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We can solve this by setting up two different equations.

x=1st number

y=2nd number

xy = 40

x-y = 3

We then use substitution.

x = 3+y

(y+3)y = 40

y^2 +3y = 40

y^2 +3y - 40 = 0

(y+8)(y-5) = 0

y = -8, 5

We then plug in the values for y.

x -(-8) = 3 x - 5 = 3

x=-5, 8

The two numbers could either be (-8,-5) or (5,8).

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14y ago

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