3.4 Real-World Applications & Word Problems with Systems

3.4 Real-World Applications & Word Problems with Systems

Quick Summary: Word problems on the Digital SAT frequently describe real-world scenarios involving two unknown quantities constrained by two conditions. By recognizing the standard word problem archetypes—Total Count / Total Value, Mixtures & Concentrations, Relative Motion (Wind/Current), and Break-Even Cost/Revenue—you can formulate and solve the system of equations immediately.

---

The 4-Step Word Problem Translation Protocol

1. Define Variables Clearly: Assign variables to the specific quantities requested in the question (e.g., let $a = \text{number of adult tickets}$, $s = \text{number of student tickets}$). Include units.
2. Identify the Two Constraints: Locate the two independent conditions described in the text (e.g., total count of tickets and total dollar revenue).
3. Formulate the System: Translate each condition into an algebraic linear equation.
4. Solve and Verify Context: Solve using elimination, substitution, or Desmos, and check that the solution makes physical sense (e.g., non-negative counts, realistic speeds).

                  +-----------------------------------------+
                  |   SAT System Word Problem Archetypes    |
                  +--------------------+--------------------+
                                       |
         +-----------------+-----------+-----+-----------------+
         |                 |                 |                 |
         v                 v                 v                 v
+-----------------+ +---------------+ +---------------+ +-----------------+
| Count & Value   | |   Mixture     | | Motion/Wind   | |   Break-Even    |
| x + y = Total   | | x + y = V_tot | | d = (r + c)t1 | | Cost = Fixed+Vx |
| P1*x+P2*y = Tot | | c1*x+c2*y=Pure| | d = (r - c)t2 | | Revenue = P*x   |
+-----------------+ +---------------+ +---------------+ +-----------------+

---

Archetype 1: Total Count & Total Value (Tickets, Coins, Purchases)

This is the most common linear system archetype on the SAT. One equation accounts for the number of items, while the second accounts for the total dollar or point value.

General Template

$\begin{cases} x + y = N_{\text{total}} & \text{(Count Equation)} \\ P_1 x + P_2 y = V_{\text{total}} & \text{(Value Equation)} \end{cases}$

Worked Example 1: Ticket Sales

Problem: A community theater sold $250$ tickets for a weekend performance, generating $3,300 in revenue. Adult tickets cost $18 each and student tickets cost $10 each. How many student tickets were sold?

• Step 1: Define variables.

  • Let $a = \text{number of adult tickets}$
  • Let $s = \text{number of student tickets}$

• Step 2: Set up equations. $\begin{cases} a + s = 250 & \text{(Total tickets sold)} \\ 18a + 10s = 3300 & \text{(Total ticket revenue)} \end{cases}$

• Step 3: Solve for $s$. From the first equation, $a = 250 - s$. Substitute into the revenue equation: $18(250 - s) + 10s = 3300$ $4500 - 18s + 10s = 3300$ $4500 - 8s = 3300$ $-8s = -1200 \implies s = 150$

There were $150$ student tickets sold (and $250 - 150 = 100$ adult tickets).

---

Archetype 2: Mixture & Solution Concentration Problems

In mixture problems, two solutions of different concentrations are combined to create a target mixture. One equation balances total volume/mass, and the other balances the pure active ingredient.

General Template

$\begin{cases} x + y = V_{\text{total}} & \text{(Total Volume)} \\ c_1 x + c_2 y = c_{\text{final}} \cdot V_{\text{total}} & \text{(Pure Ingredient Balance)} \end{cases}$

Worked Example 2: Saline Solution Mixture

Problem: A chemist mixes an $8%$ saline solution with a $20%$ saline solution to produce $30$ liters of a $12%$ saline solution. How many liters of the $8%$ solution are needed?

• Step 1: Define variables.

  • Let $x = \text{liters of 8% solution}$
  • Let $y = \text{liters of 20% solution}$

• Step 2: Set up equations. $\begin{cases} x + y = 30 & \text{(Total volume)} \\ 0.08x + 0.20y = 0.12(30) & \text{(Pure salt amount)} \end{cases}$

• Step 3: Solve the system. $0.12(30) = 3.6$ Multiply the second equation by $100$ to clear decimals: $8x + 20y = 360$. Substitute $y = 30 - x$: $8x + 20(30 - x) = 360$ $8x + 600 - 20x = 360$ $-12x = -240 \implies x = 20$

The chemist must use $20$ liters of the $8%$ solution (and $10$ liters of the $20%$ solution).

---

Archetype 3: Relative Motion (Wind and Water Currents)

When a vehicle (plane or boat) moves with or against a current, its speed relative to the ground is altered by the current's speed:

• With the current (Tailwind / Downstream): $\text{Effective Speed} = r + c$
• Against the current (Headwind / Upstream): $\text{Effective Speed} = r - c$

Using $\text{Distance} = \text{Rate} \times \text{Time}$ ($d = r \cdot t$): $\begin{cases} (r + c) \cdot t_{\text{with}} = d_{\text{with}} \\ (r - c) \cdot t_{\text{against}} = d_{\text{against}} \end{cases}$

Worked Example 3: Airplane with Tailwind/Headwind

Problem: An airplane flies $1,200$ miles from City A to City B in $2$ hours with a tailwind. On the return flight against the same wind, the trip takes $2.5$ hours. What is the speed of the wind?

• Step 1: Set up rate equations.

  • Tailwind speed: $r + w = \frac{1200}{2} = 600\text{ mph}$
  • Headwind speed: $r - w = \frac{1200}{2.5} = 480\text{ mph}$

• Step 2: Solve the system. $\begin{cases} r + w = 600 \\ r - w = 480 \end{cases}$

• Step 3: Subtract equations to isolate $w$. $(r + w) - (r - w) = 600 - 480$ $2w = 120 \implies w = 60\text{ mph}$

The wind speed is $60\text{ mph}$ (and the plane's still-air speed is $540\text{ mph}$).

---

Archetype 4: Business Economics & Break-Even Analysis

Break-even analysis models the production cost and revenue of a business:

• Cost Function: $C(x) = \text{Fixed Overhead} + (\text{Unit Variable Cost}) \cdot x$
• Revenue Function: $R(x) = (\text{Selling Price per Unit}) \cdot x$
• Break-Even Point: The number of units $x$ where $R(x) = C(x)$ (Profit $= 0$).

Worked Example 4: Break-Even Calculation

A small bakery spends $1,200 monthly on kitchen rent (fixed cost) plus $4 in ingredients per cake. If each cake sells for $19, how many cakes must the bakery sell each month to break even?

$C(x) = 1200 + 4x$ $R(x) = 19x$ $19x = 1200 + 4x \implies 15x = 1200 \implies x = 80\text{ cakes}$

---

Desmos Strategies for Real-World Word Problems

1. Enter Equations Directly: Type $a + s = 250$ and $18a + 10s = 3300$ using $x$ and $y$ (Desmos requires $x$ and $y$ for plotting).
2. Window Adjustment: For large real-world quantities, click the wrench settings icon and adjust bounds (e.g., $x \in [0, 300]$, $y \in [0, 300]$).
3. Click the Intersection: Desmos will display $(100, 150)$ directly.

---

SAT Translation Pitfalls & Exam Checklist