Question 207562
one number is two less than a second number. if you take one-half of the first number and increase it by the second number, the result is at least 41. find the least possible value for the second number.
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Let 1st be "a" ; 2nd be "b"; 3rd be "c"
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Equations:
a = b-2
(1/2)a+b >=41
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Substitute a = b-2 into the 2nd equation and solve for "b":
(1/2)(b-2)+b >= 41
Multiply thru by 2 to get:
(b-2)+2b >= 82
3b >= 84
b >= 28
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Cheers,
Stan H.