SOLUTION: How would I solve for 3x^2+10x-8/3x^2+x-2? I got 4, is that correct.

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Question 946600: How would I solve for 3x^2+10x-8/3x^2+x-2? I got 4, is that correct.
Found 2 solutions by richard1234, MathLover1:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
You cannot "solve" 3x^2+10x-8/3x^2+x-2. You can find other things, e.g. asymptotes, or domain (set of all possible x-values for which the expression is defined), but the word "solve" in this case makes no sense.

Post your question exactly as written, with all information.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
%283x%5E2%2B10x-8%29%2F%283x%5E2%2Bx-2%29 ....factor completely; write 10x in numerator as 12x-2x, and x in denominator as 3x-2x

%283x%5E2%2B12x-2x-8%29%2F%283x%5E2%2B3x-2x-2%29 ...group

%28%283x%5E2%2B12x%29-%282x%2B8%29%29%2F%28%283x%5E2%2B3x%29-%282x%2B2%29%29...factor out 3x

%283x%28x%2B4%29-2%28x%2B4%29%29%2F%283x%28x%2B1%29-2%28x%2B1%29%29

%28%283x-2%29+%28x%2B4%29%29%2F%28%283x-2%29+%28x%2B1%29%29
since denominator cannot be equal to zero, we will have %283x-2%29+%28x%2B1%29=0 if x=2%2F3 or x=-1
since we can cancel %283x-2%29) like
%28cross%28%283x-2%29%29+%28x%2B4%29%29%2F%28cross%28%283x-2%29%29+%28x%2B1%29%29

%28x%2B4%29%2F%28x%2B1%29 then x cannot be equal to -1


since numerator is %28x%2B4%29 and it will be zero if x=-4, so
solution is:
x+=+-4