You can
put this solution on YOUR website!Perform the indicated operations and simplify.
[x^2/3x^2-5x-2]-[2x/3x+1]∙[1/(x-2)]
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Factor where you can:
[x^2/(x-2)(3x+1]-[2x/3x+1]∙[1/(x-2)]
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lcd = (x-2)(3x+1)
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Write each fraction with the lcd as its denominator:
[x^2/lcd] - [2x/lcd]
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Combine the numerators over the lcd:
(x^2-2x]/[(x-2)(3x+1)]
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(x(x-2))/[(x-2)(3x+1)]
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x/(3x+1)
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Cheers,
Stan H.
You can
put this solution on YOUR website!(Note: A solution provided by another tutor has some very serious errors in it. The solution is not correct. The first part was correct. You do want to factor the denominators, use the factored denominators to find the lcd and you do want to write each fraction with the lcd for the denominator. But after that it goes very wrong. This is where I will start.)
lcd:
To avoid some common errors with subtractions, I am going to rewrite the expression as additions:
The first fraction already has the lcd as a denominator. The denominator of the second fraction is missing (x-2) so we will multiply its numerator and denominator by (x-2). The denominator of the third fraction is missing (3x+1) so we will multiply its numerator and denominator by (3x+1):
Multiply out the numerators but leave the denominators factored for now:
Now that the denominators are the same we can add:
At this point we should try to reduce the fraction. To reduce the fraction we have to factor the numerator and denominator. (We didn't multiply out the denominator earlier so that we would not have to factor it again now.) The only factoring that can be done on the numerator, however, is to factor out a -1:
As we can see, there are no common factors to cancel. So all we can do is simplify (i.e. multiply out):
(