Question

Solve the following quadratic equation; 3(7x+1/5x-3) -4 (5x-3/7x+1)= 11

Answer 1

Given:

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

On Putting 7x+1/5x-3 = y,
3y - 4/y = 11
(3y × y -4 ) / y  = 11
3y² - 4 = 11y
3y² - 11 y - 4 = 0
3y² -12y + y - 4 = 0
3y ( y - 4) +1(y - 4)= 0
(3y +1) (y - 4)= 0

(3y +1) = 0  or  (y - 4)= 0
y = -⅓   or  y = 4

Case 1. y = -⅓
7x+1/5x-3 = - ⅓
3(7x+1) = -1(5x-3)
21x +3 = -5x +3
21x +5x = 3 -3
26x = 0
x = 0/26

x = 0

Case 2. y = 4
7x+1/5x-3 = 4
7x+1 = 4(5x-3 )
7x +1 = 20x -12
7x - 20x = -12-1
-13x = -13

x = 13/13
x = 1

Hence, the value of x=  0 or x = 1.

HOPE THIS WILL HELP YOU...

Answer 2

This will help you

It is easy to understand