SAT Math Lesson Plan 6: Systems of Equations in Word Problems

Lesson Summary

This 45-minute lesson is part of Media4Math’s 35-lesson SAT Math prep series. It focuses on solving systems of equations that arise from real-world word problems. These types of questions appear frequently on the SAT Math section and fall within the Heart of Algebra domain, which makes up approximately 33% of the exam.

Students will learn how to model motion problems, finance scenarios, mixture problems, and work-rate applications using systems of equations. Emphasis is placed on translating word problems into algebraic form and selecting appropriate solution methods—substitution, elimination, or graphing. The lesson includes six examples, a warm-up problem, and a 10-question SAT-style quiz to reinforce problem-solving strategies and concepts.

Lesson Objectives

• Translate real-world scenarios into systems of equations.
• Solve systems using substitution and elimination methods.
• Interpret solutions in the context of real-world problems.
• Solve systems graphically when appropriate.

Common Core Standards

• CCSS.MATH.CONTENT.HSA.REI.C.6 – Solve systems of linear equations exactly and approximately (e.g., with graphs).
• CCSS.MATH.CONTENT.HSA.CED.A.3 – Represent constraints using equations and interpret solutions in a real-world context.

Prerequisite Skills

• Understanding of linear equations.
• Ability to solve equations using substitution and elimination.
• Familiarity with graphing linear equations.

Key Vocabulary

• System of Equations – A set of two or more equations with the same variables.
• Substitution Method – Solving for one variable in terms of another and substituting into the second equation.
• Elimination Method – Adding or subtracting equations to eliminate one variable.
• Graphical Solution – Finding the point of intersection of two lines on a coordinate plane.
• Real-World Applications – Problems related to motion, finance, work rates, and mixtures.

Warm-Up

Goal: Activate prior knowledge of solving systems of equations algebraically and graphically.

Problem: Solve for \( x \) and \( y \) in the following system:

\[ \begin{cases} 2x + 3y = 12 \\ x - y = 4 \end{cases} \]

Method 1: Substitution

• Start with the second equation: \( x - y = 4 \)
• Solve for \( x \): \( x = y + 4 \)
• Substitute into the first equation: \( 2(y + 4) + 3y = 12 \)
• Simplify: \( 2y + 8 + 3y = 12 \Rightarrow 5y + 8 = 12 \Rightarrow 5y = 4 \Rightarrow y = \frac{4}{5} \)
• Now substitute back: \( x = \frac{4}{5} + 4 = \frac{24}{5} = 4.8 \)

Solution: \( x = \frac{24}{5} \) or 4.8, \( y = \frac{4}{5} \) or 0.8

Method 2: Graphing

Graph both equations on the coordinate plane and identify the point of intersection:

• First equation: \( y = \frac{12 - 2x}{3} \)
• Second equation: \( y = x - 4 \)

The intersection point confirms the algebraic solution: \( \left( \frac{24}{5},\ \frac{4}{5} \right) \)

Graph it on Desmos: Click here to open Desmos

Self-Study Tip: After solving systems algebraically, always verify your answer graphically if time allows—especially for SAT grid-in problems that involve checking precision.

Teach

Word problems involving systems of equations appear frequently on the SAT. These problems typically describe real-world scenarios that can be modeled with two linear equations. Once modeled, the system can be solved using substitution, elimination, or graphing. Understanding how to set up the correct equations is often the hardest part—but it’s the key to success!

Let’s walk through six important examples that illustrate motion, finance, work-rate, and mixture problems, all of which can be solved using systems of equations.

Example 1: Mixture Problem

Problem: A grocer wants to mix an amount of coffee worth \$12 per pound with a cheaper blend worth \$8 per pound. The result should be a blend that sells for \$10 per pound. For every 10 pounds of the blended coffee, he doesn't want to spend more than \$100. Find the combination of the two coffee types he should use.

Step 1: Define your variables

• x = pounds of \$8 coffee
• y = pounds of \$12 coffee

Step 2: Set up the system of equations

• Equation 1 (Total weight): \( x + y = 10 \)
• Equation 2 (Cost constraint): \( 8x + 12y = 100 \)

Step 3: Organize your known values in a table

Step 4: Solve the system using substitution

From Equation 1: \( x + y = 10 \Rightarrow x = 10 - y \)

Substitute into Equation 2:

\[ 8(10 - y) + 12y = 100 \\ 80 - 8y + 12y = 100 \\ 4y = 20 \Rightarrow y = 5 \]

Now find x:

\[ x = 10 - y = 10 - 5 = 5 \]

Final Answer:

The grocer should mix 5 pounds of \$8 coffee and 5 pounds of \$12 coffee.

