Question

Solve the following quadratic equations by factorization: $3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11$; $x\neq \frac{3}{5}$, $\frac{-1}{7}$

Answer 1

SOLUTION :  

Given : 3(7x + 1/ 5x - 3)  - 4(5x - 3/7x + 1) = 11

3[(7x + 1)² - 4(5x - 3)²] / (7x + 1)(5x - 3) = 11

[ By taking LCM]

[3((7x)² + 1² + 2×7x×1 ) - 4((5x)² + 3² - 2× 5x ×3] / (35x² -21x +5x -3) = 11

[3(49x² + 1 + 14x ) - 4 (25x² +9 - 30x )] / (35x² - 16x - 3) = 11

[147x² + 3 + 42x - 100x² - 36 + 120x] / (35x² - 16x - 3) = 11

(47x² +162x -33) / (35x² - 16x - 3) = 11

(47x² +162x -33) = 11(35x² - 16x - 3)  

(47x² +162x -33) = (385x² - 176x - 33)  

385x² - 47x² - 176x - 162x - 33 + 33 = 0

338x² - 338x = 0

338x(x - 1) = 0

338x = 0  or  x - 1 = 0

x = 0/338  or  x = 1

x = 0  or  x = 1  

Hence, the roots of the quadratic equation 3(7x + 1/ 5x - 3)  - 4(5x - 3/7x + 1) = 11 are 0 & 1 .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Solution :

Given,

$3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11$; $x\neq \frac{3}{5}$, $\frac{-1}{7}$

Let a = ( 7x+1 )/( 5x-3 ) ----( 1 )

=> 3a - 4/a = 11

=> ( 3a² - 4 )/a = 11

=> 3a² - 4 = 11a

=> 3a² - 11a - 4 = 0

Splitting the middle term , we get

=> 3a² - 12a + 1a - 4 = 0

=> 3a( a - 4 ) + 1( a - 4 ) = 0

=> ( a - 4 )( 3a + 1 ) = 0

=> a - 4 = 0 or 3a + 1 = 0

=> a = 4 or 3a = -1

i ) a = 4

=>(7x + 1)/(5x - 3)=4 [from ( 1 )]

=> 7x + 1 = 4( 5x - 3 )

=> 7x + 1 = 20x - 12

=> 1 + 12 = 20x - 7x

=> 13 = 13x

=> 13/13 = x

=> x = 1

ii ) 3a = -1

=> 3( 7x + 1 )/( 5x - 3 ) = -1

[ from ( 1 ) ]

=> 21x + 3 = - ( 5x - 3 )

=> 21x + 3 = -5x + 3

=> 21x + 5x = 3 - 3

=> 26x = 0

=> x = 0

Therefore ,

x = 1 or x = 0

•••••