x = ones
y = fives
x+y = 14 total bills
rewrite as x = 14-y
1x +5y = 50
1(14-y) +5y = 50
14-1y +5y = 50
14 +4y = 50
4y = 36
y = 36/4 = 9
x = 14-y = 14-9 = 5
9+5 = 14
9x5 = 45 +5x1 = 50
he had 9 fives and 5 ones
Answer:
There were 5 bills of '1' and 9 bills of '5'.
Step-by-step explanation:
let the number of '1' bill be x
let the number of '5' bill be y.
Total number of bills = 14
x + y = 14 ..[1]
Worth of 14 bills = 50
So,[tex]x\times 1+y\times 5=\$50[/tex]
[tex]x+5y=50[/tex]..[2]
x + y = 14
x = 14 - y
Putting value of x in [2]:
[tex]14- y +5y=50[/tex]
[tex]y=\frac{50-14}{4}=9[/tex]
y = 9
x = 14-y = 14 - 9 = 5
There were 5 bills of '1' and 9 bills of '5'.
