A student has some $1 bills and $5 bills in his wallet. He has a total of 15 bills that are worth $43.How much of each type of bill does he have?

Answer 1

Let x stand for the amount of $1 bills and y stand for the amount of $5 bills.

x + y = 15

1x + 5y = 43

If you manipulate equation 1, you get: x = 15 - y

Now you can plug the x in equation 1 into the x in equation 2.

1(15 - y) + 5y = 43

15 + 4y = 43

4y = 28

y=7

Now just plug y back into your equation to find x.

x + 7 = 15

x= 8

Answer 2

The student has 15 total bills, some are worth $1 and some are worth $5. If I told you he had five $5 bills, how would you find out how many $1 bills he has? You'd say:

fifteen total bills - five $5 bills = ten $1 bills

But we don't know how many $5 bills he has. Whenever we don't know a value in math we can use a variable. Let's use x.

So he has x $5 bills. Then he has (15-x) $1 bills. If we add all the bills together, it must equal $43.

(x)($5) + (15-x)($1) = $43

We can make this a little easier to read by rearranging and dropping the $ signs:

15-x+5x=43

Now solve:

15+4x=43

4x=28

x=7

Since x is 7, he has seven $5 bills. The remainder, then are $1 bills: 15-7=8.

It's always good to check your answer:

7x$5=$35

8x$1=$8

$35+$8=$43

This checks out.

The student has seven $5 bills and eight $1 bills.