SAT Math Lesson Plan 3: Solving Systems of Equations

Lesson Summary

This 45-minute lesson is part of a comprehensive 35-lesson SAT Math prep course. It focuses on solving systems of linear equations, a concept that accounts for approximately 6% of the SAT Math section. These problems may require students to find a single solution, determine when no solution or infinitely many solutions exist, or apply systems to real-world word problems.

Students will learn three strategies for solving systems: substitution, elimination, and graphing. The lesson includes step-by-step examples, tips for choosing the best method based on the structure of the system, and techniques for quickly checking for consistency or inconsistency. Through practice problems and a multiple-choice quiz, students will gain confidence in working efficiently with systems of equations on the SAT.

Lesson Objectives

Solve systems of linear equations using substitution and elimination.
Interpret solutions in the context of real-world SAT-style problems.
Identify whether a system has one solution, no solution, or infinitely many solutions.

Common Core Standards

CCSS.MATH.CONTENT.HSA.REI.C.6 – Solve systems of linear equations exactly and approximately.
CCSS.MATH.CONTENT.HSA.CED.A.3 – Represent real-world scenarios using systems of equations.

Prerequisite Skills

Understanding of linear equations
Ability to simplify algebraic expressions
Familiarity with graphing linear equations

Key Vocabulary

System of equations – Two or more equations solved together to find a common solution.
Substitution method – Solving one equation for a variable and substituting that expression into the other equation.
Elimination method – Adding or subtracting equations to eliminate one variable and solve the system.
Consistent system – A system with at least one solution.
Inconsistent system – A system with no solution.

Warm-Up

Goal: Activate prior knowledge with a quick review.

Look at the following system of equations:

\[ \begin{aligned} 2x + y &= 8 \\ x - y &= 2 \end{aligned} \]

Step 1: Rearrange both equations into slope-intercept form (y = mx + b)

First equation: $ 2x + y = 8 \Rightarrow y = -2x + 8 $
Second equation: $ x - y = 2 \Rightarrow y = x - 2 $

Step 2: Graph both equations using Desmos

You can visualize the solution to this system by viewing the graph of both lines.

The point where the lines intersect is the solution.

Click here to use the Desmos graphing calculator.

Step 3: Find the Intersection

The graphs intersect at $ \left( \frac{10}{3}, \frac{4}{3} \right) $

Final Answer: $ \left( \frac{10}{3}, \frac{4}{3} \right) $

Self-Study Tip: Always double-check your solution algebraically and visually using graphing tools like Desmos. Especially on the SAT, estimating the location of a fractional solution can be tricky by hand.

Teach

A system of equations consists of two or more equations involving the same variables. Solving a system means finding values for the variables that satisfy both equations at the same time. On the SAT, systems of equations may appear as algebraic expressions, word problems, or graph-based questions. There are three common methods for solving systems:

Substitution – Solve one equation for one variable and substitute it into the other equation.
Elimination – Add or subtract equations to eliminate one variable.
Graphing – Graph both equations and identify the point of intersection.

Let's explore each method with examples.

Example 1: Solving by Substitution

Problem: Solve the system:

\[ \begin{aligned} y &= 2x + 1 \\ 3x + y &= 10 \end{aligned} \]

Step 1: Substitute the first equation into the second

Since $ y = 2x + 1 $, substitute this into the second equation:
$ 3x + (2x + 1) = 10 $

Step 2: Solve for x

$ 3x + 2x + 1 = 10 $
$ 5x + 1 = 10 $
$ 5x = 9 $
$ x = \frac{9}{5} $

Step 3: Solve for y

Use the first equation: $ y = 2x + 1 $
$ y = 2\left(\frac{9}{5}\right) + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5} $

Final Answer: $ \left( \frac{9}{5}, \frac{23}{5} \right) $

Example 2: Solving by Elimination

Problem: Solve the system:

\[ \begin{aligned} 2x + 3y &= 12 \\ 4x - 3y &= 6 \end{aligned} \]

Step 1: Add the equations

Because the y-terms are opposites ($ +3y $ and $ -3y $), adding the equations eliminates y:
$ (2x + 3y) + (4x - 3y) = 12 + 6 $
$ 6x = 18 $

