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x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?

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x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?  [#permalink]

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New post 20 Mar 2019, 04:43
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26% (02:25) correct 74% (02:23) wrong based on 35 sessions

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Re: x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?  [#permalink]

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New post 20 Mar 2019, 04:58
Use concept of (a+b)² on both x and y. Split 34 into 25 and 9

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x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?  [#permalink]

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New post 20 Mar 2019, 05:19
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Bunuel wrote:
x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?

(A) 1
(B) 2
(C) 3
(D) 6
(E) 8


The equation \(x^2 + y^2 = 10x - 6y - 34\) can be re-written as \(x^2 - 2(5)x + (5)^2 + y^2 + 2(3)y + (3)^2 = 0\)

-> \((x - 5)^2 + (y + 3)^2 = 0\) based on formulae \((a + b)^2 = a^2 + 2ab + b^2\) and \((a - b)^2 = a^2 - 2ab + b^2\)

Therefore, for equation to be equal to zero, x = 5 and y = -3 and the value of x + y = 2(Option B)
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Re: x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?  [#permalink]

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New post 31 Mar 2019, 16:27
pushpitkc wrote:
Bunuel wrote:
x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?

(A) 1
(B) 2
(C) 3
(D) 6
(E) 8


The equation \(x^2 + y^2 = 10x - 6y - 34\) can be re-written as \(x^2 - 2(5)x + (5)^2 + y^2 + 2(3)y + (3)^2 = 0\)

-> \((x - 5)^2 + (y + 3)^2 = 0\) based on formulae \((a + b)^2 = a^2 + 2ab + b^2\) and \((a - b)^2 = a^2 - 2ab + b^2\)

Therefore, for equation to be equal to zero, x = 5 and y = -3 and the value of x + y = 2(Option B)


pushpitkc

Hi!

Where did the 34 go?

Kind regards!
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Re: x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?  [#permalink]

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New post 01 Apr 2019, 03:56
x^2 + y^2 = 10x - 6y - 34
x^2 -10x+ y^2+6y+34=0
x^2 - 2(5)x + y^2 + 2(3)y +34 =0
To write the above expression in the form of a^2 + b^2 + 2ab, b^2 term is missing. We can split 34 into 25 and 9 inorder to get the b^2 term.
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Re: x and y satisfy the equation x^2 + y^2 = 10x - 6y - 34. What is x + y?   [#permalink] 01 Apr 2019, 03:56
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