Question

Solve the following quadratic equations by factorization: $3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5$; $x\neq \frac{3}{5}$, $\frac{-3}{2}$

Answer 1

SOLUTION :  

Given : 3(3x - 1/ 2x + 3)  - 2(2x + 3 /3x - 1) = 5

3[(3x - 1)² - 2(2x + 3)²] / (3x - 1)(2x + 3) = 5

[ By taking LCM]

[3((3x)² + 1² - 2×3x×1 ) - 2((2x)² + 3² + 2× 2x ×3] / (6x² - 2x + 9x - 3) = 5

[3(9x² + 1 - 6x ) - 2 (4x² + 9 + 12x )] / (6x² + 7x - 3) = 5

[27x² + 3 - 18x - 8x² - 18 - 24x] / (6x² + 7x - 3) = 5

(19x² - 42x - 15) /  (6x² + 7x - 3) = 5

(19x² - 42x - 15) = 5(6x² + 7x - 3)

(19x² - 42x - 15) = (30x² + 35x - 15)  

30x² - 19x² + 35x + 42x - 15 + 15 = 0

11x² + 77x = 0

11x(x + 7) = 0

11x = 0  or  x + 7 = 0

x = 0/11  or  x = - 7

x = 0  or  x = - 7  

Hence, the roots of the quadratic equation 3(3x - 1/ 2x + 3)  - 2(2x + 3 /3x - 1)= 5 are 0 & - 7 .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Solution :

Given $3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5$; $x\neq \frac{3}{5}$, $\frac{-3}{2}$

Let ( 3x-1 )/( 2x + 3 ) = a ----( 1 )

Now ,

3a - 2/a = 5

=> ( 3a² - 2 )/a = 5

=> 3a² - 2 = 5a

=> 3a² - 5a - 2 = 0

Splitting the middle term , we get

=> 3a² - 6a + 1a - 2 = 0

=> 3a( a - 2 ) + 1( a - 2 ) = 0

=> ( a - 2 )( 3a + 1 ) = 0

Now ,

a - 2 = 0 or 3a + 1 = 0

=> a = 2 or 3a = -1

i ) Take a = 2

From ( 1 ) ,

( 3x - 1 )/( 2x + 3 ) = 2

=> 3x - 1 = 2( 2x + 3 )

=> 3x - 1 = 4x + 6

=> -1 - 6 = 4x - 3x

=> -7 = x

=> x = -7

ii ) Take 3a = -1

=> 3( 3x - 1 )/( 2x + 3 ) = -1

=> 9x - 3 = -1( 2x + 3 )

=> 9x - 3 = -2x - 3

=> 9x + 2x = -3 + 3

=> 11x = 0

=> x = 0

Therefore ,

x = -7 or x = 0

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