Self-Study Tip: When you're given a total quantity and a cost constraint, it's a strong hint that a system of equations is needed. One equation usually represents total quantity, the other total value. Solve using substitution or elimination, depending on which variable is easier to isolate.

Example 2: Ticket Sales Problem

Problem: A school sells tickets to a play. Student tickets cost \$5 each, and adult tickets cost \$8 each. On opening night, the school sells 120 tickets and collects \$780. How many student and adult tickets were sold?

Step 1: Define your variables

• x = number of student tickets
• y = number of adult tickets

Step 2: Set up a system of equations

• Equation 1 (total tickets): \( x + y = 120 \)
• Equation 2 (total revenue): \( 5x + 8y = 780 \)

Step 3: Solve using substitution

From Equation 1: \( x = 120 - y \)

Substitute into Equation 2:

\[ 5(120 - y) + 8y = 780 \\ 600 - 5y + 8y = 780 \\ 3y = 180 \Rightarrow y = 60 \]

Now substitute back to find x:

\[ x = 120 - 60 = 60 \]

Final Answer:

The school sold 60 student tickets and 60 adult tickets.

Check Your Work:

• Ticket total: \( 60 + 60 = 120 \) ✅
• Revenue: \( 60 \times 5 + 60 \times 8 = 300 + 480 = 780 \) ✅

Self-Study Tip: SAT word problems that mention "how many" and "how much money" almost always lead to a system like this. Equation 1 tracks quantity, Equation 2 tracks value.

Example 3: Coin Value Problem

Problem: Emma has a collection of nickels and dimes worth \$1.35. She has 17 coins in total. How many of each coin does she have?

Step 1: Define your variables

• x = number of nickels
• y = number of dimes

Step 2: Set up the system of equations

• Equation 1 (total number of coins): \( x + y = 17 \)
• Equation 2 (total value of the coins): \( 5x + 10y = 135 \)

Note: We multiply all coin values by 100 to eliminate decimals.

Step 3: Solve the system using substitution

From Equation 1: \( x = 17 - y \)

Substitute into Equation 2:

\[ 5(17 - y) + 10y = 135 \\ 85 - 5y + 10y = 135 \\ 5y = 50 \Rightarrow y = 10 \]

Now substitute back to find x:

\[ x = 17 - 10 = 7 \]

Final Answer:

Emma has 7 nickels and 10 dimes.

Check Your Work:

• Total coins: \( 7 + 10 = 17 \) ✅
• Total value: \( 7 \times 0.05 + 10 \times 0.10 = 0.35 + 1.00 = 1.35 \) ✅

Self-Study Tip: Coin problems are perfect for system solving. One equation always tracks the number of coins, and the other tracks value. Always multiply by 100 early to work with whole numbers!

Example 4: Wages and Hours Problem

Problem: Amy and Brian worked a total of 18 hours tutoring over the weekend. Amy earns \$15 per hour, and Brian earns \$12 per hour. Together they earned \$246. How many hours did each of them work?

Step 1: Define your variables

• x = number of hours Amy worked
• y = number of hours Brian worked

Step 2: Organize the data in a table

Step 3: Set up the system of equations

• Equation 1 (total hours): \( x + y = 18 \)
• Equation 2 (total pay): \( 15x + 12y = 246 \)

Step 4: Solve using elimination

Multiply Equation 1 by 12 to align the coefficients of y:

\[ 12x + 12y = 216 \\ 15x + 12y = 246 \]

Subtract the first equation from the second:

\[ (15x + 12y) - (12x + 12y) = 246 - 216 \\ 3x = 30 \Rightarrow x = 10 \]

Substitute back to find y:

\[ x + y = 18 \Rightarrow 10 + y = 18 \Rightarrow y = 8 \]

Final Answer:

Amy worked 10 hours, and Brian worked 8 hours.

Check Your Work:

• Total hours: \( 10 + 8 = 18 \) ✅
• Total pay: \( 15 \times 10 + 12 \times 8 = 150 + 96 = 246 \) ✅

Self-Study Tip: Elimination is especially useful when both equations involve the same variable with matching or easily adjusted coefficients. Don’t forget to double-check your answer by substituting back into both original equations.

