solving a system of nonlinear equations

**vbarnett326** · November 15th 2008, 07:26 PM

14x^2-3xy+9y^2=71
2x+3y=9

I solve for x in the second equation and get x=-3y+9 /2 but when i plug it back in i am not sure how to square (-3y+9)/2 in the first equation?

**mr fantastic** · November 15th 2008, 08:19 PM

Originally Posted by vbarnett326

14x^2-3xy+9y^2=71
2x+3y=9

I solve for x in the second equation and get x=-3y+9 /2 but when i plug it back in i am not sure how to square (-3y+9)/2 in the first equation?

$14 \cdot \left( \frac{9 - 3y}{2}\right)^2 = 14 \cdot \left(\frac{3(3 - y)}{2}\right)^2 = 14 \cdot \frac{9 (3 - y)^2}{4} = \frac{63}{2} \cdot (9 - 6y + y^2)$

**vbarnett326** · November 16th 2008, 11:16 AM

i don't understand how you got 63/2.

and also -3((-3y+9)/2))y equals (9y^2 -27y)/-6y?????? is that correct

**mr fantastic** · November 16th 2008, 11:39 AM

Originally Posted by vbarnett326

i don't understand how you got 63/2. Mr F says: Did you multiply the numbers? ${\color{red}\frac{14 \cdot 9}{4} = \frac{63}{2}}$ . You're meant to do more than just look at the solution - you're meant to mess in and do it yourself, using what I posted as a guide.

and also -3((-3y+9)/2))y equals (9y^2 -27y)/-6y?????? is that correct

Mr F says: Certainly not! Why are you multiplying the bottom by -3 and y as well?? Do you honestly think that something like ${\color{red}2 \cdot \frac{4}{3} = \frac{2 \cdot 4}{2 \cdot 3}}$ ? The answer is (9y^2 -27y)/2

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