Simultaneous Equation

[TheRedAndGreenBeast]

Doing some revision for GCSE maths and this question came up. I've spent a while trying to figure it out but with little success. I've tried x = ±3 y = 0 but then that would not be true for x + y = 2.

x^2 + y^2 = 9
x + y = 2

x = ? or ?
y = ? or ?

Give answer to 2 decimal places. Please show working out so I will learn how to do it.

If you can give me the answer I'll be your best friend! =D

 

[TheRedAndGreenBeast]

Answer is 13/4 and -5/4

How did you get that?

 

[Shinichi Izumi]

x+y=2, rearrange to x=2-y, substitute this x to the other formula x^2+y^2=9.

(2-y)^2+y^2=9

Solve from there

 

[TheRedAndGreenBeast]

I think I may have worked it out but I'm not sure if I'm correct or not; can somebody check this out?

x + y = 2
x = 2 - y
(2 - y)^(2) + y^2 = 9
4 - 2y + 2y + y^2 + y^2 = 9
2y^2 = 9
y^2 = 4.5
y = ±2.12 (to 2 decimal places)

4.5 is half of 9 so x^2 must be 4.5 as well meaning x = 2.12 as well (if this is correct).

EDIT: Actually, if x + y = 2 then the answer can't be 2.12...

 

[TheRedAndGreenBeast]

x+y=2, rearrange to x=2-y, substitute this x to the other formula x^2+y^2=9.

(2-y)^2+y^2=9

Solve from there

Thanks, that reassured me and supported my idea. Though that only gives me one answer when the question implies there are two possible answers.

 

[Shinichi Izumi]

Your expansion is wrong.

It should be 4-2y-2y+y^2+y^2=9

= 2y^2-4y-5=0

x should equal 2.87 and -0.87

 

[TheRedAndGreenBeast]

Your expansion is wrong.

It should be 4-2y-2y+y^2+y^2=9

= 2y^2-4y-5=0

Ah yes, thanks for pointing that out, my bad.

 

[Flakez]

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Use quadratic formula

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Remember the +/- sign should be -/+ but the symbol doesn't exist here.

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[TheRedAndGreenBeast]

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Use quadratic formula

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Remember the +/- sign should be -/+ but the symbol doesn't exist here.

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Forgot to use this formula lol thanks for reminding me.

 

[Shinichi Izumi]

Yeh ignore my first answer of 13/4 and -5/4, wolfram alpha failed me..

It's 2.87 and -.87

 

[Flakez]

Forgot to use this formula lol thanks for reminding me.

Glad I helped. And thanks for the rep it was just enough to give me another bar

 

[Lyke]

LOL what the fvck

 

[ShiroShonen]

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Remember the +/- sign should be -/+ but the symbol doesn't exist here.

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What kind of sorcery is this? _@:

 

[Flakez]

What kind of sorcery is this? _@:

It's not as hard as it looks.

 

[Transcendence]

How... are you having problems with this? Unless you are in lower grades and I don't know what GSCE is (I'm Canadian). Isolate for y by rearranging the equation and getting y = 2-x. That is your y value. Sub it in the to the equation and then expand it and subsequently collect like terms until you have gotten a simplified expression. Most likely, the equation will stop there and you will be left with a trinomial. Since you have that, you must use the quadratic formula. Plug in the respective a b and c values and your two x values are the answers to your question.

 

[Shinichi Izumi]

How... are you having problems with this? Unless you are in lower grades and I don't know what GSCE is (I'm Canadian). Isolate for y by rearranging the equation and getting y = 2-x. That is your y value. Sub it in the to the equation and then expand it and subsequently collect like terms until you have gotten a simplified expression. Most likely, the equation will stop there and you will be left with a trinomial. Since you have that, you must use the quadratic formula. Plug in the respective a b and c values and your two x values are the answers to your question.

Remember that not everyone is as skillful as you in math ^.^

 

[EnDash]

How... are you having problems with this? Unless you are in lower grades and I don't know what GSCE is (I'm Canadian). Isolate for y by rearranging the equation and getting y = 2-x. That is your y value. Sub it in the to the equation and then expand it and subsequently collect like terms until you have gotten a simplified expression. Most likely, the equation will stop there and you will be left with a trinomial. Since you have that, you must use the quadratic formula. Plug in the respective a b and c values and your two x values are the answers to your question.

are you trying to help him or discourage him?

OP: you can prove this is correct by checking with the original formula:
1 + sqrt(14) / 2 ≈ 2.87
1 - sqrt(14) / 2 ≈ -0.87

x + y = 2 -> 2.87 + -0.87 = 2 so it's correct here

the other one is only correct if you use the full value, my calculator gives me:

x = 1 + sqrt(14) / 2 = 2.8708286933869706927918743661583
y = 1 - sqrt(14) / 2 = -0.87082869338697069279187436615827

x^2 + y^2 = 9 -> 8.2416573867739413855837487323167 + 0.75834261322605861441625126768344 = 9 so it's correct here as well

it's a little inaccurate in the calculator but you get the idea. i'm sure you know already how to do this but it's good to check whenever you can.