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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..

Ive uploaded the paper via PDF file please help me

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

Hi, I'll help you with the Task 5.

$\displaystyle x^2 - 2xy +4y = -60\\ x+y = 18 \\ y = 18-x$

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**)

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

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.

The quadratic formula is

$\displaystyle \frac{-b(\pm)\sqrt(b^2-4ac)}{2a}$

for $\displaystyle ax^2+bx+c =0 $

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

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.

The quadratic formula is

$\displaystyle \frac{-b(+-)\sqrt(b^2-4ac)}{2a}$

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 $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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