MGU Mathematics Series — Algebra I
Part III: Functions, Graphs, and Linear Models
Systems · Graphing · Substitution · Elimination · Inequalities · Feasible Regions
Chapter 8 of 14
MGU Mathematics Series  ·  Algebra I  ·  Library Textbook Edition

Chapter 8:
Systems of Linear
Equations and Inequalities

Simultaneous conditions, intersecting lines, substitution, elimination, graphing, linear inequalities in two variables, and feasible regions — the algebra of multiple constraints

■ Math Level: Algebra I ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapter 7: Linear Functions and Equations  ·  Prepares students for Part IV: Exponents, Polynomials, and Factoring

Chapter 8 — Front Matter

Purpose and Overview

Most situations in mathematics, science, and everyday life do not involve a single condition. A business must consider both cost and revenue. A traveler plans a route that satisfies both time and distance limits. A scientist studies two variables that must both satisfy a physical law. When two or more conditions hold at the same time, mathematics captures that situation as a system.

Chapter 8 teaches you to think about multiple conditions simultaneously. You will learn to find the values that satisfy two equations at once, recognize when no such values exist, and discover when infinitely many values work. You will extend this thinking to inequalities — situations where solutions are not single points but entire regions of a plane.

By the end of this chapter, you should be able to solve systems by graphing, substitution, and elimination; interpret what each type of solution means; write and solve systems from real situations; graph linear inequalities in two variables; and describe feasible regions defined by systems of inequalities.

Opening Question

A small theater sells two types of tickets: student tickets for $6 each and adult tickets for $10 each. On one evening, 200 tickets are sold and the total revenue is $1,520. How many student tickets were sold, and how many adult tickets were sold?

There are two unknowns here. You could write one equation for the total number of tickets and another for the total revenue. You need both equations to find both unknowns. That is a system of equations. By the end of this chapter, you will have the tools to solve this problem — and many others like it.

14Sections
11Objectives
12Key Terms
15+Worked Examples

Why This Chapter Matters

Linear equations in one variable give you a single condition with a single unknown. But most real problems involve at least two unknowns and at least two conditions. Systems are not just a tool for harder problems — they are a more accurate model of how the world actually works.

Understanding systems prepares you for Algebra II systems with three or more variables; geometry problems involving intersecting lines; science applications such as mixture problems and equilibrium models; economics and finance applications such as break-even analysis and budget constraints; computer science, where constraint systems are central to optimization; and statistics, where regression lines model relationships between two variables.

Learning Objectives

  1. 1Explain what a system of linear equations represents.
  2. 2Determine whether an ordered pair is a solution to a system.
  3. 3Solve a system of linear equations by graphing.
  4. 4Identify systems with one solution, no solution, and infinitely many solutions.
  5. 5Solve a system of linear equations by substitution.
  6. 6Solve a system of linear equations by elimination.
  7. 7Choose an appropriate solution method based on the structure of the system.
  8. 8Write and solve systems of equations from word problems.
  9. 9Graph a linear inequality in two variables.
  10. 10Graph a system of linear inequalities and identify the feasible region.
  11. 11Recognize and correct common mistakes when working with systems.
Chapter 8 — Contents

Sections at a Glance

Sections 8.1 – 8.5
Understanding and Graphing Systems
System Notation · Solutions · Graphing · One/No/Infinite Solutions
Section 8.1

What a System Represents

One Condition Is Not Always Enough

Throughout Chapters 3 and 7, you worked with one equation in one or two variables. A single linear equation in two variables has infinitely many solutions — every point on the line satisfies it.

But what if you know more than one thing about a situation? Suppose you know that two numbers add up to 10. That gives you x + y = 10. There are infinitely many pairs that satisfy this — for example, (3, 7), (0, 10), or (5.5, 4.5). You cannot identify a single unique answer yet.

Now suppose you also know that one number is four more than twice the other: y = 2x + 4. Each equation alone has infinitely many solutions. But together, they describe two separate conditions that must both hold at the same time. The pair (x, y) you are looking for must satisfy both equations simultaneously. When you bring two equations together and ask for a common solution, you have created a system of linear equations.

System Notation

A system is written using a left brace to indicate the equations are considered together:

{ x + y = 10

{ y = 2x + 4

The word simultaneously is critical. Solving a system means finding a pair of values that makes both equations true at the same time — not solving each equation separately.

Graphical Meaning

Every linear equation in two variables corresponds to a line in the coordinate plane. A system of two equations corresponds to two lines. The solution, if it exists, is the point where those two lines intersect — the one point that lies on both lines simultaneously.

Practice 8.1Concept Check
  1. In your own words, explain what it means for two equations to be a system. What does "simultaneously" mean?
  2. Why does a single equation in two variables have infinitely many solutions, but a system of two equations may have only one?
  3. Describe what the solution to a system of two linear equations looks like on a coordinate plane.
Section 8.2

Solutions to Systems

An ordered pair (x, y) is a solution to a system if and only if it satisfies every equation in the system. Satisfying one equation but not the other is not enough.

Example 8.2.1
Checking a proposed solution
  • ProblemDetermine whether (2, 5) is a solution to the system: 3x + y = 11 and x − y = −3.
  • ThinkSubstitute x = 2 and y = 5 into both equations and check each one.
  • Solve Check equation 1: 3(2) + 5 = 6 + 5 = 11 ✓ Check equation 2: 2 − 5 = −3 ✓
  • CheckBoth equations are satisfied when x = 2, y = 5.
Answer: Yes, (2, 5) is a solution to the system.
Example 8.2.2
A pair that fails one equation
  • ProblemDetermine whether (4, 1) is a solution to the system: 2x − y = 7 and x + 3y = 10.
  • ThinkSubstitute and check both equations. Do not stop after the first one.
  • Solve Check equation 1: 2(4) − 1 = 8 − 1 = 7 ✓ Check equation 2: 4 + 3(1) = 4 + 3 = 7 ≠ 10 ✗
  • CheckThe first equation is satisfied, but the second is not.
Answer: No, (4, 1) is not a solution. It satisfies only one equation.
Critical Caution

Do not stop checking after the first equation is satisfied. A solution to a system must satisfy every equation. Students who check only one equation often report wrong answers.

