What are the value(s) of x and y?

This came up in a Math quiz at school and I wasn't able to solve it. Can you?

Substitute X & Y

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Internet. Serious Business.

It would help if you explained a bit about the subject you are currently learning.

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X^2 + Y^2 = 49

Substitute X & Y

 

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x = 7 or -7
y= 7 or -7

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clicky

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cbf subbing in the original equations if xcutioners is correct though then

x = 7 or -7
y= 7 or -7

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How does 7^2 +7^2 = 49?

Even with imaginary numbers, you could only get +/- 98, or 0

x = sqrt(45),

y = 2.

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x^2-y^2=41
x^2+y^2=49
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2x^2 = 90
x=root(45)
so, y=2....

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Your equations don't make sense. I don't understand how you got the first equation - how did you know that x^2 - y^2 would equal 41?

If you solve your second equation it comes out as x= 22.5. If you first divide x by both sides and you get 2^2 = 90/x or 4=90/x, so x = 22.5

Could you explain?

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Quote: from martyfrmn at 2:21 am on Nov. 9, 2009

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x^2-y^2=41
 x^2+y^2=49  
 ---------------  
 2x^2 = 90
 x=root(45)  
 so, y=2....

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Your equations don't make sense. I don't understand how you got the first equation - how did you know that x^2 - y^2 would equal 41?

If you solve your second equation it comes out as x= 22.5. If you first divide x by both sides and you get 2^2 = 90/x or 4=90/x, so x = 22.5

Could you explain?

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He forgot the negative values for those numbers, but otherwise his answers are right.  Any combination of x = plus or minus root(45) and y = plus or minus 2

How he got it?  You factor the first equation

x^4-y^4 = (x^2-y^2)(x^2+y^2)

Factorization through difference of squares.  Since you know x^2+y^2, then you can solve for x^2-y^2.  And then you can solve the rest easily.