Maharashtra State BoardSSC (English Medium) 9th Standard
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Solve (I) 16 X 2 − 20 X + 9 8 X 2 + 12 X + 21 = 4 X − 5 2 X + 3 - Algebra

Sum

Solve 
`[ 16x^2 - 20x +9]/[ 8x^2 + 12x + 21] = ( 4x - 5 )/( 2x + 3)`

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Solution

`[ 16x^2 - 20x +9]/[ 8x^2 + 12x + 21] = ( 4x - 5 )/( 2x + 3)`

If x = 0, then `[ 16 xx 0 - 20 xx 0 + 9]/[ 8 xx 0 + 21 xx 0 + 21] =  [ 4 xx 0 - 5]/[ 2 xx 0 + 3] ⇒ 9/21 = -5/3`, which is not true.

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

⇒ `[ 16x^2 - 20x + 9]/[ 8x^2 + 12x +21] = [ 4x - 5]/[2x + 3] = [(16x^2 - 20x + 9) -4x( 4x - 5)]/[(8x^2 + 12x +21) - 4x( 2x + 3 )`                                                     .....( Theorem of equal ratios)

⇒ `[ 16x^2 - 20x + 9]/[ 8x^2 + 12x +21] = [ 4x - 5]/[2x + 3] = [16x^2 - 20x + 9- 16x^2 - 20x]/[8x^2 + 12x +21 - 8x^2 - 12x]`

⇒ `[ 16x^2 - 20x + 9]/[ 8x^2 + 12x +21] = [ 4x - 5]/[2x + 3] = 9/ 21`

`therefore { 4x - 5}/{ 2x + 3} = 9/21`

⇒ `{ 4x - 5}/{ 2x + 3} = 3/7`
⇒ `28x - 35 = 6x + 9`
⇒ `28x - 6x = 35 + 9`
⇒ `22x = 44`
⇒ `x = 2`

Thus, the solution of the given equation is x = 2.

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
Practice Set 4.4 | Q 4.1 | Page 74
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