Maharashtra State BoardSSC (English Medium) 9th Standard
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Solve : 12 X 2 + 18 X + 42 18 X 2 + 12 X + 58 = 2 X + 3 3 X + 2 - Algebra

Sum

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

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Solution

`[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.

Concept: Theorem on Equal Ratios
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APPEARS IN

Balbharati Mathematics 1 Algebra 9th Standard Maharashtra State Board
Chapter 4 Ratio and Proportion
Problem Set 4 | Q 11 | Page 79
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