The least common multiple for the denominator is
LCM=2(x-1)(x+1). Multiply each term by the LCM to get...
4(x-1)-(x-1)(x+1)=8
4x-4-x2+1=8......distribute and expand terms
-x2+4x+5=8......combine terms, rewrite equation
LCM=2(x-1)(x+1). Multiply each term by the LCM to get...
4(x-1)-(x-1)(x+1)=8
4x-4-x2+1=8......distribute and expand terms
-x2+4x+5=8......combine terms, rewrite equation
-x2+4x-3=0......subtract 8 both sides
(x-1)(x-3)=0......factor
x-1=0 -or- x-3=0.......let each term equal to zero, solve for "x"
x=1 x=3
Now from the problem statement equation, we can conclude that
the value "x=1" cannot be a solution because the first term would
then have zero in the denominator, division by zero is undefined.
So "x=3" is the single solution To the equation. Substitute back into the original equation and check for truth...
2/(3-2)-1/2=4/(32-1)
(2/2)-(1/2)=4/(9-1)
1-1/2=4/8
1/2=1/2..........√check, solution is correct. Always check
your solution, work