Question 248047
the difference of two positive numbers is 3 and the sum of their squares is 45. Find the numbers.
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Equations:
x - y = 3
x^2 + y^2 = 45
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x = y+3
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Substitute for "x" and solve for "y":
(y+3)^2 + y^2 = 45
y^2 + 6y + 9 + y^2 = 45
2y^2 + 6y - 36 = 0
y^2 + 3y - 18 = 0
(y+6)(y-3) = 0
y = 3 or y = -6
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If y = 3, x-3 = 3, x = 6
If y = -6, x --6 = 3, x = -3
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Checking x^2 + y^2 = 45
3^2 + 6^2 = 9 + 36 = 45
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Solutions: 
x=6 and y = 3
OR
x = -3 and y = -6
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Cheers,
Stan H.