Practice 8.2Checking Solutions

Determine whether each ordered pair is a solution to the given system. Show all substitution steps.

  1. (3, 2): x + y = 5, x − y = 1
  2. (1, 4): 2x + y = 6, x − y = −3
  3. (0, 0): 3x − y = 0, 2x + 5y = 0
  4. (−1, 3): x + 2y = 5, 3x − y = −6
  5. (5, −2): 4x + 3y = 14, x − y = 7
  6. (2, 2): y = x, x + y = 4
Section 8.3

Solving Systems by Graphing

One natural way to solve a system is to graph both lines and look for their intersection. The point of intersection is the solution — the one point that lies on both lines. The graphing method is especially useful for visualizing the relationship between two equations, or for estimating a solution quickly. It is less precise when the intersection involves fractional coordinates.

Procedure: Solving by Graphing

1. Write both equations in slope-intercept form y = mx + b if not already done.

2. Graph both lines on the same coordinate plane.

3. Identify the point where the lines intersect.

4. Check the intersection point in both original equations.

5. Write the solution as an ordered pair.

Example 8.3.1
Solving by graphing — slope-intercept form
  • ProblemSolve the system: y = 2x − 1 and y = −x + 5.
  • ThinkBoth equations are already in slope-intercept form. Graph each line using the y-intercept and slope, then find where they cross.
  • Solve y = 2x − 1: y-intercept (0, −1), slope 2 → also (1, 1) y = −x + 5: y-intercept (0, 5), slope −1 → also (1, 4)
  • Check At (2, 3): y = 2(2) − 1 = 3 ✓ y = −(2) + 5 = 3 ✓
Answer: (2, 3). Both x = 2 and y = 3 satisfy both equations simultaneously.
Diagram Instruction — Example 8.3.1

Draw a coordinate plane. Plot the line y = 2x − 1 using points (0, −1) and (1, 1), extending in both directions. Label it. Plot y = −x + 5 using points (0, 5) and (1, 4), extending in both directions. Label it. Mark the intersection at (2, 3) with a filled dot and label it "solution: (2, 3)".

Example 8.3.2
Rewriting before graphing
  • ProblemSolve: 2x + y = 6 and x − y = 0.
  • ThinkRewrite both in slope-intercept form before graphing.
  • Solve 2x + y = 6 → y = −2x + 6 x − y = 0 → y = x y = −2x + 6: y-intercept (0, 6), slope −2 y = x: y-intercept (0, 0), slope 1 Lines intersect at (2, 2).
  • Check 2(2) + 2 = 6 ✓ 2 − 2 = 0 ✓
Answer: (2, 2).
Always Check Algebraically

A graph gives a visual estimate, but coordinates can be hard to read precisely. Always verify the answer you read from the graph by substituting into both original equations.

Practice 8.3Solving by Graphing

For each system: rewrite in slope-intercept form if needed, graph both lines, identify the intersection, and check your answer algebraically.

  1. y = x + 2 and y = −x + 6
  2. y = 3x − 4 and y = x + 2
  3. y = 2x and x + y = 6
  4. 2x − y = 3 and x + y = 6
  5. y = −x + 5 and y = 2x − 1
  6. 3x + y = 7 and x − y = 1
Section 8.4

Intersecting Lines

When the two lines in a system have different slopes, they will always intersect at exactly one point. This is the most common situation in Algebra I systems, and it is called an independent, consistent system. Consistent means a solution exists. Independent means the solution is unique.

Geometrically, two lines with different slopes cannot be parallel. They must cross at one and only one point in the entire coordinate plane.

Example 8.4.1
Confirming an intersection from slopes
  • ProblemWithout graphing, determine whether the system y = 3x + 1 and y = −2x + 6 has a unique solution.
  • ThinkCompare the slopes of both equations.
  • Solve First equation: slope = 3 Second equation: slope = −2 The slopes are different.
  • CheckDifferent slopes → the lines are not parallel → they must intersect at exactly one point.
Answer: The system has exactly one solution. The lines intersect at one point in the coordinate plane.
Section 8.5

No Solution and Infinitely Many Solutions

Parallel Lines: No Solution

When two lines have the same slope but different y-intercepts, they are parallel. Parallel lines never intersect. There is no ordered pair that lies on both lines simultaneously. A system with no solution is called inconsistent.

If you try to solve an inconsistent system algebraically, you will reach a contradiction — a false statement such as 0 = 5. That contradiction is the algebraic signal that no solution exists.

Example 8.5.1
Recognizing no solution
  • ProblemDetermine the number of solutions for: y = 2x + 3 and y = 2x − 4.
  • ThinkCompare slopes and y-intercepts.
  • Solve Both equations have slope 2. y-intercepts are 3 and −4 — different. → The lines are parallel.
  • CheckSubstituting: 2x + 3 = 2x − 4 → 3 = −4. This is a contradiction.
Answer: No solution. The system is inconsistent. The lines are parallel and never intersect.

Coincident Lines: Infinitely Many Solutions

When two equations describe the same line — the same slope and the same y-intercept — every point on that line is a solution to both equations. The system has infinitely many solutions and is called a dependent system.

Algebraically, the process produces a statement that is always true, such as 0 = 0, meaning the two equations are equivalent.

Example 8.5.2
Recognizing infinitely many solutions
  • ProblemDetermine the number of solutions for: y = 3x − 2 and 6x − 2y = 4.
  • ThinkRewrite the second equation in slope-intercept form and compare.
  • Solve 6x − 2y = 4 −2y = −6x + 4 y = 3x − 2
  • CheckBoth equations are y = 3x − 2. They describe the same line. Substituting: 4 = 4. Always true.
Answer: Infinitely many solutions. The equations describe the same line. Every point on y = 3x − 2 is a solution.

