The Student Room Group
Reply 1
ok
2x + 3y = 13 so 2x = 13 - 3y and so x=(13-3y)/2
put into equation
x^2 + y^2 = 78
((13-3y)/2)^2 + y^2 = 78
(169 + 9y^2 - 78y)/4 + y^2 = 78

multiply by 4:
169 + 9y^2 - 78y + 4y^2 = 312
13y^2 - 78y - 143 = 0
divide by 13:
y^2 - 6y- 11 = 0

use quadratic equation .... i'll put that up in a sec
Reply 2
blah
Reply 3
y = [ -b +or- sqr rt (b^2 -4ac) ] / 2a

using y^2 - 6y - 11 = 0
a = 1
b= -6
c= -11

so y = [ 6 +or- sqr rt (36 - 4x1x-11) ] / 2x1
= [ 6 +or- sqr rt (80) ] / 2
y = either (6 + sqr rt 80) /2 = 3 + (2 root5)
or (6 - sqr rt 80) /2 = 3 - (2 root5)

then stick them into original equation to find x
Reply 4
blah to u too
slop8
Solve The Simultaneous Equations: 2x + 3y = 13
x^2+y^2=78

2x + 3y = 13
---> x = (13-3y)/2

Sub into other eq:
---> [(13-3y)/2]^2 + y^2 = 78
---> [(13-3y)(13-3y)]/4 + y^2 = 78
---> (13 - 3y)(13 - 3y) + 4y^2 - 312 = 0
---> 9y^2 - 78y + 169 + 4y^2 - 312 = 0
---> 13y^2 - 78y - 143 = 0
---> y^2 - 6y - 11 = 0
---> y = {6(+/-)Sqrt[36 - 4(-11)]}/2
= {6(+/-)Sqrt[80]}/2
= {6(+/-)4Sqrt5}/2
= 3(+/-)2Sqrt5

When y = 3 + 2Sqrt5:
---> 2x + 3(3 + 2Sqrt5) = 13
---> 2x + 9 + 6Sqrt5 = 13
---> 2x = 4 - 6Sqrt5
---> x = 2 - 3Sqrt5

When y = (3 - 2Sqrt5):
---> 2x + 3(3 - 2Sqrt5) = 13
---> 2x + 9 - 6Sqrt5 = 13
---> 2x = 4 + 6Sqrt5
---> x = 2 + 3Sqrt5

Hence the Solutions are:
y = (3 + 2Sqrt5) and x = (2 - 3Sqrt5) OR:
y = (3 - 2Sqrt5) and x = (2 + 3Sqrt5)
Reply 6
so I reckon the answers are:
when y=3 + 2 root5
x=2 - 3 root5

and when y= 3- 2 root 5
x = 2 + 3 root5

better get someone to check that ;p
Reply 7
fight, fight, fight!


hehe woooooooooo ! :biggrin:
Reply 8
real thanks to franks and nimia!!
Reply 9
hehe/.. care to step outside?

lol joking. I did all those trivial manipulation but missed one single god damn minus sign and got the whole thing wrong.
Reply 10
I did all those trivial manipulation but missed one single god damn minus sign and got the whole thing wrong.

looooser !!! :biggrin: