SOLUTION: been researching this problem and i believe it is an inconsistent equation after I have factored all the denominators and try to multiply through to get rid of the fractions and c

Algebra ->  Linear-equations -> SOLUTION: been researching this problem and i believe it is an inconsistent equation after I have factored all the denominators and try to multiply through to get rid of the fractions and c      Log On


   

Question 93381: been researching this problem and i believe it is an inconsistent equation
after I have factored all the denominators and try to multiply through to get rid of the fractions and continue working the problem get stuck.All the other problems I have done before have the same combinations once factored not the case here.
4/(x^2+3x-10)-1/(x^2+x-6)=3/(x^2-x-12)
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
From your answer I gathered that you understand the methods that need to be applied to solve
the problem, but you are probably making a minor math mistake somewhere that is leading you astray.
.
There is a solution for x and here's how you go about getting it.
.
Start with the given problem:
.
4%2F%28x%5E2%2B3x-10%29+-+1%2F%28x%5E2%2Bx-6%29+=+3%2F%28x%5E2-x-12%29
.
Factor each of the three denominators. When you do you get:
.

.
Then identify each of the individual factors. The list is:
.
x-2
x%2B5
x%2B3
x-4
.
When you multiply these together, you get a common denominator that can be used for each
term. As an abbreviation, use CD for this common denominator.
.
To put a term over this common denominator, you must multiply its numerator by those factors
in the common denominator that were not in its original denominator. Putting the terms over
the common denominator results in:
.

.
Now you can get rid of the CD by multiplying both sides of this equation (all terms) by
the CD. When you do that the CD multiplier cancels with the CD in the denominator
and you are left with just the numerators of the terms in the problem. That is, you are
left with:
.
4%28x%2B3%29%28x-4%29+-+%28x%2B5%29%28x-4%29+=+3%28x-2%29%28x%2B5%29
.
In each of the three groupings, multiply the two sets of parentheses to get:
.
4%28x%5E2+-x+-12%29-+%28x%5E2%2Bx-20%29=+3%28x%5E2%2B3x-10%29
.
Now multiply by the constants in front of the parentheses (note that in the second term
on the left side the constant is -1). When you do those multiplications you get:
.
4x%5E2+-4x+-48+-x%5E2+-+x+%2B20+=+3x%5E2+%2B9x+-30
.
Combine the like terms on each side of the equation. On the left side you have 4x%5E2 and
-x%5E2 which combine to 3x%5E2. You also have -4x and -x which combine
to -5x. And finally you have -48 and 20 which combine to -28.
So the left
side is reduced to 3x%5E2-5x-28 and the equation becomes:
.
3x%5E2+-5x-28+=+3x%5E2+%2B+9x+-+30
.
If you subtract 3x%5E2 from both sides, the 3x%5E2 terms on the left and the right
side are eliminated and you are left with:
.
-5x-28+=+9x-30
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Subtract 9x from both sides and the equation becomes:
.
-14x-28+=+-30
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Add +28 to both sides and you get:
.
-14x+=+-2
.
Finally solve for x by dividing both sides by -14 and you have:
.
x+=+%28-2%29%2F%28-14%29+=+1%2F7
.
And that's the answer ... x+=+1%2F7
.
Hope this helps you to find your math error.