Summary: The Three Cases

Slopesy-InterceptsLinesSolutionsSystem Type
DifferentAnyIntersectingExactly oneConsistent, independent
SameDifferentParallelNoneInconsistent
SameSameCoincidentInfinitely manyConsistent, dependent
Do Not Confuse the Signals

A contradiction like 3 = −4 or 0 = 5 means no solution. A true statement like 0 = 0 or 4 = 4 means infinitely many solutions. These signals go in opposite directions. Check your algebra when you get either result — but if the work is correct, trust the signal.

Practice 8.4 & 8.5Identifying the Number of Solutions

Without solving, determine whether each system has one solution, no solution, or infinitely many solutions. Justify your answer by comparing slopes and y-intercepts.

  1. y = 5x + 2 and y = 5x + 7
  2. y = 3x − 1 and y = −x + 3
  3. y = 2x − 4 and 4x − 2y = 8
  4. y = −4x and y = −4x + 1
  5. 3x + y = 6 and 6x + 2y = 12
  6. y = x + 1 and y = −x − 1
Sections 8.6 – 8.9
Algebraic Methods and Applications
Substitution · Elimination · Choosing a Method · Word Problems
Section 8.6

Solving Systems by Substitution

Graphing shows you the solution visually, but it has limits. If the intersection involves fractions or decimals, reading it from a graph is imprecise. The substitution method uses one equation to express one variable in terms of the other, then substitutes that expression into the second equation. The result is a single equation in one variable, which you already know how to solve.

Procedure: Solving by Substitution

1. Choose one equation and solve it for one variable.

2. Substitute the expression into the other equation.

3. Solve the resulting one-variable equation.

4. Substitute the value back to find the other variable.

5. Write the solution as an ordered pair.

6. Check in both original equations.

Example 8.6.1
Basic substitution — equation already solved for y
  • ProblemSolve: y = 3x − 1 and 2x + y = 9.
  • ThinkThe first equation is already solved for y. Substitute 3x − 1 in place of y in the second equation.
  • Solve 2x + (3x − 1) = 9 5x − 1 = 9 5x = 10 x = 2 Find y: y = 3(2) − 1 = 5
  • Check Equation 1: 5 = 3(2) − 1 = 5 ✓ Equation 2: 2(2) + 5 = 9 ✓
Answer: (2, 5).
Example 8.6.2
Solving for a variable first
  • ProblemSolve: 3x + y = 7 and x − 2y = −4.
  • ThinkNeither equation is solved for a variable. Solve the first equation for y — it has a coefficient of 1 on y.
  • Solve From equation 1: y = 7 − 3x Substitute into equation 2: x − 2(7 − 3x) = −4 x − 14 + 6x = −4 7x = 10 x = 10/7 y = 7 − 3(10/7) = 49/7 − 30/7 = 19/7
  • Check Equation 1: 3(10/7) + 19/7 = 30/7 + 19/7 = 49/7 = 7 ✓ Equation 2: 10/7 − 2(19/7) = 10/7 − 38/7 = −28/7 = −4 ✓
Answer: (10/7, 19/7).
Example 8.6.3
Substitution revealing no solution
  • ProblemSolve: y = 4x + 2 and 8x − 2y = 3.
  • ThinkSubstitute the first equation into the second.
  • Solve 8x − 2(4x + 2) = 3 8x − 8x − 4 = 3 −4 = 3
  • CheckThis is a contradiction. The algebra is correct.
Answer: No solution. The system is inconsistent; the two lines are parallel.
Choosing Which Variable to Isolate

Choose the variable that is easiest to isolate — the one with no coefficient or with a coefficient of 1 or −1. This minimizes fractions in your work and keeps the algebra cleaner.

Practice 8.6Substitution Practice
  1. y = 5x − 2 and 3x + y = 10
  2. x = 2y + 1 and 3x − 4y = 7
  3. 2x + y = 8 and y = x − 1
  4. x + 3y = 10 and 2x − y = 5
  5. y = −3x + 4 and 6x + 2y = 8
  6. 4x − y = 11 and y = 2x + 1
Section 8.7

Solving Systems by Elimination

The elimination method — also called the addition method — removes one variable by adding or subtracting the two equations. If the coefficients of one variable are opposites, adding the equations eliminates that variable entirely. This method is especially powerful when both equations are in standard form Ax + By = C.

