SOLUTION: Find 3 #'s in which the second # is 3 times the first # and the third # is 2 less than 5 times the first #. The sum of three #'s is 115. Find the three #'s. What I have tried:

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Question 162804: Find 3 #'s in which the second # is 3 times the first # and the third # is 2 less than 5 times the first #. The sum of three #'s is 115. Find the three #'s.
What I have tried:
3(x+2)+2(5x+1)=115
3x+6+10x+1=115
13x+7=115
-7 -7
13x=108
/13 /13
x=8.31
I know this isn't correct. Please help

Found 2 solutions by gonzo, checkley77:
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
you have 3 numbers.
call them a,b,c
a is the first number
b is the second number
c is the third number
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the second number is 3 times the first.
b = 3*a
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the third number is 2 less than 5 times the first.
since it's 5 * the first.
c = 5*a
if it's 2 less than that, then
c = 5*a - 2
-----
sum of the numbers is 115
a + b + c = 115
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you have 3 equations to work with.
a + b + c = 115
b = 3*a
c = 5*a - 2
-----
you can solve these simultaneously or you can use substitution.
substitution looks like it might be more direct.
we'll try that.
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you work with the equation
a + b + c = 115
you substitute 3*a for b
you get
a + 3*a + c = 115
you substitute 5*a-2 for c
you get
a + 3*a + 5*a - 2 = 115
you combine like terms to get
9*a - 2 = 115
you add 2 to both sides of the equation to get
9*a = 117
you divide both sides of the equation by 9 to get
a = 117/9 = 13.
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if a = 13, then 3*a = 39, and 5*a -2 = 63
original equation of
a + b + c = 115
becomes
13 + 39 + 63 = 115
this becomes
115 = 115
this means the answer is correct.
-----
answer is
a = 13
b = 39 = 3 * 13
c = 63 = 5 * 13 - 2

Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
S=3F
T=5F-2
F+S+T=115
F+3F+(5F-2)=115
9F=115+2
9F=117
F=117/9
F=13 FOR THE FIRST NUMBER.
3*13=39 FOR THE SECOND NUMBER.
5*13-2=65-2=63 FOR THE THIRD NUMBER.
PROOF:
13+39+63=115
115=115