## College Algebra 7th Edition

$15$ nickels. $10$ dimes. $5$ quarters.
Write a system of equations to solve for the three variables, $n$, $d$, and $q$ based on the problem. $5n+10d+25q=300$ $d=2q$ $5+d=n$ And now we will manipulate the third equation. $5+d=n$ $d=n-5=2q$ $n=2q+5$ Plug these into the first equation to solve for $q$. $5(2q+5)+10(2q)+25q=300$ $(2q+5)+2(2q)+5q=60$ $11q+5=60$ $q=\frac{60-5}{11}=\frac{55}{11}=5$ Substitute this value for $q$ into the other equations. $n=2q+5=2(5)+5=15$ $d=2q=2(5)=10$