SOLUTION: Please help me solve this equation: numeric word problem using 3 equations. The sum of three number is 105. The third is 11 less than 10 times the second. Twice the first

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Question 168836This question is from textbook Prentice Hall Algebra 2 with Trigonometry
: Please help me solve this equation:

numeric word problem using 3 equations.
The sum of three number is 105. The third is 11 less than 10 times the second. Twice the first is 7 more than three times the second. Find the Numbers This question is from textbook Prentice Hall Algebra 2 with Trigonometry

Found 2 solutions by Mathtut, gonzo:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
lets call the numbers a,b,and c
a+b+c=105....eq 1
c=10b-11.....eq 2
2a=3b+7......eq 3
:
lets solve for a in eq 3. a=(3b+7)/2
:
now lets take c's value from eq 2 and a's value from eq 3 and plug them into eq 1
:
((3b+7)/2)+b+(10b-11)=105 multiply all terms by 2
:
3b+7+2b+20b-22=210---> now combine like terms
:
25b=225:
highlight%28b=9%29
highlight%28c=10%289%29-11=79%29
highlight%28a=%283%289%29%2B7%29%2F2=17%29
a,b,c........17,9,79 respectively

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
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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