# Any chance of some help? please:(

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• April 16th 2008, 12:41 PM
nialldhee
Any chance of some help? please:(
been stuck on this for ages...

x²+4y²=10
2y+3x=10

cant get the answer... can any1 do it please... its my gcse pratice paper :(
• April 16th 2008, 12:46 PM
o_O
Looking at the first question, note the blue:
$x^{2} + 4y^{2} = 10$
$x^{2} + {\color{blue} (2y)^{2}} = 10$

From your second equation:
${\color{blue}2y} + 3x = 10$
${\color{blue}2y} = 10 - 3x$

So plug 2y from the second equation to your first one:
$x^{2} + \left(10 - 3x\right)^{2} = 10$

Expand, move everything to one side, and solve the quadratic ;)
• April 16th 2008, 12:55 PM
Moo
Hello o_O...

Unfortunately, it's x²+4y², not x²-4y² :)

If we continue it, we finally get :

$x^{2} {\color{red} +} \left(10 - 3x\right)^{2} = 10$
• April 16th 2008, 12:57 PM
o_O
Merci, mon amie française ;)
• April 16th 2008, 12:58 PM
Moo
You're welcome dear :)