SOLUTION: Would you please verify that I have correctly done the below problem?
3a/a+4 - 2a^2 - 16/a^2 -16 = Below is how I did the problem:
3a/a + 4 + 2a^2 - 16a/(a-4)(a-4) = 3a(a-4)
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-> SOLUTION: Would you please verify that I have correctly done the below problem?
3a/a+4 - 2a^2 - 16/a^2 -16 = Below is how I did the problem:
3a/a + 4 + 2a^2 - 16a/(a-4)(a-4) = 3a(a-4)
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Question 1169222: Would you please verify that I have correctly done the below problem?
3a/a+4 - 2a^2 - 16/a^2 -16 = Below is how I did the problem:
3a/a + 4 + 2a^2 - 16a/(a-4)(a-4) = 3a(a-4) + (2a^2 - 16a)/ (a-4)(a+4) = 3A^2 - 12A+ 2A^2 -16/ (A-4)(A+4) = ANSWER 5A^2 - 28A / (A-40(A+4) or 5a^2 -28a/a^2 -16
Thank you Found 2 solutions by Edwin McCravy, MathLover1:Answer by Edwin McCravy(20054) (Show Source):
You should follow the rules of PEMDAS. I'll try to figure out what you
intended although you ignored the use of parentheses around entire
numerators and entire denominators, which PEMDAS requires.
There is a discrepancy in the last term on the second fraction. On some
steps you have -16 and on others you have -16a.
I think it should be -16a.
Your error in the previous step was failing to change
the sign when removing parentheses preceded by a minus.
But you didn't even have a parentheses there to do that
to, which vitiated your answer. Maybe you just wrote a +
when it should have been a -.
Edwin
