Question

Solve the following quadratic equations by factorization:
(3/x+1)-1/2=(2/3x-1),x≠-1,1/3

Answer 1

The values of x in the given quadratic equation is 1 and 3

Step-by-step explanation:

Given equation is $\frac{3}{x-1}-\frac{1}{2}=\frac{2}{3x-1}$ and $x\neq 1,\frac{1}{3}$

To solve the given equation by Factorization method :

• $\frac{3}{x-1}-\frac{1}{2}=\frac{2}{3x-1}$
• $\frac{3(2)-1(x+1)}{(x+1)(2)}=\frac{2}{3x-1}$
• $\frac{6-x-1}{x(2)+x(2)}=\frac{2}{3x-1}$  

• $\frac{5-x}{2x+2}=\frac{2}{3x-1}$
• $(5-x)(3x-1)=2(2x+2)$
• $5(3x)+5(-1)-x(3x)-x(-1)=2(2x)+2(2)$
• $15x-5-3x^2+x=4x+4$
• $16x-3x^2-5=4x+4$

• $16x-3x^2-5-(4x-4)=4x+4-(4x+4)$
• $16x-3x^2-5-4x-4=0$
• $12x-3x^2-9=0$
• $-3x^2+12x-9=0$
• $-3(x^2-4x+3)=0$

Dividing by 3 on both sides

• $\frac{-3(x^2-4x+3)}{3}=\frac{0}{3}$
• $x^2-4x+3=0$
• $(x-1)(x-3)=0$
• x-1=0 or x-3=0
• x=1 or x=3
• Therefore x=1,3

The values of x in the given quadratic equation is 1 and 3