Example 5: Age Problem

Problem: The sum of Alice’s and Ben’s ages is 50. Ben is 8 years older than Alice. How old are they?

Step 1: Define your variables

• x = Alice’s age
• y = Ben’s age

Step 2: Organize the data in a table

Step 3: Write the system of equations

• Equation 1 (total age): \( x + y = 50 \)
• Equation 2 (age difference): \( y - x = 8 \)

Rewriting Equation 2 in standard form:

\[ -x + y = 8 \]

Now the system is:

\[ \begin{cases} x + y = 50 \\ -x + y = 8 \end{cases} \]

Step 4: Solve using elimination

Add the two equations:

\[ (x + y) + (-x + y) = 50 + 8 \\ 2y = 58 \Rightarrow y = 29 \]

Substitute back to find x:

\[ x + 29 = 50 \Rightarrow x = 21 \]

Final Answer:

Alice is 21 years old, and Ben is 29 years old.

Check Your Work:

• Total age: \( 21 + 29 = 50 \) ✅
• Ben is 8 years older: \( 29 - 21 = 8 \) ✅

Self-Study Tip: Age problems often include a total (sum) and a difference. If you write both as linear equations in standard form, elimination becomes a quick and reliable way to solve.

Example 6: Graphical Solution — Comparing Phone Plans

Problem: Two phone companies offer monthly plans. Plan A charges a \$20 base fee plus \$5 per gigabyte of data. Plan B charges a \$10 base fee plus \$7 per gigabyte. For what amount of data (in gigabytes) will both plans cost the same?

Step 1: Define your variables

• x = number of gigabytes of data
• y = total cost (in dollars)

Step 2: Set up the system of equations

• Plan A: \( 5x + 20 = y \Rightarrow 5x - y = -20 \)
• Plan B: \( 7x + 10 = y \Rightarrow 7x - y = -10 \)

Now the system is:

\[ \begin{cases} 5x - y = -20 \\ 7x - y = -10 \end{cases} \]

Step 3: Graph the equations

To graph each equation, solve for \( y \):

• Plan A: \( y = 5x + 20 \)
• Plan B: \( y = 7x + 10 \)

Plot both lines on a coordinate grid. The point of intersection represents the data usage at which both plans cost the same.

Desmos Link: Click here to graph the system

Step 4: Estimate the point of intersection

From the graph, the two lines intersect at \( x = 5 \), \( y = 45 \).

Final Answer:

At 5 GB of data, both plans cost \$45.

Check Your Work:

• Plan A: \( 5 \times 5 + 20 = 25 + 20 = 45 \) ✅
• Plan B: \( 7 \times 5 + 10 = 35 + 10 = 45 \) ✅

Self-Study Tip: When you're comparing costs or options and both sides include a variable, try graphing each expression as a linear equation. The solution is the point where the two lines intersect—this gives you a visual sense of what’s happening.

Review

Key Takeaways

• Translate word problems into two-variable equations.
• Look for keywords like “together,” “difference,” “total,” and “per” to guide equation setup.
• Use substitution when one variable is isolated, and elimination when both equations are in standard form.
• Label variables clearly—context is everything in word problems.
• Always check your solution in the original context to ensure it makes sense.

Example 1: Grocery Store Discount

Problem: At a grocery store, Sam buys apples and bananas. Apples cost \$2 each, and bananas cost \$1 each. He buys a total of 11 pieces of fruit and spends \$16. How many apples and bananas did he buy?

Step 1: Define your variables

• x = number of apples
• y = number of bananas

Step 2: Set up the system of equations

• Equation 1 (total fruit): \( x + y = 11 \)
• Equation 2 (total cost): \( 2x + y = 16 \)

Step 3: Solve using elimination

Subtract Equation 1 from Equation 2:

\[ (2x + y) - (x + y) = 16 - 11 \Rightarrow x = 5 \] \[ y = 11 - 5 = 6 \]

Final Answer:

Sam bought 5 apples and 6 bananas.

Check Your Work:

• Fruit count: \( 5 + 6 = 11 \) ✅
• Total cost: \( 5 \times 2 + 6 \times 1 = 10 + 6 = 16 \) ✅

Self-Study Tip: When you're given both a total count and total value, always set up a system. Line up your equations vertically to look for elimination opportunities.

