Solve: 12x2+18x+4218x2+12x+58=2x+33x+2 - Algebra

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प्रश्न

Solve:

`[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2]`

बेरीज

उत्तर

`[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2]`

If x = 0, then `[12 xx 0 + 18 xx  0 + 42]/[ 18 xx 0 + 12 xx 0 +58 ] = [ 2 xx 0+ 3]/[ 3 xx 0 + 2] ⇒ 42/58 = 3/2`, which is not true.

So, x = 0 is not a solution of the given equation.

Now,

`[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2] = [6x ( 2x + 3)]/[6x( 3x + 2)]` ....(by Theorem of equal ratios)

⇒ `[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2]  = [(12x^2 + 18x + 42) - 6x ( 2x + 3)]/[(18 x^2 + 12x +58) - 6x( 3x + 2)]` 

⇒ `[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2]  = (12x^2 + 18x + 42 - 12x^2 + 18x )/(18 x^2 + 12x +58 - 18x^2 - 12x )`

⇒ `[12x^2 + 18x + 42]/[ 18 x^2 + 12x +58 ] = [ 2x + 3]/[ 3x + 2] = 42/58`

`therefore [ 2x + 3]/[ 3x + 2] = 42/58`

⇒ `[ 2x + 3]/[ 3x + 2] = 21/29`

⇒ `58x +87 = 63x + 42`

⇒ `63x - 58x = 87 - 42`

⇒ `5x = 45`

⇒ `x = 9`

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

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Theorem on Equal Ratios
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Ratio and Proportion - Problem Set 4 [पृष्ठ ७९]

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बालभारती Algebra (Mathematics 1) [English] 9 Standard Maharashtra State Board
पाठ 4 Ratio and Proportion
Problem Set 4 | Q (11) | पृष्ठ ७९
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