Solve each of the following quadratic equations:
3x+1−12=23x−1,x≠−1,13
3x+1−12=23x−1
3x+1−23x−1=12
[3(3x−1)−2(x+1)](x+1)(3x−1)=12
(9x−3−2x−2)(3x2−x+3x−1)=12
2(7x−5)=(3x2+2x−1)
3x2+2x−1−14x+10=0
3x2−12x+9=0
x2−4x+3=0
x2−3x−x+3=0
x(x−3)−1(x−3)=0
(x−3)(x−1)=0
Either
x−3=0
x=3
or
x−1=0
x=1
So x = 1, 3