Question 351755
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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y = x-3
Substitute for "y" and solve for "x":
x^2 + (x-3)^2 = 45
2x^2-6x- 36 = 0
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x^2-3x-18 = 0
(x-6)(x+3) = 0
x = 6 or x = -3
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Substitute for "x" and solve for "y"
If x = 6, y = 3
If x = -3, y = -6
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{{{graph(400,300,-10,10,-10,10,x-3,(45-x^2)^(1/2),-(45-x^2)^(1/2))}}}
Cheers,
Stan H.