Answer
System of equations:
$x+y+z+p=26$
$x-2y=-1$
$y-4z=0$
Number of each bill denomination:
15 one dollar bills, 8 five dollar bills, 2 ten dollar bill, and 1 twenty dollar bills
Work Step by Step
Firstly, We are told that there are four kinds of bills: 1 dollar, 5 dollars, 10 dollars, and 20 dollar bills
Let x= number of 1 dollar bills, y= number of 5 dollar bills, z= number of 10 dollar bills, and p= number of 20 dollar bills.
Secondly, we are told that the number of 5 dollar bills is four times the number of 10 dollar bills:
$y=4z$
$y-4z=0$.... Equation 1
Thirdly, we are told that the number of one dollar bills is one less than twice the number of 5 dollar bills:
$x=2y-1$
$x-2y=-1$ ...Equation 2
Fourthly, we are told that the total number of bills is 26:
$x+y+z+p=26$... Equation 3
System of equations:
$x+y+z+p=26$
$x-2y=-1$
$y-4z=0$
Augmented Matrix:
(See the image)
Elementary Row Operations:
(See image)
Parameter: p=s
$z+\frac{1}{13}s=\frac{27}{13}$
$z=\frac{27}{13}-\frac{1}{13}s$
$y=-27+17(\frac{27}{13}-\frac{1}{13}s)+s$
$y=\frac{108}{13}-\frac{17}{13}s+s$
$y=\frac{108}{13}-\frac{4}{13}s$
$x=26-(\frac{108}{13}-\frac{4}{13}s)-(\frac{27}{13}-\frac{1}{13}s)-s$
$x=\frac{203}{13}-\frac{8}{13}s$
When s=1: z=2, y=8, x=15
Total amount=(1×15)+(5×8)+(2×10)+(1×20)
Total amount=95 dollars
Therefore the number of each bill denomination:
15 one dollar bills, 8 five dollar bills, 2 ten dollar bill, and 1 twenty dollar bills
