Question 168836
let a = first number.
let b = second number.
let c = third number.
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sum of 3 numbers is 105
a + b + c = 105
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third number is 11 less than 10 times the second.
c = 10*b - 11
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twice the first number is 7 more than 3 times the second.
2*a = 3*b + 7
a = (3*b+7)/2
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start with:
a + b + c = 105
substitute for a:
(3*b+7)/2 + b + c = 105
substitute for c:"
(3*b+7)/2 + b + 10*b - 11 = 105
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equation is now in terms of b and can be solved.
expand all terms by removing parentheses:
3*b/2 + 7/2 + b + 10*b - 11 = 105
multiply both sides of equation by 2:
3*b + 7 + 2*b + 20*b - 22 = 210
combine like terms:
25*b - 15 = 210
add 15 to both sides of equation:
25*b = 225
divide both sides of equation by 25:
b = 9
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if b = 9, then:
(3*b+7)/2 = (27+7)/2 = 34/2 = 17 = a
and:
10*b - 11 = 90 - 11 = 79 = c
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you have:
a = 17
b = 9
c = 79
total = 105
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total = a + b + c = 79 + 9 + 17 = 88 + 17 = 105 (ok)
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c = 10*b - 11 becomes:
c = 10*9 - 11 = 90 - 11 = 79 (ok) 
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a = (3*b+7)/2 becomes:
a = (3*9+7)/2 = (27+7)/2 = 35/2 = 17 (ok)
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