SOLUTION: I am so confused...Please help me how to figure out setting up the following equation to solve: The cash drawer of the market contains $227 in bills. There are six more $5 bill

Question 400267: I am so confused...Please help me how to figure out setting up the following equation to solve:
The cash drawer of the market contains $227 in bills. There are six more $5 bills than $10 bills. The number of $1 bills is two more than 24 times the number of $10 bills. How many bills of each kind are there?
I appreciate any help I can get!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The cash drawer of the market contains $227 in bills.
There are six more $5 bills than $10 bills.
The number of $1 bills is two more than 24 times the number of $10 bills.
How many bills of each kind are there?
:
From the information we assume their are 1, 5, and 10 bills (no 20's)
let c = no. of $1 bills
let f = no. of $5
let t = no. of $10
:
Write an equation for each statement:
:
"The cash drawer of the market contains $227 in bills."
1c + 5f + 10t = 227
:
"There are six more $5 bills than $10 bills."
f = t+6
:
"The number of $1 bills is two more than 24 times the number of $10 bills."
c = 24t + 2
:
Use the 1st equation, substitute (t+6) for f, and (24t+2) for c
1(24t+2) + 5(t+6) + 10t = 227
24t + 2 + 5t + 30 + 10t = 227
combine like terms
24t + 5t + 10t + 2 + 30 = 227
39t + 32 = 227
39t = 227 - 32
39t = 195
t =
t = 5 ten dollar bills
:
Find the ones and fives
c = 24t + 2
c = 24(5) + 2
c = 120 + 2
c = 122 one dollar bills
:
f = t + 6
f = 5 + 6
f = 11 five dollar bills
;
:
See if this is true:
1(122) + 5(11) + 10(5) =
122 + 55 + 50 = 227, confirm our solutions
:
:
Did his unconfuse you somewhat?

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