A cash register contains only five dollar and ten dollar bills. It contains twice as many five's as ten's and the total amount of money in the cash register is 580 dollars. How many ten's are in the cash register?
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So x = 5, y = 10, z= # of 5's have, and a= # of 10's have
2xz +ya =580
2 (5) z + 10a = 580
10z +10a = 580
10(z+a) = 580
We have twice as many fives than tens, so
z = 2a
Substitution Property of Equality
2a +a = 58
3a = 58
a = 19.333
So z= 38.667
So we have 19 bill of 10's and 38 bill of 5's, but the last one? 1/3 for 10's and 2/3 for 5's, so 19 1/3 bill of 10's and 38 2/3 bill of 5's
Sorry, I not totally sure if I did it right. But I hope that I'm right since I checked many time before I sent this.
what if the money was 680 would it be the same formula