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2x Maths Questions (3 Viewers)

Varin9

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Simutaneous Equations
Find real numbers for p,q,r-

p^2 + q^2 + r^2 = 3
p + q + r =1
p + 2q - r = 5

(just a thought, but i you square each term in the second one, it doesnt seem to work)

Solve for x and y

x + y * root2
------------- = root2
root2 - 1

Thanks in advance to everyone appreciate it
 

Alkanes

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umm is the answer for the second question x = 2 and y = -1? ill post up the solution if its right =)
 

SpiralFlex

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Disregard question fixing it soon

Let's add equation 2 and 3

p + q + r + p + 2q - r = 1 + 5

2p + 3q = 6 [Equation 4]

Now, let's minus equation 3 from 2

p + q + r - (p + 2q - r) = -4 Equation 5

-q + 2r = -4 [6]

Let's focus on equation [4] and [6]

2p + 3q = 6

- q + 2r = 4

Let's times [6] by 3

-3q + 6r = -12 [7]

[4] + [7]

2p + 6r = -6

Hence, p + 2r = -3

Change the subject for 2r: 2r = -3 - p [8]

Substitute this into equation [6]

-q + 2(-3-p) = -4

-q - 6 - 2p = -4

-q - 2p = 2 [9]

[4] + [9]

2q = 8

q = 4

Sub for the rest: you should get q = 1, p =

To be completed soon

Give me a second to finish this.

2nd one: times all sides by root 2.

2 - sqrt(2) = x + y sqrt 2

By the theorem of surds, x must be 2, y = -1 no matter what.
 
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Varin9

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回复: Re: 2x Maths Questions

Still working on first one.

2nd one: times all sides by root 2.

2 - sqrt(2) = x + y sqrt 2

By the theorem of surds, x must be 2, y = -1 no matter what.
Yeah i though that, thx, but the one that has me stumped is the first one.
 

SpiralFlex

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Re: 回复: Re: 2x Maths Questions

Hold on mate, i'm trying to do it, it's taking long.
 

jyu

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p^2 + q^2 + r^2 = 3
p + q + r =1
p + 2q - r = 5

Please check the equations
 

AAEldar

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Just solved it.



Or



Will post working up in a sec.
 

SpiralFlex

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I'm sure I have seen this in one of the Cambridge books. Can you check if you typed equations right?
 

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