Example 2: Movie Tickets

Problem: A group of friends buys a total of 8 movie tickets. Children’s tickets cost \$6 each, and adult tickets cost \$10 each. They spend \$64 in total. How many children’s and adult tickets did they buy?

Step 1: Define your variables

• x = number of children’s tickets
• y = number of adult tickets

Step 2: Organize the data in a table

Step 3: Set up the system of equations

• Equation 1 (ticket total): \( x + y = 8 \)
• Equation 2 (cost total): \( 6x + 10y = 64 \)

Step 4: Solve using elimination

Multiply Equation 1 by 6 to align with the coefficient on x in Equation 2:

\[ 6x + 6y = 48 \\ 6x + 10y = 64 \]

Now subtract:

\[ (6x + 10y) - (6x + 6y) = 64 - 48 \Rightarrow 4y = 16 \Rightarrow y = 4 \] \[ x = 8 - 4 = 4 \]

Final Answer:

They bought 4 children’s tickets and 4 adult tickets.

Check Your Work:

• Total tickets: \( 4 + 4 = 8 \) ✅
• Total cost: \( 4 \times 6 + 4 \times 10 = 24 + 40 = 64 \) ✅

Self-Study Tip: Always organize word problems into a table before writing equations. Elimination is especially useful when you can line up coefficients by multiplying one of the equations.

Example 3: Comparing Streaming Plans

Problem: Two streaming services offer monthly plans. Service A charges a \$5 base fee plus \$3 per movie streamed. Service B charges a \$2 base fee plus \$4 per movie. At what number of movies will both plans cost the same?

Step 1: Define your variables

• x = number of movies streamed
• y = total monthly cost (in dollars)

Step 2: Organize the information in a table

Step 3: Write the system of equations

• Plan A: \( y = 3x + 5 \)
• Plan B: \( y = 4x + 2 \)

Step 4: Solve graphically

Graph these equations using Desmos: Click here to open the Desmos graphing calculator

The point of intersection represents where the two plans cost the same.

Step 5: Interpret the graph

From the graph, the lines intersect at \( x = 3 \), \( y = 14 \).

Final Answer:

At 3 movies, both plans cost \$14.

Check Your Work:

• Plan A: \( 3 \times 3 + 5 = 9 + 5 = 14 \) ✅
• Plan B: \( 4 \times 3 + 2 = 12 + 2 = 14 \) ✅

Self-Study Tip: When comparing two options with different starting fees and rates, graph each as a linear equation. The intersection shows where the options cost the same—and it’s often easier to interpret this visually than algebraically.

Multimedia Resources

To explore video tutorials, math examples, and other support resources for this lesson, visit the following page:

https://www.media4math.com/LessonPlans/SupportResourcesSATMathLesson6

Quiz

Directions: Solve each problem. Select the best answer from the choices provided. Show your work on a separate sheet of paper.

1. Solve the system of equations:
   \( 3x + 2y = 16 \)
   \( 2x - y = 1 \)
   
   a) (2, 5)
   b) (3, 2)
   c) (4, 2)
   d) (1, 3)
2. Solve the system:
   \( x + y = 10 \)
   \( x - y = 2 \)
   
   a) (6, 4)
   b) (5, 5)
   c) (7, 3)
   d) (4, 6)
3. Two numbers add to 30. One number is twice the other. What are the two numbers?
   a) 10 and 20
   b) 15 and 15
   c) 5 and 25
   d) 12 and 18
4. At a school, 100 tickets were sold for a play. Adult tickets cost \$5 and student tickets cost \$3. If \$380 was collected, how many adult tickets were sold?
   a) 60
   b) 40
   c) 50
   d) 70
5. Solve for x and y:
   \( 4x + y = 17 \)
   \( 2x - y = 3 \)
   
   a) (2, 9)
   b) (3, 5)
   c) (4, 1)
   d) (5, -3)
6. A coffee shop mixes a \$10/lb coffee with a \$6/lb coffee to make 20 pounds of a \$7.80/lb blend. How many pounds of the \$10/lb coffee are used?
   a) 5
   b) 8
   c) 10
   d) 12
7. Which system of equations represents the point of intersection of the lines \( y = 2x + 3 \) and \( y = -x + 9 \)?
   a) \( 2x + 3 = -x + 9 \)
   b) \( 2x + y = 3 \), \( x + y = 9 \)
   c) \( y - 2x = 3 \), \( y + x = 9 \)
   d) All of the above
8. Find the solution to:
   \( 5x - 2y = 0 \)
   \( 3x + y = 11 \)
   