Procedure: Solving by Elimination

1. Align both equations with like terms in the same columns.

2. Multiply one or both equations by constants so the coefficients of one variable are opposites.

3. Add the equations to eliminate that variable.

4. Solve the resulting one-variable equation.

5. Substitute back to find the other variable.

6. Write the solution. Check in both original equations.

Example 8.7.1
Direct elimination — coefficients already opposite
  • ProblemSolve: 3x + 2y = 12 and 3x − 2y = 0.
  • ThinkThe y-coefficients are +2 and −2. They are already opposites. Add the equations directly.
  • Solve (3x + 2y) + (3x − 2y) = 12 + 0 6x = 12 x = 2 Substitute: 3(2) + 2y = 12 → 2y = 6 → y = 3
  • Check 3(2) + 2(3) = 6 + 6 = 12 ✓ 3(2) − 2(3) = 6 − 6 = 0 ✓
Answer: (2, 3).
Example 8.7.2
Multiplying one equation before eliminating
  • ProblemSolve: 2x + 3y = 13 and 5x + y = 17.
  • ThinkEliminate y. Multiply the second equation by −3 to make the y-coefficients opposites (+3 and −3).
  • Solve Multiply equation 2 by −3: −3(5x + y) = −3(17) → −15x − 3y = −51 Add to equation 1: (2x + 3y) + (−15x − 3y) = 13 + (−51) −13x = −38 x = 38/13 Find y: 5(38/13) + y = 17 190/13 + y = 17 y = 221/13 − 190/13 = 31/13
  • Check 2(38/13) + 3(31/13) = 76/13 + 93/13 = 169/13 = 13 ✓ 5(38/13) + 31/13 = 190/13 + 31/13 = 221/13 = 17 ✓
Answer: (38/13, 31/13).
Example 8.7.3
Multiplying both equations
  • ProblemSolve: 3x + 4y = 10 and 5x − 6y = 12.
  • ThinkTo eliminate y, use LCM(4, 6) = 12. Multiply equation 1 by 3 and equation 2 by 2.
  • Solve Equation 1 × 3: 9x + 12y = 30 Equation 2 × 2: 10x − 12y = 24 Add: 19x = 54 → x = 54/19 3(54/19) + 4y = 10 162/19 + 4y = 190/19 4y = 28/19 → y = 7/19
  • Check 3(54/19) + 4(7/19) = 162/19 + 28/19 = 190/19 = 10 ✓ 5(54/19) − 6(7/19) = 270/19 − 42/19 = 228/19 = 12 ✓
Answer: (54/19, 7/19).
Example 8.7.4
Elimination revealing infinitely many solutions
  • ProblemSolve: 4x − 2y = 8 and −2x + y = −4.
  • ThinkMultiply the second equation by 2 to create opposite coefficients.
  • Solve Equation 2 × 2: −4x + 2y = −8 Add: (4x − 2y) + (−4x + 2y) = 8 + (−8) 0 = 0
  • Check0 = 0 is always true. The equations are equivalent.
Answer: Infinitely many solutions. Every point on 4x − 2y = 8 is a solution.
Multiply Every Term

When multiplying an equation by a constant, multiply every term — including the constant on the right side. A common error is multiplying only the left side. For example, multiplying 3x − y = 7 by −2 must give −6x + 2y = −14, not −6x + 2y = 7.

Practice 8.7Elimination Practice
  1. x + y = 8 and x − y = 2
  2. 3x + 2y = 14 and 3x − y = 5
  3. 2x + 5y = 3 and 4x − y = 11
  4. 5x + 3y = 7 and 2x + 6y = 10
  5. 4x − 3y = 2 and −8x + 6y = −4
  6. 6x + 4y = 20 and 3x − 2y = 4
Section 8.8

Choosing a Solution Method

You now have three methods for solving systems: graphing, substitution, and elimination. Each has advantages. Part of mathematical fluency is choosing the right tool for a given situation.

SituationRecommended Method
One equation is already solved for x or ySubstitution
One variable has coefficient 1 or −1Substitution
Coefficients are already oppositesElimination
Both equations in standard form Ax + By = CElimination
You need a visual estimate or want to check reasonablenessGraphing
Solution involves large or fractional coordinatesElimination or Substitution (not graphing)
Example 8.8.1
Choosing and solving efficiently
  • ProblemSolve: y = −5x + 3 and 2x + 3y = 1.
  • ThinkThe first equation is already solved for y. Use substitution.
  • Solve 2x + 3(−5x + 3) = 1 2x − 15x + 9 = 1 −13x = −8 x = 8/13 y = −5(8/13) + 3 = −40/13 + 39/13 = −1/13
  • Check −1/13 = −5(8/13) + 3 = −40/13 + 39/13 = −1/13 ✓ 2(8/13) + 3(−1/13) = 16/13 − 3/13 = 13/13 = 1 ✓
Answer: (8/13, −1/13).
Practice 8.8Choosing a Method

For each system, state which method you would choose and briefly explain why. Then solve.

  1. y = 2x + 1 and 3x − y = 4    [state method first]
  2. 4x + 3y = 10 and −4x + y = 6    [state method first]
  3. x − y = 2 and 2x + y = 7    [state method first]
  4. 5x + 2y = 11 and 3x − 4y = 1    [state method first]
Section 8.9

Systems from Word Problems

Word problems involving two unknowns often have two distinct conditions built in. Your job is to define your variables clearly, write one equation for each condition, solve the system, and check the solution in context. This follows the Read → Represent → Solve → Check → Explain method used throughout this course.

Example 8.9.1
The ticket problem (opening question)
  • ReadTwo unknowns: student tickets and adult tickets. Two conditions: 200 tickets total; $1,520 revenue total.
  • Represent Let s = number of student tickets ($6 each) Let a = number of adult tickets ($10 each) s + a = 200 (total tickets) 6s + 10a = 1520 (total revenue)
  • Solve From equation 1: s = 200 − a Substitute: 6(200 − a) + 10a = 1520 1200 − 6a + 10a = 1520 4a = 320 a = 80 s = 200 − 80 = 120
  • Check Total tickets: 120 + 80 = 200 ✓ Total revenue: 6(120) + 10(80) = 720 + 800 = 1520 ✓
Answer: 120 student tickets and 80 adult tickets were sold.
Example 8.9.2
A mixture problem
  • ReadMix a 20% acid solution and a 50% acid solution to make 300 mL of 30% acid solution.
  • Represent Let x = mL of 20% solution Let y = mL of 50% solution x + y = 300 (total volume) 0.20x + 0.50y = 90 (acid: 30% of 300 = 90 mL)
  • Solve x = 300 − y 0.20(300 − y) + 0.50y = 90 60 − 0.20y + 0.50y = 90 0.30y = 30 y = 100 x = 300 − 100 = 200
  • Check Total: 200 + 100 = 300 ✓ Acid: 0.20(200) + 0.50(100) = 40 + 50 = 90 ✓
Answer: 200 mL of 20% solution and 100 mL of 50% solution.
Example 8.9.3
A rate problem
  • ReadTwo trains leave the same station traveling in opposite directions at 55 mph and 75 mph. After how many hours will they be 390 miles apart?
  • Represent Let t = hours of travel (same for both trains). Train 1 travels 55t miles; Train 2 travels 75t miles. 55t + 75t = 390
  • Solve 130t = 390 t = 3
  • Check 55(3) + 75(3) = 165 + 225 = 390 ✓
Answer: The trains will be 390 miles apart after 3 hours.
Strategy for Word Problems

Identify what you do not know and name your variables explicitly. Look for exactly two conditions — typically a total count and a total value, a total amount and a total concentration, or two rate conditions. Write one equation per condition. Solve and check in context.

