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

Given:

Given quadratic equation is $3(\frac{7x+1}{5x-3})-4(\frac{5x-3}{7x+1})=11, x≠\frac{3}{5}, \frac{-1}{7}$.

To do:

We have to solve the given quadratic equation.

Solution:

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

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

$\frac{3(49x^2+7x+7x+1)-4(25x^2-15x-15x+9)}{35x^2-21x+5x-3}=11$

$\frac{147x^2+42x+3-100x^2+120x-36}{35x^2-16x-3}=11$

$47x^2+162x-33=11(35x^2-16x-3)$

$47x^2+162x-33=385x^2-176x-33$

$(385-47)x^2+(-176-162)x-33+33=0$

$338x^2-338x=0$

$338(x^2-x)=0$

$x^2-x=0$

$x(x-1)=0$

$x=0$ or $x-1=0$

$x=0$ or $x=1$

The values of $x$ are $0$ and $1$.

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Updated on: 10-Oct-2022

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