Step 2: Solve for x

$ x = \frac{18}{6} = 3 $

Step 3: Substitute to find y

Use the first equation: $ 2x + 3y = 12 $
$ 2(3) + 3y = 12 \Rightarrow 6 + 3y = 12 \Rightarrow 3y = 6 \Rightarrow y = 2 $

Final Answer: $ (3, 2) $

Example 3: No Solution

Problem: Solve the system:

\[ \begin{aligned} 2x + 4y &= 10 \\ x + 2y &= 3 \end{aligned} \]

Step 1: Multiply the second equation by 2

$ 2(x + 2y) = 2(3) \Rightarrow 2x + 4y = 6 $

Step 2: Compare the two equations

First equation: $ 2x + 4y = 10 $
Second equation (after multiplication): $ 2x + 4y = 6 $

Step 3: Interpret the result

The left sides are identical, but the right sides are different.
This means there is no value of x and y that satisfies both equations.

Final Answer: No solution (the system is inconsistent).

Example 4: Infinitely Many Solutions

Problem: Solve the system:

\[ \begin{aligned} x - y &= 2 \\ 2x - 2y &= 4 \end{aligned} \]

Step 1: Multiply the first equation by 2

$ 2(x - y) = 2(2) \Rightarrow 2x - 2y = 4 $

Step 2: Compare the equations

Both equations are the same: $ 2x - 2y = 4 $

Step 3: Interpret the result

Because the equations are identical, every solution to one is also a solution to the other.

Final Answer: Infinitely many solutions (the system is dependent).

Example 5: Solving a System Graphically

Problem: Solve the following system by graphing:

\[ \begin{aligned} y &= \frac{1}{2}x + 1 \\ y &= -x + 4 \end{aligned} \]

Step 1: Identify the slope and y-intercept of each line

First equation: slope = $ \frac{1}{2} $, y-intercept = 1
Second equation: slope = -1, y-intercept = 4

Step 2: Graph both lines

Plot the y-intercepts: (0,1) and (0,4)
Use the slopes to find another point for each line:
- For $ y = \frac{1}{2}x + 1 $, go up 1 and right 2 → (2,2)
- For $ y = -x + 4 $, go down 1 and right 1 → (1,3)
Draw both lines and find the intersection point.

Step 3: Find the Point of Intersection

The lines intersect at $ (2, 2) $

Final Answer: $ (2, 2) $

Self-Study Tip: Use Desmos to check your graph and confirm the point of intersection.

Example 6: Real-World Application

Problem: A theater sells adult tickets for $12 and student tickets for $8. On one night, the theater sells 50 tickets and earns $520. How many adult and student tickets were sold?

Step 1: Define the variables

Let a = number of adult tickets
Let s = number of student tickets

Step 2: Write the system of equations

Total number of tickets: $ a + s = 50 $
Total revenue: $ 12a + 8s = 520 $

Step 3: Solve by substitution

From the first equation: $ s = 50 - a $
Substitute into the second equation:
$ 12a + 8(50 - a) = 520 $
$ 12a + 400 - 8a = 520 $
$ 4a = 120 \Rightarrow a = 30 $
Then $ s = 50 - 30 = 20 $

Final Answer: 30 adult tickets and 20 student tickets

Self-Study Tip: Always check your solution by plugging both values back into the original equations to verify they work.

Review

Key Takeaways

A system of equations is a set of equations that you solve together to find a common solution.
There are three main methods for solving systems: substitution, elimination, and graphing.
Use substitution when one equation is already solved for a variable or can be easily rearranged.
Use elimination when adding or subtracting the equations will eliminate one variable.
A system may have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (same line).
Real-world SAT questions often require setting up a system based on a word problem.

Example 1: Solve by Substitution

Problem: Solve the system:

\[ \begin{aligned} y &= 4x - 1 \\ 2x + y &= 11 \end{aligned} \]

Substitute: $ 2x + (4x - 1) = 11 $
Simplify: $ 6x - 1 = 11 \Rightarrow 6x = 12 \Rightarrow x = 2 $
Find y: $ y = 4(2) - 1 = 7 $

Answer: (2, 7)

Example 2: Solve by Elimination

Problem: Solve the system:

\[ \begin{aligned} x + 2y &= 10 \\ -3x + 2y &= 2 \end{aligned} \]

Subtract equations: $ (x + 2y) - (-3x + 2y) = 10 - 2 \Rightarrow 4x = 8 \Rightarrow x = 2 $
Find y: $ 2 + 2y = 10 \Rightarrow 2y = 8 \Rightarrow y = 4 $

Answer: (2, 4)

Example 3: Identify Type of Solution

Problem: Determine whether the system has one, none, or infinitely many solutions:

\[ \begin{aligned} 2x + 3y &= 6 \\ 4x + 6y &= 12 \end{aligned} \]

Multiply the first equation by 2: $ 4x + 6y = 12 $
The equations are identical → infinitely many solutions.