Practice 8.9Systems Word Problems

Define your variables, write the system, solve, check, and answer the question in context.

  1. The sum of two numbers is 45. Their difference is 9. Find the two numbers.
  2. One number is three times another. Their sum is 48. Find both numbers.
  3. Adult movie tickets cost $12 and student tickets cost $8. A group of 15 people paid a total of $148. How many adults and students were in the group?
  4. A store sells small gift bags for $4 and large gift bags for $7. A customer buys 9 bags for a total of $48. How many of each size?
  5. A farmer plants two crops. Crop A takes 3 labor-hours per acre, Crop B takes 5. The farmer has 90 labor-hours and plants 22 acres total. How many acres of each crop?
  6. A pharmacist mixes a 15% saline solution and a 35% saline solution to make 200 mL of a 25% solution. How many mL of each?
Sections 8.10 – 8.13
Linear Inequalities and Feasible Regions
Inequalities in Two Variables · Graphing · Systems of Inequalities · Feasible Regions
Section 8.10

Linear Inequalities in Two Variables

Everything so far in this chapter involved equations — statements of equality. But many real situations do not require exact equality. A budget says "spend at most $500." A building code says "allow at least 2 square feet per person." These are inequalities.

A linear inequality in two variables compares a linear expression in x and y to a number using <, >, ≤, or ≥. Unlike a linear equation, which corresponds to a line, a linear inequality corresponds to a half-plane — one entire side of a line in the coordinate plane.

Boundary Lines and Half-Planes

Every linear inequality has a corresponding equation called the boundary line. For y ≤ 2x + 3, the boundary line is y = 2x + 3. The boundary line divides the plane into two half-planes. One half-plane satisfies the inequality; the other does not.

≤ or ≥: boundary line IS included → draw a solid line

< or >: boundary line is NOT included → draw a dashed line

Testing a Point

To determine which half-plane satisfies the inequality, test a point not on the boundary line. The point (0, 0) is usually the easiest to test, unless the boundary line passes through the origin — in that case, use (1, 0) or (0, 1) instead.

Section 8.11

Graphing Linear Inequalities

Procedure: Graphing a Linear Inequality

1. Replace the inequality symbol with = and graph the boundary line (solid or dashed).

2. Choose a test point not on the boundary line.

3. Substitute the test point into the original inequality.

4. If true, shade the half-plane containing the test point. If false, shade the other half-plane.

5. Label the boundary line with its equation.

Example 8.11.1
Graphing a non-strict inequality
  • ProblemGraph: y ≤ 2x − 1.
  • ThinkBoundary line: y = 2x − 1. Symbol is ≤, so use a solid line. Test (0, 0).
  • Solve Graph y = 2x − 1 as a solid line. y-intercept: (0, −1), slope: 2 Test (0, 0): 0 ≤ 2(0) − 1 = −1 → False Shade the half-plane NOT containing (0, 0): shade below and to the right of the line.
  • CheckTest a point in the shaded region, e.g. (2, 0): 0 ≤ 2(2) − 1 = 3. True. ✓
Answer: The solution is all points on or below the line y = 2x − 1, shown by a solid boundary with shading below.
Diagram Instruction — Example 8.11.1

Draw a coordinate plane. Draw a solid line for y = 2x − 1, passing through (0, −1) and (1, 1). Shade the region below the line. Mark (0, 0) with an open circle labeled "not in solution." Label the shaded region "solution region."

Example 8.11.2
Graphing a strict inequality
  • ProblemGraph: 3x + 2y > 6.
  • ThinkBoundary line: 3x + 2y = 6. Symbol is >, so use a dashed line. Test (0, 0).
  • Solve Intercepts of boundary line: x = 0: y = 3 → point (0, 3) y = 0: x = 2 → point (2, 0) Draw a dashed line through (0, 3) and (2, 0). Test (0, 0): 3(0) + 2(0) = 0 > 6 → False Shade the half-plane NOT containing (0, 0): shade above the line.
  • CheckTest (0, 4): 3(0) + 2(4) = 8 > 6. True. ✓
Answer: All points strictly above the line 3x + 2y = 6. The boundary is dashed — points on the line are not included.
Diagram Instruction — Example 8.11.2

Draw a coordinate plane. Draw a dashed line through (0, 3) and (2, 0). Shade the region above the line. Mark (0, 0) as not in the solution. Label the shaded region "solution region."

Solid vs. Dashed — A Common Error

Using a solid line for a strict inequality (< or >) is one of the most frequent graphing mistakes. Points on the boundary do not satisfy a strict inequality. Draw a dashed line to show that the boundary is excluded.

Practice 8.11Graphing Inequalities

Graph each inequality. Show the boundary line (solid or dashed), your test point, and the shaded region.

  1. y > 3x − 2
  2. y ≤ −x + 4
  3. 2x + y < 6
  4. 3x − 4y ≥ 12
  5. y ≥ −2x + 1
  6. x − y < 3
Section 8.12

Systems of Linear Inequalities

A system of linear inequalities consists of two or more inequalities that must be satisfied at the same time. A solution to such a system is any ordered pair that satisfies every inequality simultaneously. Graphically, you graph each inequality as a shaded half-plane. The solution is the region where all shadings overlap — the intersection of the individual solution regions.

