Apply the seven-step problem-solving strategy and the total value of coins model to a coin mixture problem in which the number of each coin type is unknown but the coin types are related by a stated condition.

Problem: Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

1. Read the problem and identify the coin types involved: dimes (worth $0.10 each) and nickels (worth $0.05 each). The total value of all coins is $2.25.
2. Identify what to find: the number of dimes and the number of nickels.
3. Name the unknowns using a single variable. Let $d$ = the number of dimes. Because there are nine more nickels than dimes, the number of nickels is $d + 9$. Organize everything in a table:

Type	Number	Value ($)	Total Value ($)
Dimes	$d$	0.10	$0.10d$
Nickels	$d + 9$	0.05	$0.05(d + 9)$
			2.25

4. Translate into an equation by adding the total values for each coin type and setting the sum equal to the overall total:

$0.10d + 0.05(d + 9) = 2.25$

5. Solve the equation:

• Distribute $0.05$: $0.10d + 0.05d + 0.45 = 2.25$
• Combine like terms: $0.15d + 0.45 = 2.25$
• Subtract $0.45$ from both sides: $0.15d = 1.80$
• Divide both sides by $0.15$: $d = 12$

Find the number of nickels: $d + 9 = 12 + 9 = 21$.

6. Check: $12$ dimes are worth $12(0.10) = 1.20$, and $21$ nickels are worth $21(0.05) = 1.05$. Adding: $1.20 + 1.05 = 2.25$ $\checkmark$
7. Answer: Adalberto has twelve dimes and twenty-one nickels.

This example extends the total value model from situations with known coin counts to ones where the counts are unknown. The relationship "nine more nickels than dimes" allows both quantities to be expressed through a single variable ($d$ and $d + 9$). Multiplying each variable expression by the coin's unit value and then summing produces an equation with decimal coefficients that requires distributing, combining like terms, and dividing — standard linear-equation techniques applied in a real-world context.