Focus on the wording "4 fewer than twice the number of pennies." Start by writing an expression for "twice the number of pennies" and then subtract 4 from that expression.

Once you have an expression for the number of nickels, use the phrase "3 more than the number of nickels" to write an expression for the number of dimes in terms of $p$.

Add your expressions for pennies, nickels, and dimes together, and set that sum equal to 47. Then look for the answer choice that matches this equation.

In Desmos, enter expressions for the three coin types in terms of $p$:

• One line for pennies as $p$.
• One line for nickels using the phrase "4 fewer than twice the number of pennies".
• One line for dimes using the phrase "3 more than the number of nickels". This helps you see all quantities in terms of the same variable.

On a new line, create an expression for the total number of coins by adding your three expressions (pennies + nickels + dimes). Desmos will display the simplified algebraic form of this sum.

Compare the simplified total-coins expression that Desmos shows with each answer choice and select the option whose left-hand side matches that expression and is set equal to 47.

The problem tells you that $p$ is the number of pennies.

You must now:

• Express the number of nickels in terms of $p$.
• Express the number of dimes in terms of $p$.
• Then add pennies + nickels + dimes to equal 47.

"The number of nickels is 4 fewer than twice the number of pennies, $p$."

• "Twice the number of pennies" is $2p$.
• "4 fewer than" means subtract 4 from that result.

So the number of nickels is $2p - 4$ (not $2(p - 4)$).

"The number of dimes is 3 more than the number of nickels."

• You just found nickels as $2p - 4$.
• "3 more than" that is $(2p - 4) + 3$, which simplifies to $2p - 1$.

So the number of dimes is $2p - 1$.

Now add all three types of coins and set the sum equal to 47:

• Pennies: $p$
• Nickels: $2p - 4$
• Dimes: $2p - 1$

So the equation must be:

which matches choice D.