Procedure: Graphing a System of Inequalities

1. Graph the first inequality (boundary, shading).

2. On the same coordinate plane, graph the second inequality.

3. Identify the region where both shadings overlap.

4. Test a point in the overlap region in both inequalities to verify.

5. Shade the overlap clearly and label it the solution region.

Example 8.12.1
A system of two inequalities
  • ProblemGraph the system: y ≥ x + 1 and y < −2x + 6.
  • ThinkGraph each inequality separately, then find the overlapping region.
  • Solve y ≥ x + 1: Boundary y = x + 1, solid line, y-int (0,1), slope 1 Test (0,0): 0 ≥ 1 → False → shade above the line y < −2x + 6: Boundary y = −2x + 6, dashed line, y-int (0,6), slope −2 Test (0,0): 0 < 6 → True → shade below the line Solution: region above the solid line AND below the dashed line.
  • Check Test point (1, 3): y ≥ x + 1: 3 ≥ 1 + 1 = 2 ✓ y < −2x + 6: 3 < −2 + 6 = 4 ✓
Answer: The solution region is the area above the solid line y = x + 1 and below the dashed line y = −2x + 6.
Diagram Instruction — Example 8.12.1

Draw a coordinate plane. Draw a solid line through (0, 1) with slope 1 and shade above it. Draw a dashed line through (0, 6) with slope −2 and shade below it. Mark the overlapping region distinctly and label it "solution region."

Example 8.12.2
A system with three inequalities — bounded region
  • ProblemGraph: x ≥ 0, y ≥ 0, and x + y ≤ 5.
  • Thinkx ≥ 0 and y ≥ 0 together restrict the solution to the first quadrant. The third inequality puts an upper bound on their sum.
  • Solve x ≥ 0: shade right of the y-axis (solid line) y ≥ 0: shade above the x-axis (solid line) x + y ≤ 5: Boundary x + y = 5, solid line: (0,5) to (5,0) Test (0,0): 0 ≤ 5 → True → shade below-left of line Solution: triangular region with vertices (0,0), (5,0), (0,5).
  • CheckTest (1, 2): x ≥ 0 ✓, y ≥ 0 ✓, 1 + 2 = 3 ≤ 5 ✓.
Answer: The solution is the triangular region in the first quadrant, bounded by the axes and the line x + y = 5, including all boundary edges.
Diagram Instruction — Example 8.12.2

Draw a coordinate plane. Shade the first quadrant. Draw a solid line from (0, 5) to (5, 0) and shade below it within the first quadrant. The solution region is the triangle with vertices (0, 0), (5, 0), and (0, 5). Label the three vertices and shade the interior.

Practice 8.12Systems of Inequalities

Graph each system of inequalities. Shade the solution region and verify with a test point.

  1. y ≥ 2x − 1 and y < −x + 4
  2. y ≤ x + 3 and y ≥ −2x + 1
  3. x ≥ 0, y ≥ 0, and 2x + 3y ≤ 12
  4. y > x − 1 and y < −x + 5
  5. x + y ≤ 6 and x − y ≥ 0
  6. y ≥ 0, x ≥ 1, and x + y ≤ 4
Section 8.13

Feasible Regions

In many real-world problems — business, engineering, logistics, resource planning — you must satisfy several constraints at once. The set of all points satisfying all constraints is called the feasible region. The word "feasible" means possible or workable. Every point inside the feasible region represents a combination of values that satisfies every constraint.

A feasible region may be bounded (enclosed within a finite polygon), unbounded (extending infinitely in one or more directions), or empty (the shaded regions do not overlap). The corners of a bounded feasible region are called vertices. In optimization problems, maximum or minimum values of a goal often occur at one of these vertices — a topic you will study formally in later courses.

Example 8.13.1
Identifying a feasible region in context
  • ProblemA school store sells notebooks for $2 and pens for $1. A student wants to spend at most $10 and buy at least 3 items. Write and describe the feasible region.
  • Represent Let n = number of notebooks, p = number of pens. Spending constraint: 2n + p ≤ 10 Item count constraint: n + p ≥ 3 Non-negativity: n ≥ 0, p ≥ 0
  • Solve Graph all four inequalities on the same plane. The feasible region is the overlapping area. Test point (2, 3): 2(2) + 3 = 7 ≤ 10 ✓ 2 + 3 = 5 ≥ 3 ✓ n ≥ 0 ✓, p ≥ 0 ✓
  • CheckBuying 2 notebooks and 3 pens: 5 items, cost $7. Both constraints satisfied.
Answer: The feasible region contains all (n, p) combinations the student can purchase within both constraints. Every point in the region represents a valid purchase plan.
Diagram Instruction — Example 8.13.1

Draw a coordinate plane with n on the horizontal axis and p on the vertical axis. Draw the boundary 2n + p = 10 (from (0,10) to (5,0)) and shade below it. Draw n + p = 3 (from (0,3) to (3,0)) and shade above it. Restrict to the first quadrant. The feasible region is the overlapping area. Label the boundary lines and the feasible region.

Non-Negativity Constraints

Whenever variables represent physical quantities — counts of items, hours worked, gallons used — they cannot be negative. Always include x ≥ 0 and y ≥ 0 when the context requires it. These constraints keep the feasible region in the first quadrant.

MGU Library Connection
  • Feasible regions and linear programming: Applied Mathematics and Business Mathematics (later MGU courses)
  • Constraint systems in computing: Computer Science Foundations
  • Data-based constraint modeling: Introduction to Statistics
  • Algebra II, Chapter 2: Linear Systems, Matrices, and Modeling
  • MGU Dictionary entries: feasible region, linear programming, constraint, half-plane
Practice 8.13Feasible Regions
  1. A budget allows at most $20. Small items cost $3, large items cost $5. You need at least 3 items. Write and graph the system of inequalities. Identify the feasible region.
  2. A farmer has 40 acres and at most 120 labor-hours available per week. Crop A requires 4 labor-hours per acre; Crop B requires 2 labor-hours per acre. Write and graph the system. Is the region bounded or unbounded?
  3. Graph the system x ≥ 0, y ≥ 0, x + y ≤ 8, and 2x + y ≤ 12. Identify all vertices of the feasible region.
Section 8.14

Common System Mistakes

Learning to identify, understand, and avoid common errors is part of developing algebraic fluency. Below are the most frequent mistakes students make when working with systems of equations and inequalities.

