Evil simultaneous equation!

Hey, i'm looking for some pointers in solving this simultaneous equation:

x^2+2y^2=11
x=3y

I've managed to get to the point of

3y^2+2y^2=11

but I have no idea where to go from here... Surely its not as simple as 5y^2=11?
Any suggestions are really appreciated thankyou!

(3y)^2 \neq 3y^2

Reply 2

Confusedddd.

how many answers thingys are there? :L like x= y= 1 for each or two for each? x

Reply 3

Xiomara

18

Original post by Mr M

(3y)^2 \neq 3y^2

I wish you were my Maths teacher.

Reply 4

MostCompetitive

2

Original post by kpritchx

Hey, i'm looking for some pointers in solving this simultaneous equation:

x^2+2y^2=11
x=3y

I've managed to get to the point of

3y^2+2y^2=11

but I have no idea where to go from here... Surely its not as simple as 5y^2=11?
Any suggestions are really appreciated thankyou!

(3y)^2 = (3)(3)(y)(y) = 9y^2

Reply 5

username360639

14

Original post by kpritchx

Hey, i'm looking for some pointers in solving this simultaneous equation:

x^2+2y^2=11
x=3y

(3y)^2=9y^2

so you have 9y^2+2y^2=11

11y^2=11

y=1, sub back to get x, which should be 3 (3,1)

Reply 6

username360639

14

Original post by MostCompetitive

(3y)^2 = (3)(3)(y)(y) = 9y^2

Dude your doing 7 AS levels wtf, legend!

Reply 7

hrickards

2

[QUOTE=",,,,;32048626"]

Original post by kpritchx

Hey, i'm looking for some pointers in solving this simultaneous equation:

x^2+2y^2=11
x=3y

(3y)^2=9y^2

so you have 9y^2+2y^2=11

11y^2=11

y=1, sub back to get x, which should be 3 (3,1)

Is 1 really the only square root of 1? :tongue:

Reply 8

username360639

14

[QUOTE="hrickards;32048728"]

Original post by ,,,,

Is 1 really the only square root of 1? :tongue:

Hey? Did i make a mistake :s, and yes it is, sqaure root of 1=1 :d

Reply 9

muffingg

18

Original post by Mr M

(3y)^2 \neq 3y^2

This. if x = 3x then x^2 doesn't equal to 3y^2, but (3y)^2.

In other words. You need to square the 3 as well, so it becomes 9y^2 in fact.

Reply 10

OwenP277

[QUOTE=",,,,;32048771"]

Original post by hrickards

Hey? Did i make a mistake :s, and yes it is, sqaure root of 1=1 :d

what's

(-1)^2

?

Reply 11

username360639

14

[QUOTE="OwenP277;32048864"]

Original post by ,,,,

what's

(-1)^2

?

1...LOL

Was the answer suppose to be + or- 1...Am i missing the point :s

(edited 12 years ago)

Reply 12

OwenP277

[QUOTE=",,,,;32048914"]

Original post by OwenP277

1...LOL

thus the square root of one is +1 or -1?

Reply 13

The Mr Z

13

Original post by kpritchx

Hey, i'm looking for some pointers in solving this simultaneous equation:

x^2+2y^2=11
x=3y

I've managed to get to the point of

3y^2+2y^2=11

but I have no idea where to go from here... Surely its not as simple as 5y^2=11?
Any suggestions are really appreciated thankyou!

x^2 = (3y)^2 = (3y)(3y) = 9y^2

9y^2 + 2y^2 = 11

11y^2 = 11, y^2 = 1, y= 1 or y= -1, x= 3 or -3

Reply 14

username360639

14

[QUOTE="OwenP277;32048957"]

Original post by ,,,,

thus the square root of one is +1 or -1?

Ops yes your right, haha, cheers for that :biggrin:

would have lost solutions

(edited 12 years ago)

Reply 15

pineapples01

2

How is this even evil in the slightest?

Reply 16

ginister

11

Squared numbers always turn positive, no matter what they were before. So yes, its plus 1

Reply 17

username582574

OP

Original post by pineapples01

How is this even evil in the slightest?

Its allot less evil now I understand that (3y)^2 doesn't equal 3y^2!
Thankyou everyone, got my GCSE Maths paper tomorrow and this has really helped!

Reply 18

mzfemalecupid

Original post by kpritchx

Thankyou everyone, got my GCSE Maths paper tomorrow and this has really helped!

me tooo
but i like simultaneous equations :biggrin:

i just hate everything else :s-smilie:

Reply 19

yaschaeffer

1

Original post by ginister

Squared numbers always turn positive, no matter what they were before. So yes, its plus 1

3i is a number. Imaginary granted, but still a number. Squaring it gives:
(3i)^2 = (3^2).(i^2) = 9.(-1) = -9 which is negative.
Not all numbers squared are positive. Only if they are positive or negative. You may have been ignoring this for sake of clarity, but the 'no matter what it was' bit isn't true.
For those who look at this and think 'Wuh!?', the square root of negative one (required in a lot of things from solving quadratic equations to engineering) is written as a lower case i.
If you take the n-th root of a number, there will be n answers. That's why there's the +/- root when you look at square (n=2) roots, ie 2 roots.
For example, there are four 4-th roots of one: 1, -1, i and -i.
If you haven't dealt with solutions of cubic functions, or quadratics where b^2-4ac is negative, then you probably haven't come across this stuff yet. You will.
Enjoy.

(edited 12 years ago)

Evil simultaneous equation!

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