Number of pennies:P
Number of nickels:N
Number of dollar bills:D
P=N+12
N=D+5
1P+5N+100D=996
^given
Therefore
P=(D+5)+12
P=D+17
Plug back into main equation
(D+17)+5(D+5)+100D=996
Combine like terms
106D+42=996
106D=954
D=954/106=9

9 Dollar bills

N=D+5
N=(9)+5=14

14 Nickels

P=N+12
P=(14)+12=26

26 Pennies

26*1+14*5+9*100=996

Final answer:
9 Dollars
14 Nickels
26 Pennies

8 0

4 years ago

A bookstore can purchase several calculators for a total cost of $120. If each calculators cost $1 less, the bookstore could pur

Tresset [83]

$x-number\ of\ calculators\ sold\ in\ regular\ price\\\\
y-\ \ first\ price\\\\
(y-1)\ \ -second\ price\\\\x=\frac{120}{y}\\
x+10=\frac{120}{y-1}\\\\Substitude\ x=\frac{120}{y}\ into\ the\ second\\\\\
\frac{120}{y}+10=\frac{120}{y-1}\ \ |-\frac{120}{y-1}-10\\\\
\frac{120}{y}-\frac{120}{y-1}=-10\\\\Make\ a\ common\ denominator\\\\
\frac{120(y-1)}{y(y-1)}-\frac{120y}{(y-1)y}=-10
\frac{120y-120)}{y(y-1)}-\frac{120y}{(y-1)y}=-10\\\\
\frac{120y-120-120y}{y(y-1)}=-10$ <u /> $Assumptions:\\y \neq 0\ \ \ and\ \ y \neq 1\\\\ \frac{-120}{y(y-1)}=-10\ \ \ | *y(y-1) \\\\
-120=-10y(y-1)\\\\
-120=-10y^2+10y\\\\
10y^2-10y-120=0\ \ \ |:10\\\\
y^2-y-12=0\\\\Factor\ equation\\\\ y^2+3y-4y-12=0\\\\
(y-4)(y+3)=0\\\\Solutions:\ \ y=4\ \ and\ \ y=-3$