# Simultaneous equation

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• Sep 30th 2011, 06:17 AM
andyboy179
Simultaneous equation
hi,

i need to know how to complete this Simultaneous equation:

x+y=2
x^2+2y^2=11

could you please explain it in detail so i fully understand please!

thanks!
• Sep 30th 2011, 06:36 AM
Plato
Re: Simultaneous equation
Quote:

Originally Posted by andyboy179
i need to know how to complete this Simultaneous equation:
x+y=2
x^2+2y^2=11
could you please explain it in detail so i fully understand please!

You have over 800 postings. So by now you should understand that we do not provide tutorial services. Please post some of your own work on this problem and then explain what you do not understand about the question.
Begin by solving for y in the first equation and substitute in the second.
• Sep 30th 2011, 06:40 AM
andyboy179
Re: Simultaneous equation
well thats the thing, im not sure how to start it. you said begin by solving y in the first equation, im not sure how to do that!
• Sep 30th 2011, 06:42 AM
Quacky
Re: Simultaneous equation
$\displaystyle x=2-y$
...or...
$\displaystyle y=2-x$
Substitute either (but not both!) into the second equation.

Editted due to my misreading of Plato's post.
• Sep 30th 2011, 06:47 AM
andyboy179
Re: Simultaneous equation
okay, would i then do:

(2-y)^2 +2y^2=11
= 4-3y=11
= 4-3y+2y^2=11
=-3y+2y^2=7
• Sep 30th 2011, 06:49 AM
Quacky
Re: Simultaneous equation
Quote:

Originally Posted by andyboy179
okay, would i then do:

(2-y)^2 +2y^2=11
= 4-3y=11
= 4-3y+2y^2=11
=-3y+2y^2=7

You're right up to:
$\displaystyle (2-y)^2 +2y^2=11$
But how do you expand $\displaystyle (2-y)^2$?
Remember that it can be rewritten $\displaystyle (2-y)(2-y)$

P.s. Be careful where you put equals signs.
• Sep 30th 2011, 06:54 AM
andyboy179
Re: Simultaneous equation
i did:
2 times 2=4
2 times -y=-2y
2 times -y=-2y
-y times-y= Y

then i added them to get 4-3y
• Sep 30th 2011, 06:55 AM
Quacky
Re: Simultaneous equation
Quote:

Originally Posted by andyboy179
i did:
2 times 2=4
2 times -y=-2y
2 times -y=-2y
-y times-y= y^2

Ok, it was just a small mistake then.
• Sep 30th 2011, 07:00 AM
andyboy179
Re: Simultaneous equation
oh yeah, damn! so continuing on i would do:

4-4y+y^2
= 4-4y+y^2+2y^2=11
= 4-4y+3y^2=11
• Sep 30th 2011, 07:07 AM
Quacky
Re: Simultaneous equation
Quote:

Originally Posted by andyboy179
oh yeah, damn! so continuing on i would do:

4-4y+y^2
= 4-4y+y^2+2y^2=11
= 4-4y+3y^2=11

Yeah, but again, be careful where you put equal signs. Lay it out like this:

$\displaystyle 4-4y+y^2+2y^2=11$

$\displaystyle 4-4y+3y^2=11$
• Sep 30th 2011, 07:09 AM
andyboy179
Re: Simultaneous equation
what you have just posted is what i have written down on my paper at home!

so i would then do:

=-4y+3y^2=7

but now i stuck after this!
• Sep 30th 2011, 07:12 AM
Quacky
Re: Simultaneous equation
Quote:

Originally Posted by andyboy179
what you have just posted is what i have written down on my paper at home!

so i would then do:

=-4y+3y^2=7

but now i stuck after this!

No, you would do:

$\displaystyle -4y+3y^2=7$ perhaps.

Even better, though would be to do:

$\displaystyle 3y^2-4y-7=0$
• Sep 30th 2011, 07:14 AM
andyboy179
Re: Simultaneous equation
i really can't do the brackets after the last part you put!

i know one bracket would have 3y and the other Y
• Sep 30th 2011, 07:20 AM
Quacky
Re: Simultaneous equation
Quote:

Originally Posted by andyboy179
i really can't do the brackets after the last part you put!

i know one bracket would have 3y and the other Y

You could always complete the square or use the formula in such a situation. Remember that not everything will factor.

We have $\displaystyle (3y\pm~)(y\mp~)$

7 is prime, the only factors are 7 and 1.
There are just 4 possibilities. Surely you can manage? You need to find a way of making -4.

Edit: Corrected.
• Sep 30th 2011, 07:23 AM
andyboy179
Re: Simultaneous equation
so would i use 7 and 1, for example would i do 7x-1= -7 (i know that wouldn't fit but is it the correct way of doing it?)
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