Long Simultaneous Equation (1 Viewer)

bored.of.u · Feb 10, 2009

ok guys i have this long simultaneous equation which i cannot get. the answer is probably simple...

p^2 y - a = p^3 x + 2ap^2
x + py = ap^3 + 2ap

thanx in advance =D

bored.of.u · Feb 10, 2009

Exphate said:
Yeah. Maybe try without the random smilies?

haha sory i dunno why when i copy and paste it from word it does that

bored.of.u · Feb 10, 2009

Exphate said:
Post the word doc then

i edited it above

tommykins · Feb 10, 2009

bored.of.u said:
ok guys i have this long simultaneous equation which i cannot get. the answer is probably simple...

p^2 y - a = p^3 x + 2ap^2
x + py = ap^3 + 2ap

thanx in advance =D

Theres 4 variables but only 2 equations, unless p and a are constants?

bored.of.u · Feb 10, 2009

tommykins said:
Theres 4 variables but only 2 equations, unless p and a are constants?

haha sorry yeh i forget to mention than p and a are constants

Trebla · Feb 10, 2009

$$To find $y$ multiply second equation by $p^3\\(1)\,\, p^2y-a=p^3x+2ap^2\\(2)\,\, p^4y-ap^6=-p^3x+2ap^4\\(1)+(2):\\ p^2y(1+p^2)-a(1+p^6)=2ap^2(1+p^2)\\y=\dfrac{2ap^2(1+p^2)+a(1+p^6)}{p^2(1+p^2)}\\$To find $x$ multiply second equation by $p$ rather than sub $y$\\(1)\,\, p^2y-a=p^3x+2ap^2\\(3)\,\, p^2y-ap^4=-px+2ap^2\\(1)-(3):\\ -a+ap^4 =px(1+p^2)\\x=\dfrac{ap^4-a}{p(1+p^2)}\\=\dfrac{a(p^2-1)}{p}$

Check the working, might've made a mistake...

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Long Simultaneous Equation (1 Viewer)

bored.of.u

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bored.of.u

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bored.of.u

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tommykins

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