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

1. ## 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..

2. ## Re: Hi not sure if this right area but really need help with this

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

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

From there on you can use the quadratic formula

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

as for the graph part, Draw the 2 fonction they'll meet in these value of X. (*Unless I messed up*)

3. ## 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

4. ## 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.

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

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

5. ## Re: Hi not sure if this right area but really need help with this

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

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

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

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

Also [TEX] \frac{-b\pm)\sqrt{b^2-4ac}}{2a}[/TEX] gives $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$