Question 154119
Find two integers whose product is 105 such that one of the integers is one more twice the other integer.
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Let x = one of two integers
and y = second of two integers
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Since we have two unknowns, we'll need two equations.
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Equation 1 comes from:"whose product is 105 "
xy = 105
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Equation 2 comes from:"one of the integers is one more twice the other integer"
x = 2y + 1
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Using the definition of 'x' from equation 2, we substitute it into equation 1 a nd solve for 'y':
xy = 105
(2y + 1)y = 105
2y^2 + y = 105
2y^2 + y - 105 = 0
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Since it can't easily be factored, we can use the "quadratic equation" to solve.  Doing so yields (See reference below):
y = {7, -7.5}
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If y = 7, we can find 'x' by substituting it into equation 1 and solve for 'x':
xy = 105
x(7) = 105
x = 105/7
x = 15
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If y = -7.5
xy = 105
x(-7.5) = 105
x = 105/(-7.5)
x = -14
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Conclusion -- there are two possible answers:
(x,y) = (15,7)
(x,y) = (-14,-7.5)
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For reference, here is the details of the quadratic:
*[invoke quadratic "x", 2, 1, -105 ]