# Don't understand this problem.

• Nov 4th 2008, 03:58 PM
Rheanna
Don't understand this problem.
I do kinda of, but for someone reason I doubt I'm getting the answer.

A cashier has \$645 in 10 dollar and 5 dollar bills. There are 90 bills in all. How many of each does the cashier have?

10x + 5x = 645
15x=645
645-90=555

I know , I know this. but for some reason I can'tfigure it out. lol

This is just 2 point for extra credit. got the other 14 questions right. Only had to do 8 of 15.

Thanks
• Nov 4th 2008, 04:03 PM
skeeter
Quote:

A cashier has \$645 in 10 dollar and 5 dollar bills. There are 90 bills in all. How many of each does the cashier have?
let x = number of \$10 bills
y = number of \$5 bills

A cashier has \$645 in 10 dollar and 5 dollar bills.

10x + 5y = 645

There are 90 bills in all.

x + y = 90

solve the system of equations using your favorite method.
• Nov 4th 2008, 04:12 PM
euclid2
Quote:

Originally Posted by skeeter
let x = number of \$10 bills
y = number of \$5 bills

A cashier has \$645 in 10 dollar and 5 dollar bills.

10x + 5y = 645

There are 90 bills in all.

x + y = 90

solve the system of equations using your favorite method.

let x be the amount of 10 dollar bills
let y be the amount of 5 dollar bills

(x+y=90)-5
10x+5y=645
-5x-5y=-450
5x=195
x=39

sub into original

39+y=90
y=90-39
y=51

therefore there were 39 10 dollar bills and 51 5 dollar bills
• Nov 4th 2008, 05:59 PM
Rheanna
sorry, I figured it out, once I did the x and y .. before I did it , I was doing x and x, soon as I did x and y I got it. Thank You. I was confusing myself. (Clapping)

This is way I set it up to figure it out.

10x + 5 (90-x) = 625

Then took 90-39 = 51