Question

Define a system and solve with a complete algebraic approach.
Beth has some nickels and dimes worth 80 cents. Billy has 6 more nickels than Beth and half as many dimes as Beth. His coins are worth 90 cents. How many of each type of coin does Beth have? How many does Billy have?

Answer 1

Let b represent Beth and d represent Billy.
This first equation represents dimes + nickels = amount of money Beth has.
\[10b_{1} + 5b_{2} = 80\]
Similarly for Billy.
\[10d_{1} + 5d_{2} = 90\]
Billy has six more nickels than Beth.
\[b_{1} + 6= d_{1}\]
Bill has half as many dimes as Beth.
\[b_{2} = \frac{d_{2}}{2}\]

Now you have four equations and four unknowns.
There's the matrix way to solve this and the substitution method. I don't know if you've learned the matrix method, so I'll use substitution.
Take Beth's equation and substitution as followed:
\[10b_{1} + 5b_{2} = 80\]
\[b_{1} = d_{1} - 6\]
\[b_{2} = \frac{d_{2}}{2}\]
\[10(d_{1} - 6) + 5(\frac{d_{2}}{2}) = 80\]
simplify and you get \[10d_{1} - 60 + \frac{5d_{2}}{2} = 80\]
Now you have a system of two equations and two unknowns with \[10d_{1} + 5d_{2} = 90\]

with those two equations, you should be able to solve for \[d_{1}\] and \[d_{2}\] and plug those values into the first two equations to get \[b_{1}\] and \[b_{2}\]