# Hi not sure if this right area but really need help with this

• Jan 17th 2013, 08:28 AM
gary147
Hi not sure if this right area but really need help with this
ive not done maths for many years, been given 2 papers to learn. ive been trying for ages to work these out but reallt have no idea,
can someone help me with this 1 and then i will try second on own..

• Jan 17th 2013, 08:44 AM
Barioth
Re: Hi not sure if this right area but really need help with this

$\displaystyle x^2 - 2xy +4y = -60\\ x+y = 18 \\ y = 18-x$
from there on we'll replace every y with (18-x)
$\displaystyle x^2-2x(18-x)+4(18-x)=-60\\x^2 -2x*18-2x*(-x)+4*18+4*(-x)\\ = x^2 -36x +2x^2+72-4x\\=3x^2-40x+72 = -60$

Then
$\displaystyle 3x^2-40x+132 = 0$

From there on you can use the quadratic formula

end up with $\displaystyle x = 6 \\ x=\frac{22}{3}$

as for the graph part, Draw the 2 fonction they'll meet in these value of X. (*Unless I messed up*)
• Jan 17th 2013, 08:48 AM
gary147
Re: Hi not sure if this right area but really need help with this
thank you so much for your time and trying to help.. could you possible explain each step and how you do it.. that way i can learn and possible do the other paper myself
thanks again
• Jan 17th 2013, 11:00 AM
Barioth
Re: Hi not sure if this right area but really need help with this
mhm after replacing y, I just added everything together. then I used the quadratic formula.

$\displaystyle \frac{-b(\pm)\sqrt(b^2-4ac)}{2a}$
for $\displaystyle ax^2+bx+c =0$

Maybe you could tell me witch part you didn't understand?
• Jan 17th 2013, 11:17 AM
Plato
Re: Hi not sure if this right area but really need help with this
Quote:

Originally Posted by Barioth
mhm after replacing y, I just added everything together. then I used the quadratic formula.

$\displaystyle \frac{-b(+-)\sqrt(b^2-4ac)}{2a}$
for $\displaystyle ax^2+bx+c =0$

Maybe you could tell me witch part you didn't understand?

[TEX]\pm[/TEX] gives $\displaystyle \pm$.

Also [TEX] \frac{-b\pm)\sqrt{b^2-4ac}}{2a}[/TEX] gives $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
• Jan 19th 2013, 07:37 PM
JacobOCD
Re: Hi not sure if this right area but really need help with this
BTW, Barioth, for further reference: LaTeX/Mathematics - Wikibooks, open books for an open world

:)