Learn on PengiOpenstax Elementary Algebra 2EChapter 5: Systems of Linear Equations

Lesson 1: Solve Systems of Equations by Graphing

Students learn how to solve systems of linear equations by graphing, including how to determine whether an ordered pair is a solution to a system and how to identify the number of solutions a linear system has. This lesson from Chapter 5 of Openstax Elementary Algebra 2E covers substituting ordered pairs to verify solutions and finding the intersection point of two lines on a coordinate plane. It provides the foundation for later methods such as substitution and elimination.

Section 1

πŸ“˜ Solve Systems of Equations by Graphing

New Concept

A system of linear equations groups two or more lines. We'll find the single ordered pair (x,y)(x, y) that solves both equations by graphing them and locating their intersection point. This skill helps classify systems and solve real-world problems.

What’s next

Ready to start? You'll begin by checking potential solutions and then move on to interactive examples that show you how to find solutions by graphing.

Section 2

Solutions of a System of Equations

Property

When two or more linear equations are grouped together, they form a system of linear equations. Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x,y)(x, y). To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Examples

  • Is (2,3)(2, 3) a solution to the system {3xβˆ’y=3x+2y=8\begin{cases} 3x - y = 3 \\ x + 2y = 8 \end{cases}? For the first equation, 3(2)βˆ’3=33(2) - 3 = 3 is true. For the second, 2+2(3)=82 + 2(3) = 8 is true. Since it makes both true, (2,3)(2, 3) is a solution.
  • Is (βˆ’1,5)(-1, 5) a solution to the system {5x+y=02x+y=4\begin{cases} 5x + y = 0 \\ 2x + y = 4 \end{cases}? For the first equation, 5(βˆ’1)+5=05(-1) + 5 = 0 is true. For the second, 2(βˆ’1)+5=3β‰ 42(-1) + 5 = 3 \neq 4 is false. Therefore, (βˆ’1,5)(-1, 5) is not a solution.
  • Is (4,βˆ’2)(4, -2) a solution to the system {x+3y=βˆ’2βˆ’2xβˆ’5y=βˆ’2\begin{cases} x + 3y = -2 \\ -2x - 5y = -2 \end{cases}? For the first equation, 4+3(βˆ’2)=βˆ’24 + 3(-2) = -2 is true. For the second, βˆ’2(4)βˆ’5(βˆ’2)=βˆ’8+10=2β‰ βˆ’2-2(4) - 5(-2) = -8 + 10 = 2 \neq -2 is false. Therefore, (4,βˆ’2)(4, -2) is not a solution.

Explanation

Think of a system's solution as a secret meeting point. It's the single ordered pair (x,y)(x, y) that exists on both lines at the same time. If a point only satisfies one equation, it hasn't arrived at the right spot.

Section 3

Solve a System by Graphing

Property

To solve a system of linear equations by graphing, follow these steps:

  1. Graph the first equation.
  2. Graph the second equation on the same rectangular coordinate system.
  3. Determine whether the lines intersect, are parallel, or are the same line.
  4. Identify the solution to the system. If the lines intersect, the point of intersection is the solution. Check the point in both equations to verify.

Examples

  • Solve the system {y=x+2y=βˆ’x+4\begin{cases} y = x + 2 \\ y = -x + 4 \end{cases} by graphing. The lines intersect at the point (1,3)(1, 3). Checking this point: 3=1+23=1+2 (true) and 3=βˆ’1+43=-1+4 (true). The solution is (1,3)(1, 3).
  • Solve the system {x+y=62xβˆ’y=3\begin{cases} x + y = 6 \\ 2x - y = 3 \end{cases} by graphing. The lines intersect at (3,3)(3, 3). Checking this point: 3+3=63+3=6 (true) and 2(3)βˆ’3=32(3)-3=3 (true). The solution is (3,3)(3, 3).
  • Solve the system {y=2x+y=5\begin{cases} y = 2 \\ x + y = 5 \end{cases} by graphing. The line y=2y=2 is horizontal. The line x+y=5x+y=5 intersects it at (3,2)(3, 2). The solution is (3,2)(3, 2).

Explanation

Graphing turns algebra into a visual treasure hunt. The two lines are paths, and the solution is the 'X' that marks the spot where they cross. This single point of intersection is the only ordered pair that satisfies both equations.

Section 4

Number of Solutions

Property

A system of equations can be classified by the number of solutions, which is determined by the relationship between the lines.

  • Intersecting Lines (One Solution): The lines have different slopes. The system is consistent and the equations are independent.
  • Parallel Lines (No Solution): The lines have the same slope and different yy-intercepts. The system is inconsistent.
  • Coincident Lines (Infinitely Many Solutions): The lines have the same slope and the same yy-intercept. The system is consistent and the equations are dependent.

