<PRE><B>Question 1105525</B>
1) x^2 + y^ 2 = 65
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2) (x + y)^2 = 121
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using equation 2, square the left side of = 
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3) x^2 + 2xy +y^2 = 121
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use equation 1 and substitute for x^2 + y^2 in equation 3
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2xy + 65 = 121
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2xy = 56
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xy = 28
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x = 28/y
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substitute for x in equation 1
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28^2/y^2 + y^2 = 65
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784 + y^4 = 65y^2
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let u = y^2, then
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u^2 -65u +784 = 0
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(u-16) * (u-49) = 0
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u = 16 or u = 49
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Note that u = y^2, so y = 4,-4 or 7,-7
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the numbers are 4 and 7 or -4 and -7
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