SOLUTION: 3x+y=16 y=16-3x x(16-3x)=16x-3x^2 x^2 +(16x-3x^2)= x^2+16x+9x^2 10x^2 +256 I get 10x^2-144x+256........the answer is 10x^2-96x+256 How do they get -96x?? This is

Question 267713: 3x+y=16
y=16-3x
x(16-3x)=16x-3x^2
x^2 +(16x-3x^2)= x^2+16x+9x^2
10x^2 +256
I get 10x^2-144x+256........the answer is 10x^2-96x+256
How do they get -96x??
This is a minimization/maximization of quadratic functions problem
Thanks for any/all help!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
3x+y=16
y=16-3x
---------- Are these the 2 equations? It's the same one twice.
There's no point in working further.
--> dependent, infinite # of solutions.
------------------
x(16-3x)=16x-3x^2 ?? Where did this come from?
x^2 +(16x-3x^2)= x^2+16x+9x^2
10x^2 +256
I get 10x^2-144x+256........the answer is 10x^2-96x+256
How do they get -96x??
-------------------
Restate the problem, if it's different than that above.

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