Examples

  • The system {y=3x+2y=3xβˆ’1\begin{cases} y = 3x + 2 \\ y = 3x - 1 \end{cases} has two lines with the same slope (m=3m=3) but different yy-intercepts. The lines are parallel, so there is no solution. The system is inconsistent.
  • The system {y=4xβˆ’1y=βˆ’2x+5\begin{cases} y = 4x - 1 \\ y = -2x + 5 \end{cases} has two lines with different slopes (m=4m=4 and m=βˆ’2m=-2). The lines intersect at one point, so there is one solution. The system is consistent and independent.
  • The system {x+y=52x+2y=10\begin{cases} x + y = 5 \\ 2x + 2y = 10 \end{cases} represents the same line, as the second equation is double the first. There are infinitely many solutions. The system is consistent and dependent.

Explanation

The relationship between two lines determines how many solutions exist. Intersecting lines cross once (one solution). Parallel lines never cross (no solution). Coincident lines are the same line, so they overlap everywhere (infinite solutions).

Section 5

Solving Applications by Graphing

Property

To solve application problems using systems of equations by graphing:

  1. Read the problem to identify what you are looking for.
  2. Name the unknowns by choosing variables to represent them.
  3. Translate the problem's conditions into a system of equations.
  4. Solve the system by graphing the equations.
  5. Check that the solution makes sense in the context of the problem.
  6. Answer the question with a complete sentence.

Examples

  • Two numbers have a sum of 12 and a difference of 2. Let the numbers be xx and yy. The system is {x+y=12xβˆ’y=2\begin{cases} x + y = 12 \\ x - y = 2 \end{cases}. The solution is (7,5)(7, 5), so the numbers are 7 and 5.
  • A store sells two types of coffee beans, one for 8 dollars per pound and another for 12 dollars per pound. A customer buys 10 pounds for a total of 92 dollars. The system is {x+y=108x+12y=92\begin{cases} x + y = 10 \\ 8x + 12y = 92 \end{cases}. The solution is (7,3)(7, 3), meaning 7 pounds of the first type and 3 of the second.
  • The perimeter of a rectangle is 20 inches. The length is 4 inches more than the width. The system is {2L+2W=20L=W+4\begin{cases} 2L + 2W = 20 \\ L = W + 4 \end{cases}. The solution is (7,3)(7, 3), so the length is 7 inches and the width is 3 inches.

Explanation

Word problems provide clues to create a system of equations. Each fact or relationship often translates into its own equation. Graphing these equations reveals the one pair of values that makes every part of the story true at the same time.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 5: Systems of Linear Equations

  1. Lesson 1Current

    Lesson 1: Solve Systems of Equations by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems of Equations by Substitution

  3. Lesson 3

    Lesson 3: Solve Systems of Equations by Elimination

  4. Lesson 4

    Lesson 4: Solve Applications with Systems of Equations

  5. Lesson 5

    Lesson 5: Solve Mixture Applications with Systems of Equations

  6. Lesson 6

    Lesson 6: Graphing Systems of Linear Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Solve Systems of Equations by Graphing

New Concept

A system of linear equations groups two or more lines. We'll find the single ordered pair (x,y)(x, y) that solves both equations by graphing them and locating their intersection point. This skill helps classify systems and solve real-world problems.

What’s next

Ready to start? You'll begin by checking potential solutions and then move on to interactive examples that show you how to find solutions by graphing.

Section 2

Solutions of a System of Equations

Property

When two or more linear equations are grouped together, they form a system of linear equations. Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x,y)(x, y). To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Examples

  • Is (2,3)(2, 3) a solution to the system {3xβˆ’y=3x+2y=8\begin{cases} 3x - y = 3 \\ x + 2y = 8 \end{cases}? For the first equation, 3(2)βˆ’3=33(2) - 3 = 3 is true. For the second, 2+2(3)=82 + 2(3) = 8 is true. Since it makes both true, (2,3)(2, 3) is a solution.
  • Is (βˆ’1,5)(-1, 5) a solution to the system {5x+y=02x+y=4\begin{cases} 5x + y = 0 \\ 2x + y = 4 \end{cases}? For the first equation, 5(βˆ’1)+5=05(-1) + 5 = 0 is true. For the second, 2(βˆ’1)+5=3β‰ 42(-1) + 5 = 3 \neq 4 is false. Therefore, (βˆ’1,5)(-1, 5) is not a solution.
  • Is (4,βˆ’2)(4, -2) a solution to the system {x+3y=βˆ’2βˆ’2xβˆ’5y=βˆ’2\begin{cases} x + 3y = -2 \\ -2x - 5y = -2 \end{cases}? For the first equation, 4+3(βˆ’2)=βˆ’24 + 3(-2) = -2 is true. For the second, βˆ’2(4)βˆ’5(βˆ’2)=βˆ’8+10=2β‰ βˆ’2-2(4) - 5(-2) = -8 + 10 = 2 \neq -2 is false. Therefore, (4,βˆ’2)(4, -2) is not a solution.

Explanation

Think of a system's solution as a secret meeting point. It's the single ordered pair (x,y)(x, y) that exists on both lines at the same time. If a point only satisfies one equation, it hasn't arrived at the right spot.

