1. ## 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!

2. ## Re: Simultaneous equation

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.

3. ## 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!

4. ## 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.

5. ## 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

6. ## Re: Simultaneous equation

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.

7. ## 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

8. ## Re: Simultaneous equation

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.

9. ## 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

10. ## Re: Simultaneous equation

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:

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

$\displaystyle 4-4y+3y^2=11$

11. ## 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!

12. ## Re: Simultaneous equation

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:

$\displaystyle 3y^2-4y-7=0$

13. ## 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

14. ## Re: Simultaneous equation

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.

15. ## 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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