Question

solve the quadratic equations:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}$
$where \: \: x ≠ - 1 \: \: or\: \: \: \frac{1}{3}$
​

Answer 1

EXPLANATION.

⇒ 3/(x + 1) - 1/2 = 2/(3x - 1).

As we know that,

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

⇒ 6 - x - 1/(2x + 2) = 2/(3x - 1).

⇒ (5 - x)/(2x + 2) = 2/(3x - 1).

⇒ (5 - x)(3x - 1) = 2(2x + 2).

⇒ 15x - 5 - 3x² + x = 4x + 4.

⇒ 16x - 5 - 3x² = 4x + 4.

⇒ 4x + 4 - 16x + 5 + 3x² = 0.

⇒ 3x² - 12x + 9 = 0.

⇒ 3(x² - 4x + 3) = 0.

⇒ x² - 4x + 3 = 0.

Factorizes the equation into middle term splits, we get.

⇒ x² - 3x - x + 3 = 0.

⇒ x(x - 3) - 1(x - 3) = 0.

⇒ (x - 1)(x - 3) = 0.

⇒ x = 1  and  x = 3.

Answer 2

Given :-

$\sf\dfrac{3}{x+1} - \dfrac{1}{2} = \dfrac{2}{3x-1}$

To Find :-

Quardatic equation

Solution :-

$\sf\dfrac{3}{x+1}-\dfrac{1}{2} = \dfrac{2}{3x-1}$

$\sf \dfrac{3}{x+1}-\dfrac{2}{3x-1}=\dfrac{1}{2}$

$\sf\dfrac{3(3x-1)-2(x+1)}{(x+1)(3x-1)}=\dfrac{1}{2}$

$\sf\dfrac{(9x-3)-(2x+2)}{(x+1)(3x-2)}=\dfrac{1}{2}$

$\sf\dfrac{7x-5}{3x^{2} +2x-1}=\dfrac{1}{2}$

$\sf 2(7x-5)=3x^{2} +2x-1$

$\sf 14x - 10=3x^{2} +2x-1$

$\sf 3x^{2} + 2x-14x+10-9=0$

$\sf 3x^{2} -12x+9=0$

$\sf 3x^{2} -9x-3x+9=0$

$\sf 3x(x-3)-3(x-3)=0$

$\sf (x-1)(x-3)=0$

Either

x - 1 = 0

x = 0 - 1

x = -1

or

x - 3 = 0

x = 0 - 3

x = -3