SOLUTION: 25: Solve for the variable in each equation below.
[x / (x + 3)] + [1 / (x - 1)] = [4 / (x2 + 2x - 3)]
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Question 80588: 25: Solve for the variable in each equation below.
[x / (x + 3)] + [1 / (x - 1)] = [4 / (x2 + 2x - 3)]
Answer by tutorcecilia(2152) (Show Source): You can put this solution on YOUR website!
[x / (x + 3)] + [1 / (x - 1)] = [4 / (x2 + 2x - 3)]
(x / (x + 3)] + [1 / (x - 1)] = [4 / (x+3)(x-1)] [factor as much as possible and wherever possible]
.
(x / (x + 3)] + [1 / (x - 1)] = [4 / (x+3)(x-1)][find the GCF: (x+3)(x-1) ]
((x+3)(x-1)(x / (x + 3)] + [( (x+3)(x-1) (1 / (x - 1)] = [( (x+3)(x-1) (4 / (x+3)(x-1))] [multiply each term by the GCF and cancel wherever possible]
.
x(x-1)+1(x+3)=4
.
=4 [simplify]
=0 [set the equation equal to zero by subtracting (4) from both sides]
.
=0 [combine like-terms]
=0 [combine like-terms]
(x+1)(x-1)=0 [try to factor or simplify as much as possible]
.
x+1=0 [set each factor equal to zero and solve for the x-term
x=-1
and x-1=0
x=1
.
Check each value of x by plugging them back into the original equation.
[x / (x + 3)] + [1 / (x - 1)] = [4 / (x2 + 2x - 3)]
Let x=-1
[(-1) / (-1 + 3)] + [1 / (-1 - 1)] = [4 / ((-1)^2 + 2(-1) - 3)]
(-1/2)+(1/-2)=(4/-4)
(-2/2)=(-4/4)
-1=-1 [checks out]
.
[x / (x + 3)] + [1 / (x - 1)] = [4 / (x2 + 2x - 3)]
Let x=1
[1 / (1 + 3)] + [1 / (1 - 1)] = [4 / ((1)^2 + 2(1) - 3)]
(1/4)+(1/0)=(4/0)
This (x=1)will not work because zero cannot be in the denumerator.
So, the only answer is x=-1
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