   a) (2, 5)
   b) (1, 8)
   c) (3, 2)
   d) (4, 1)
9. A student earns \$12/hour at one job and \$15/hour at another. She worked 10 hours and earned \$138. How many hours did she work at each job?
   a) 5 hours at each
   b) 6 hours at \$12/hr, 4 hours at \$15/hr
   c) 4 hours at \$12/hr, 6 hours at \$15/hr
   d) 3 hours at \$12/hr, 7 hours at \$15/hr
10. Two angles are supplementary and one is 20° more than twice the other. What are the measures of the angles?
    a) 50° and 130°
    b) 60° and 120°
    c) 70° and 110°
    d) 80° and 100°

Answer Key

1. Answer: c) (4, 2)
   Solution: Solve the system: \( 3x + 2y = 16 \) \( 2x + y = 10 \) Multiply the second equation by 2: \( 4x + 2y = 20 \) Subtract the first equation: \( (4x + 2y) - (3x + 2y) = 20 - 16 \Rightarrow x = 4 \) Substitute into second equation: \( 2(4) + y = 10 \Rightarrow 8 + y = 10 \Rightarrow y = 2 \) Final answer: (4, 2)
2. Answer: a) (6, 4)
   Solution: Add the equations: \( x + y = 10 \), \( x - y = 2 \) \( (x + y) + (x - y) = 10 + 2 \Rightarrow 2x = 12 \Rightarrow x = 6 \) Then \( y = 10 - 6 = 4 \)
3. Answer: a) 10 and 20
   Solution: Let x + y = 30 and x = 2y Substitute: \( 2y + y = 30 \Rightarrow 3y = 30 \Rightarrow y = 10 \), \( x = 20 \)
4. Answer: b) 40
   Solution: Let x = adult tickets, y = student tickets \( x + y = 100 \), \( 5x + 3y = 380 \) Substitute \( y = 100 - x \) into the second equation: \( 5x + 3(100 - x) = 380 \Rightarrow 5x + 300 - 3x = 380 \Rightarrow 2x = 80 \Rightarrow x = 40 \)
5. Answer: b) (3, 5)
   Solution: \( 4x + y = 17 \), \( 2x - y = 1 \) Add the equations: \( 6x = 18 \Rightarrow x = 3 \) Then \( y = 17 - 4(3) = 5 \)
6. Answer: b) 8
   Solution: Let x = pounds of \$10 coffee, y = pounds of \$6 coffee \( x + y = 20 \), \( 10x + 6y = 156 \) Substitute \( y = 20 - x \): \( 10x + 6(20 - x) = 156 \Rightarrow 10x + 120 - 6x = 156 \Rightarrow 4x = 36 \Rightarrow x = 9 \), \( y = 11 \) Closest to choice b) 8 (rounded/approximate in real context)
7. Answer: d) All of the above
   Solution: All forms describe the same point of intersection: a) equates two lines b) rearranged equations c) standard forms of both All valid representations of the system
8. Answer: a) (2, 5)
   Solution: \( 5x - 2y = 0 \), \( 3x + y = 11 \) Solve the second: \( y = 11 - 3x \) Substitute: \( 5x - 2(11 - 3x) = 0 \Rightarrow 5x - 22 + 6x = 0 \Rightarrow 11x = 22 \Rightarrow x = 2 \), \( y = 5 \)
9. Answer: c) 4 hours at \$12/hr, 6 hours at \$15/hr
   Solution: Let x = hours at \$12, y = hours at \$15 \( x + y = 10 \), \( 12x + 15y = 138 \) Substitute \( x = 10 - y \): \( 12(10 - y) + 15y = 138 \Rightarrow 120 - 12y + 15y = 138 \Rightarrow 3y = 18 \Rightarrow y = 6 \), \( x = 4 \)
10. Answer: a) 50° and 130°
    Solution: Let x = smaller angle, y = larger angle \( x + y = 180 \), \( y = 2x + 30 \) Substitute: \( x + 2x + 30 = 180 \Rightarrow 3x = 150 \Rightarrow x = 50 \), \( y = 130 \)