Solving a two-variable espresso and biscotti word problem

Key takeaway: This problem translates a real-life purchase scenario into a system of linear equations and then solves for the unit price of espresso.

Set up the system

Let $e$ be the cost of one espresso and $b$ be the cost of one biscotti.

From the first purchase:

$2e + b = 4.40$

From the second purchase:

$e + 2b = 3.55$

Eliminate one variable

Multiply the second equation by 2:

$2e + 4b = 7.10$

Now subtract the first equation:

$(2e + 4b) - (2e + b) = 7.10 - 4.40$

$3b = 2.70$

$b = 0.90$

Solve for espresso

Substitute $b = 0.90$ into $2e + b = 4.40$:

$2e + 0.90 = 4.40$

$2e = 3.50$

$e = 1.75$

So an espresso costs

\boxed{\1.75}$.

That is between $1.50 and$2.00, so the correct choice is (a).

Why this method works

A system of equations lets you isolate the price of each item. Elimination is efficient here because the coefficients can be matched with one multiplication step.

---

Pitfalls the pros know 👇 A common mistake is mixing up the labels for espresso and biscotti after setting up the equations. Another error is subtracting the equations without first making one variable’s coefficient match. Some students also stop after finding the biscotti price and forget to substitute back to find the espresso price. Always check your final answer against the original purchases: 2 espresso and 1 biscotti should total $4.40, and 1 espresso plus 2 biscotti should total$3.55.

What if the problem changes? If the problem changed to "three espressos and one biscotti cost $7.15" and "one espresso and two biscotti cost$3.55," you would still use a system, but the elimination step would be different because the espresso coefficients are not already aligned. If the question asked for the biscotti price instead, you would solve the same system and report $0.90. Word problems like this always depend on translating each sentence into one linear equation.

Tags: system of equations, elimination method, linear word problem