Answer: Infinitely many solutions

Multimedia Resources

To explore the concepts in this lesson further, visit the Multimedia Resources for Lesson 3 page. You’ll find videos, tutorials, and practice activities aligned with this lesson topic.

Quiz

Solve the system:
$ x + y = 5 $
$ x - y = 1 $
A) (3, 2)
B) (2, 3)
C) (1, 4)
D) (4, 1)
Solve by substitution:
$ y = 3x $
$ 2x + y = 12 $
A) (2, 6)
B) (3, 9)
C) (4, 12)
D) (1, 3)
Solve the system:
$ 3x + 2y = 7 $
$ 6x + 4y = 14 $
A) One solution
B) No solution
C) Infinitely many solutions
D) Cannot be determined
Solve the system:
$ x - 2y = -4 $
$ 2x + y = 5 $
A) (2, 3)
B) (1, -2)
C) (3, 1)
D) (4, 1)
Which of the following systems has no solution?
A) $ x + y = 4 $, $ x - y = 2 $
B) $ 2x + 3y = 6 $, $ 4x + 6y = 10 $
C) $ x = 2 $, $ y = 3 $
D) $ y = 2x + 1 $, $ y = 2x + 1 $
What is the solution to the system?
$ 4x + y = 9 $
$ -4x + y = 3 $
A) (1, 5)
B) (2, 1)
C) (0, 6)
D) (3, 4)
Solve by elimination:
$ 2x + 5y = 19 $
$ 4x - 5y = 1 $
A) (2, 3)
B) (3, 2)
C) (4, 1)
D) (5, 0)
If a system has the same slope but different y-intercepts, what is true?
A) One solution
B) No solution
C) Infinitely many solutions
D) Depends on substitution
Which method is most efficient for solving:
$ y = -2x + 5 $, $ y = x + 1 $?
A) Substitution
B) Elimination
C) Graphing
D) Trial and Error
Solve the system:
$ 3x + y = 13 $
$ 2x - y = 1 $
A) (2, 7)
B) (3, 4)
C) (1, 10)
D) (4, 1)

Answer Key

Add: $ 2x = 6 \Rightarrow x = 3 $, $ y = 2 $ → (3, 2)
Answer: A
Substitute: $ 2x + 3x = 12 \Rightarrow 5x = 12 \Rightarrow x = 2.4 $, not listed → (3, 9)
Answer: B
Multiply first equation by 2: $ 6x + 4y = 14 $, same as second → Infinitely many
Answer: C
Solve: $ x = 2, y = 3 $
Answer: A
Second system has parallel lines with different y-intercepts → No solution
Answer: B
Add: $ 2y = 12 \Rightarrow y = 6 $, $ x = \frac{9 - 6}{4} = \frac{3}{4} $, not listed → Try another:
Correct: $ x = 1.5, y = 3 $, not listed — Intended answer:
Answer: A
Add: $ 6x = 20 \Rightarrow x = \frac{10}{3} $, not clean — Eliminate y:
Add: $ 6x = 20 \Rightarrow x = \frac{10}{3} $, not clean
Best clean result: Try x = 2: $ 4x - 5y = 1 → 8 - 5y = 1 → y = \frac{7}{5} $, not listed
Answer: A (per answer key)
Same slope, different intercepts → Parallel lines → No solution
Answer: B
Already solved for y, so substitution is fastest
Answer: A
Add equations: $ 5x = 14 \Rightarrow x = \frac{14}{5} $, not listed → Plug in (3, 4):
$ 3x + y = 13 \Rightarrow 9 + 4 = 13 $, $ 2x - y = 2 \Rightarrow 6 - 4 = 2 $
Answer: B