Section 3

Solve a System by Graphing

Property

To solve a system of linear equations by graphing, follow these steps:

  1. Graph the first equation.
  2. Graph the second equation on the same rectangular coordinate system.
  3. Determine whether the lines intersect, are parallel, or are the same line.
  4. Identify the solution to the system. If the lines intersect, the point of intersection is the solution. Check the point in both equations to verify.

Examples

  • Solve the system {y=x+2y=βˆ’x+4\begin{cases} y = x + 2 \\ y = -x + 4 \end{cases} by graphing. The lines intersect at the point (1,3)(1, 3). Checking this point: 3=1+23=1+2 (true) and 3=βˆ’1+43=-1+4 (true). The solution is (1,3)(1, 3).
  • Solve the system {x+y=62xβˆ’y=3\begin{cases} x + y = 6 \\ 2x - y = 3 \end{cases} by graphing. The lines intersect at (3,3)(3, 3). Checking this point: 3+3=63+3=6 (true) and 2(3)βˆ’3=32(3)-3=3 (true). The solution is (3,3)(3, 3).
  • Solve the system {y=2x+y=5\begin{cases} y = 2 \\ x + y = 5 \end{cases} by graphing. The line y=2y=2 is horizontal. The line x+y=5x+y=5 intersects it at (3,2)(3, 2). The solution is (3,2)(3, 2).

Explanation

Graphing turns algebra into a visual treasure hunt. The two lines are paths, and the solution is the 'X' that marks the spot where they cross. This single point of intersection is the only ordered pair that satisfies both equations.

Section 4

Number of Solutions

Property

A system of equations can be classified by the number of solutions, which is determined by the relationship between the lines.

  • Intersecting Lines (One Solution): The lines have different slopes. The system is consistent and the equations are independent.
  • Parallel Lines (No Solution): The lines have the same slope and different yy-intercepts. The system is inconsistent.
  • Coincident Lines (Infinitely Many Solutions): The lines have the same slope and the same yy-intercept. The system is consistent and the equations are dependent.

Examples

  • The system {y=3x+2y=3xβˆ’1\begin{cases} y = 3x + 2 \\ y = 3x - 1 \end{cases} has two lines with the same slope (m=3m=3) but different yy-intercepts. The lines are parallel, so there is no solution. The system is inconsistent.
  • The system {y=4xβˆ’1y=βˆ’2x+5\begin{cases} y = 4x - 1 \\ y = -2x + 5 \end{cases} has two lines with different slopes (m=4m=4 and m=βˆ’2m=-2). The lines intersect at one point, so there is one solution. The system is consistent and independent.
  • The system {x+y=52x+2y=10\begin{cases} x + y = 5 \\ 2x + 2y = 10 \end{cases} represents the same line, as the second equation is double the first. There are infinitely many solutions. The system is consistent and dependent.

Explanation

The relationship between two lines determines how many solutions exist. Intersecting lines cross once (one solution). Parallel lines never cross (no solution). Coincident lines are the same line, so they overlap everywhere (infinite solutions).

Section 5

Solving Applications by Graphing

Property

To solve application problems using systems of equations by graphing:

  1. Read the problem to identify what you are looking for.
  2. Name the unknowns by choosing variables to represent them.
  3. Translate the problem's conditions into a system of equations.
  4. Solve the system by graphing the equations.
  5. Check that the solution makes sense in the context of the problem.
  6. Answer the question with a complete sentence.

Examples

  • Two numbers have a sum of 12 and a difference of 2. Let the numbers be xx and yy. The system is {x+y=12xβˆ’y=2\begin{cases} x + y = 12 \\ x - y = 2 \end{cases}. The solution is (7,5)(7, 5), so the numbers are 7 and 5.
  • A store sells two types of coffee beans, one for 8 dollars per pound and another for 12 dollars per pound. A customer buys 10 pounds for a total of 92 dollars. The system is {x+y=108x+12y=92\begin{cases} x + y = 10 \\ 8x + 12y = 92 \end{cases}. The solution is (7,3)(7, 3), meaning 7 pounds of the first type and 3 of the second.
  • The perimeter of a rectangle is 20 inches. The length is 4 inches more than the width. The system is {2L+2W=20L=W+4\begin{cases} 2L + 2W = 20 \\ L = W + 4 \end{cases}. The solution is (7,3)(7, 3), so the length is 7 inches and the width is 3 inches.

Explanation

Word problems provide clues to create a system of equations. Each fact or relationship often translates into its own equation. Graphing these equations reveals the one pair of values that makes every part of the story true at the same time.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 5: Systems of Linear Equations

  1. Lesson 1Current

    Lesson 1: Solve Systems of Equations by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems of Equations by Substitution

  3. Lesson 3

    Lesson 3: Solve Systems of Equations by Elimination

  4. Lesson 4

    Lesson 4: Solve Applications with Systems of Equations

  5. Lesson 5

    Lesson 5: Solve Mixture Applications with Systems of Equations

  6. Lesson 6

    Lesson 6: Graphing Systems of Linear Inequalities