Mistake 1 — Checking in Only One Equation
Checked equation 1 only, declared solution found.

A solution to a system must satisfy every equation. Always check the proposed solution in all equations in the system.

Mistake 2 — Forgetting to Multiply the Constant in Elimination
Multiply 3x − y = 7 by −2: result written as −6x + 2y = 7

Correct: −2(3x − y) = −2(7) gives −6x + 2y = −14. Every term on both sides must be multiplied.

Mistake 3 — Stopping After One Variable
Solves for x = 3 and writes "The solution is x = 3."

A solution to a system in two variables is an ordered pair (x, y). Always find both values and write the complete ordered pair.

Mistake 4 — Shading the Wrong Half-Plane
Shades the half-plane that does NOT satisfy the inequality.

Always test a specific point in the region you plan to shade. If the point satisfies the inequality, shade that side. If not, shade the other side.

Mistake 5 — Solid Line for a Strict Inequality
Draws a solid line for y < 3x + 1.

Correct: use a dashed line for < and >. Points on the boundary do not satisfy a strict inequality. Use a solid line only for ≤ and ≥.

Mistake 6 — Swapping Variables in Word Problem Setup
In the ticket problem, writes 10s + 6a = 1520 (swapping the prices).

Always define variables explicitly before writing equations. Then re-read the problem to confirm each equation matches a specific condition.

Mistake 7 — Misreading 0 = 0 as No Solution
Elimination gives 0 = 0; student concludes "no solution."

Correct: 0 = 0 is always true — this signals infinitely many solutions (dependent system). A false statement like 0 = 5 signals no solution (inconsistent system).

Mistake 8 — Substituting Without Parentheses
In y = 4x and 2x + y = 18, substitutes to get 2x + 4 = 18.

Correct: substitute the entire expression: 2x + 4x = 18. The variable expression replaces y, not just the coefficient.

Practice 8.14Error Analysis

Each solution below contains an error. Identify the mistake and write the correct solution.

  1. A student solves the system 3x + y = 7 and x − y = 1 by adding and getting 4x = 8, so x = 2. Then the student writes "The solution is x = 2." What is missing?
  2. A student graphs y > 2x − 3 and draws a solid boundary line. What is wrong and how should it be corrected?
  3. A student solves: y = 4x and 2x + y = 18. The student writes 2x + 4 = 18 and gets x = 7. Identify the error and find the correct solution.
  4. When solving a system, a student reaches 0 = 0 and writes "No solution." Explain why this conclusion is wrong and what the correct interpretation is.
  5. A student multiplies the equation 5x − 2y = 8 by 3 and writes 15x − 2y = 8. Identify the error and write the correct result.
Chapter 8 — Back Matter

Chapter Summary

Key Terms Review

system of linear equations
Two or more linear equations in the same variables, considered together — requiring a solution that satisfies all equations simultaneously.
solution to a system
An ordered pair (x, y) that satisfies every equation in the system. Graphically, the point where all lines intersect.
consistent system
A system that has at least one solution — either exactly one (independent) or infinitely many (dependent).
inconsistent system
A system with no solution. Graphically, the lines are parallel. Algebraically, elimination or substitution produces a false statement.
independent system
A consistent system with exactly one solution. The two lines have different slopes and intersect at one point.
dependent system
A consistent system with infinitely many solutions. The two equations describe the same line. Algebraically produces a true statement like 0 = 0.
substitution method
A strategy for solving a system by isolating one variable and substituting that expression into the other equation to get a one-variable equation.
elimination method
A strategy for solving a system by multiplying equations to create opposite coefficients, then adding to eliminate one variable.
linear inequality in two variables
A statement comparing a linear expression in x and y to a number using <, >, ≤, or ≥. Its solution set is a half-plane.
boundary line
The line corresponding to the equation form of an inequality. Solid for ≤ or ≥; dashed for < or >. Divides the plane into two half-planes.
half-plane
One of the two regions on either side of a boundary line. One half-plane satisfies the inequality; the other does not.
feasible region
The set of all points satisfying every inequality in a system simultaneously — the intersection of all individual solution regions. May be bounded, unbounded, or empty.

Concept Review Questions

Answer in complete sentences.

  1. What does it mean for an ordered pair to be a solution to a system of equations?
  2. What is the graphical interpretation of a solution to a system of two linear equations?
  3. How can you recognize an inconsistent system algebraically? What equation result signals it?
  4. How can you recognize a dependent system algebraically? What equation result signals it?
  5. Describe when you would prefer substitution over elimination, and vice versa.
  6. What is a boundary line, and how do you decide whether to draw it solid or dashed?
  7. How do you determine which half-plane to shade when graphing an inequality?
  8. What is a feasible region, and what does a point inside it represent in a real-world problem?
  9. Why must you check a proposed system solution in both equations, not just one?

Skill Practice

Part AGraphing Systems

Graph each system, identify the intersection, and state whether it has one solution, no solution, or infinitely many solutions.

  1. y = x + 2 and y = −x + 6
  2. y = 3x − 4 and y = 3x + 1
  3. y = 2x and x + y = 6
  4. 2x − y = 3 and 4x − 2y = 6
Part BSubstitution
  1. y = 5x − 2, 3x + y = 10
  2. x = 2y + 1, 3x − 4y = 7
  3. 2x + y = 8, y = x − 1
  4. x + 3y = 10, 2x − y = 5
  5. y = −3x + 4, 6x + 2y = 8
Part CElimination
  1. x + y = 8, x − y = 2
  2. 3x + 2y = 14, 3x − y = 5
  3. 2x + 5y = 3, 4x − y = 11
  4. 5x + 3y = 7, 2x + 6y = 10
  5. 4x − 3y = 2, −8x + 6y = −4
Part DGraphing Inequalities
  1. y > 3x − 2
  2. y ≤ −x + 4
  3. 2x + y < 6
  4. 3x − 4y ≥ 12
Part ESystems of Inequalities
  1. y ≥ 2x − 1 and y < −x + 4
  2. x ≥ 0, y ≥ 0, 2x + 3y ≤ 12
  3. y ≤ x + 3 and y ≥ −2x + 1

Word Problems

For each problem: define your variables, write the system, solve, check, and answer the question in context.

