Best answer:
You correctly solved for a ----> a = (7b²−14b)/6
This is a quadratic expression, that has minimum value when b = 1
So for b > 1, as b increases, then so does a.
However, we have the following restrictions on integers a and b:
a > 0, b > 0
a = (7b²−14b)/6 > 0 ----> b < 0 or b > 2 ---->...
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Best answer:
You correctly solved for a ----> a = (7b²−14b)/6
This is a quadratic expression, that has minimum value when b = 1
So for b > 1, as b increases, then so does a.
However, we have the following restrictions on integers a and b:
a > 0, b > 0
a = (7b²−14b)/6 > 0 ----> b < 0 or b > 2 ----> b > 2
Since a increases as b increases (for b > 4/7, and therefore for b > 2), we must find smallest integer value of b > 2 such that a = (7b²−14b)/6 is an integer.
Since a is an integer, 6 divides 7b²−14b = 7b(b−2) ----> 6 divides b(b−2)
----> 2 divides b(b−2) and 3 divides b(b−2)
b and b−2 are both even or both odd, so they must be both even to be divisible by 2
One of b or b−2 is divisible by 3. So either b = 6 or b−2 = 6 ----> smallest value of b = 6
b = 6
a = 7(6)(6−2)/6 = 28
Smallest value of a + b = 28 + 6 = 34
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1 day ago