# 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]

#### 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