  1. The sum of two numbers is 45. Their difference is 9. Find both numbers.
  2. One number is three times another. Their sum is 48. Find both numbers.
  3. Adult tickets cost $12 and student tickets cost $8. A group of 15 people paid $148 total. How many adults and how many students?
  4. A store sells small gift bags for $4 and large gift bags for $7. A customer buys 9 bags for $48. How many of each size?
  5. Two cars leave the same city in opposite directions, one at 60 mph and one at 45 mph. After how many hours are they 315 miles apart?
  6. A family invests $5,000 total: part at 4% annual interest and part at 6%. After one year, the total interest earned is $240. How much was invested at each rate?
  7. A pharmacist mixes a 15% saline solution with a 35% saline solution to make 200 mL of a 25% solution. How many mL of each solution are needed?
  8. A field has length 3x meters and width (x + 5) meters. A fence around the perimeter requires 50 meters of fencing. Find x and the dimensions of the field.

Real-World Applications

  1. Finance. A student earns $9/hr at one job and $12/hr at another. One week, the student works 20 hours total and earns $213. How many hours at each job?
  2. Nutrition. Food A has 200 calories per serving and Food B has 150 calories per serving. A meal plan includes 3 servings totaling 500 calories. How many servings of each food?
  3. Budget Constraints. A design student needs fonts ($15 each) and image packs ($25 each). The student has at most $150 and needs at least 4 items. Write a system of inequalities and describe the feasible region. Give two specific purchase combinations that work.
  4. Agriculture. A farmer has 40 acres and 100 labor-hours per week. Crop A needs 3 hours/acre; Crop B needs 2 hours/acre. Write and graph the system of inequalities. Is the feasible region bounded?

Error Analysis

Each problem shows a student's work. Identify the error and write the correct solution.

Problem 1
System: 3x + y = 7 and x − y = 1. Student adds and gets 4x = 8, x = 2. Writes: "The solution is x = 2."

What is missing? What is the complete, correct solution?

Problem 2
Student graphs y > 2x − 3 with a solid boundary line.

What is wrong, and how should it be corrected?

Problem 3
System: y = 4x and 2x + y = 18. Student writes 2x + 4 = 18, gets x = 7.

Identify the substitution error. What is the correct solution?

Problem 4
Elimination gives 0 = 0. Student writes "No solution."

Explain why this is wrong. What is the correct conclusion?

Problem 5
Student multiplies 4x − 2y = 10 by 3 and writes 12x − 2y = 10.

Identify the error and write the correct result.

Chapter 8 Checkpoint

Chapter 8 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 8
Part A — Concepts
  1. Explain in your own words what a solution to a system of linear equations represents.
  2. A student says: "I solved the first equation and got x = 4. That's the answer." What is wrong with this statement?
  3. Describe the algebraic signal that tells you a system has no solution.
  4. Describe the algebraic signal that tells you a system has infinitely many solutions.
  5. How do you decide whether to draw a solid or dashed boundary line when graphing an inequality?
Part B — Skills: Solving Systems
  1. Solve by any method: y = 2x − 3 and y = −x + 6. Check your answer.
  2. Solve by substitution: 4x + y = 10 and 2x − 3y = 12.
  3. Solve by elimination: 3x + 5y = 15 and 6x + 10y = 30. What does the result tell you?
  4. Solve by elimination: 2x − y = 4 and 3x + 2y = 13.
Part C — Inequalities
  1. Graph the inequality y ≥ −2x + 3.
  2. Graph the system: y < x + 4 and y ≥ −x. Shade the solution region and verify with a test point.
Part D — Applications
  1. A school store sells pencils for $0.50 and notebooks for $2.50. A student buys 12 items for $16.00. How many pencils and notebooks did the student buy?
  2. Two hikers start from the same trailhead and walk in opposite directions. One hiker walks at 3 mph and the other at 4 mph. After how many hours are they 14 miles apart?
  3. Write a real-world situation that could be modeled by: x + y = 50 and 8x + 12y = 480. Define your variables, explain each equation, and solve the system.
Bridge Note — Looking Ahead
Part IV: Exponents, Polynomials, and Factoring

Chapter 8 closes Part III of Algebra I. You have now built a strong foundation in linear relationships — equations, inequalities, functions, graphs, and systems. Every key skill in these chapters has been tied to meaning: slope as rate of change, systems as simultaneous conditions, inequalities as ranges of possibility.

Part IV opens a new kind of algebra. Instead of constant rates of change, you will study exponential patterns — situations where a quantity is multiplied by a fixed factor over and over. Here is what to watch for:

  • Chapter 9: Exponent rules follow directly from repeated multiplication — not arbitrary memorization.
  • Chapter 10: Polynomial multiplication extends the distributive property you first learned in Chapter 1.
  • Chapter 11: Factoring is multiplication in reverse — recognize patterns, undo products, rewrite in useful forms.
  • Chapter 12: All of these ideas converge in quadratic equations — the first major nonlinear function family of Algebra I.

The reasoning skills you have developed here — checking solutions, translating words into algebra, working systematically, and interpreting results in context — will continue to serve you throughout the course.

MGU Library Connection
  • Linear equation forms and graphing reference — Appendix H and the Algebra I Formula Reference
  • Graphing technology reference for systems — Appendix N: Calculator and Graphing Technology Skills
  • Algebra II, Chapter 2: Linear Systems, Matrices, and Modeling (three-variable systems; matrix methods)
  • Algebra II, Chapter 14: Probability, Statistics, and Decision-Making (regression models as systems)
  • MGU Dictionary entries: system of equations, feasible region, elimination method, substitution method, half-plane