Learn on PengienVision, Algebra 2Chapter 1: Linear Functions and Systems

Lesson 6: Linear Systems

In this Grade 11 enVision Algebra 2 lesson, students learn to solve systems of linear equations and systems of linear inequalities using graphing, substitution, and elimination methods. The lesson extends to three-variable systems and introduces matrix concepts including augmented matrices and coefficient matrices. Students also explore inconsistent systems and interpret solution regions on coordinate planes in real-world contexts.

Section 1

Solve Systems by Substitution

Property

To solve a system of linear equations by substitution: solve one equation for one variable in terms of the other, then substitute this expression into the second equation to create a single-variable equation.

Examples

Section 2

Solve Systems by Elimination

Property

To solve a system of equations by elimination:

Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
Step 2. Make the coefficients of one variable opposites by multiplying one or both equations by appropriate numbers.
Step 3. Add the equations to eliminate one variable.
Step 4. Solve for the remaining variable.
Step 5. Substitute back into one of the original equations to solve for the other variable.
Step 6. Write the solution as an ordered pair and check it.

Examples

Section 3

Solve a system with three variables

Property

To solve a system of linear equations with three variables**

Step 1. Write the equations in standard form. If any coefficients are fractions, clear them.
Step 2. Eliminate the same variable from two equations.
Step 3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
Step 4. The two new equations form a system of two equations with two variables. Solve this system.
Step 5. Use the values of the two variables found in Step 4 to find the third variable.
Step 6. Write the solution as an ordered triple.
Step 7. Check that the ordered triple is a solution to all three original equations.

Examples

Solve $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ x+2y-z=2 \end{cases}$ . Adding (1) and (3) gives $2x+3y=8$ . Adding (2) and (3) gives $3x+y=5$ , or $y=5-3x$ . Substituting into $2x+3y=8$ gives $2x+3(5-3x)=8$ , so $x=1$ . Then $y=2$ and $z=3$ . The solution is $(1, 2, 3)$ .

Section 4

Augmented Matrix

Property

A matrix is a rectangular array of numbers arranged in rows and columns.
To represent a system of linear equations with a matrix, we write each equation in standard form. The coefficients of the variables and the constant of each equation become a row in the matrix. A vertical line replaces the equal signs. This is the augmented matrix for the system.
For the system:

\begin{cases} 3x - y = -3 \\ 2x + 3y = 6 \end{cases}

The augmented matrix is:

\left[ \begin{array}{cc|c} 3 & -1 & -3 \\ 2 & 3 & 6 \end{array} \right]

Examples

The system $\begin{cases} 2x - 7y = 4 \\ x = 3y + 1 \end{cases}$ is first rewritten with the second equation in standard form as $x - 3y = 1$ . The augmented matrix is $\left[\begin{array}{cc|c} 2 & -7 & 4 \\ 1 & -3 & 1 \end{array}\right]$ .

The system $\begin{cases} x + 2y - z = 5 \\ 3x + y + 4z = 8 \\ 2x - y + 9z = -2 \end{cases}$ corresponds to the augmented matrix $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 5 \\ 3 & 1 & 4 & 8 \\ 2 & -1 & 9 & -2 \end{array}\right]$ .

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solve Systems by Substitution

Property

Examples

Section 2

Solve Systems by Elimination

Property

To solve a system of equations by elimination:

Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
Step 2. Make the coefficients of one variable opposites by multiplying one or both equations by appropriate numbers.
Step 3. Add the equations to eliminate one variable.
Step 4. Solve for the remaining variable.
Step 5. Substitute back into one of the original equations to solve for the other variable.
Step 6. Write the solution as an ordered pair and check it.

Examples

Section 3

Solve a system with three variables

Property

To solve a system of linear equations with three variables**

Step 1. Write the equations in standard form. If any coefficients are fractions, clear them.
Step 2. Eliminate the same variable from two equations.
Step 3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
Step 4. The two new equations form a system of two equations with two variables. Solve this system.
Step 5. Use the values of the two variables found in Step 4 to find the third variable.
Step 6. Write the solution as an ordered triple.
Step 7. Check that the ordered triple is a solution to all three original equations.

Examples

Solve $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ x+2y-z=2 \end{cases}$ . Adding (1) and (3) gives $2x+3y=8$ . Adding (2) and (3) gives $3x+y=5$ , or $y=5-3x$ . Substituting into $2x+3y=8$ gives $2x+3(5-3x)=8$ , so $x=1$ . Then $y=2$ and $z=3$ . The solution is $(1, 2, 3)$ .

Section 4

Augmented Matrix

Property

\begin{cases} 3x - y = -3 \\ 2x + 3y = 6 \end{cases}

The augmented matrix is:

\left[ \begin{array}{cc|c} 3 & -1 & -3 \\ 2 & 3 & 6 \end{array} \right]

Examples

The system $\begin{cases} 2x - 7y = 4 \\ x = 3y + 1 \end{cases}$ is first rewritten with the second equation in standard form as $x - 3y = 1$ . The augmented matrix is $\left[\begin{array}{cc|c} 2 & -7 & 4 \\ 1 & -3 & 1 \end{array}\right]$ .

The system $\begin{cases} x + 2y - z = 5 \\ 3x + y + 4z = 8 \\ 2x - y + 9z = -2 \end{cases}$ corresponds to the augmented matrix $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 5 \\ 3 & 1 & 4 & 8 \\ 2 & -1 & 9 & -2 \end{array}\right]$ .

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Solve Systems by Substitution

Property

Examples

Section 2

Solve Systems by Elimination

Property

To solve a system of equations by elimination:

Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
Step 2. Make the coefficients of one variable opposites by multiplying one or both equations by appropriate numbers.
Step 3. Add the equations to eliminate one variable.
Step 4. Solve for the remaining variable.
Step 5. Substitute back into one of the original equations to solve for the other variable.
Step 6. Write the solution as an ordered pair and check it.

Examples

Section 3

Solve a system with three variables

Property

To solve a system of linear equations with three variables**

Step 1. Write the equations in standard form. If any coefficients are fractions, clear them.
Step 2. Eliminate the same variable from two equations.
Step 3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
Step 4. The two new equations form a system of two equations with two variables. Solve this system.
Step 5. Use the values of the two variables found in Step 4 to find the third variable.
Step 6. Write the solution as an ordered triple.
Step 7. Check that the ordered triple is a solution to all three original equations.

Examples

Solve $\begin{cases} x+y+z=6 \\ 2x-y+z=3 \\ x+2y-z=2 \end{cases}$ . Adding (1) and (3) gives $2x+3y=8$ . Adding (2) and (3) gives $3x+y=5$ , or $y=5-3x$ . Substituting into $2x+3y=8$ gives $2x+3(5-3x)=8$ , so $x=1$ . Then $y=2$ and $z=3$ . The solution is $(1, 2, 3)$ .

Section 4

Augmented Matrix

Property

\begin{cases} 3x - y = -3 \\ 2x + 3y = 6 \end{cases}

The augmented matrix is:

\left[ \begin{array}{cc|c} 3 & -1 & -3 \\ 2 & 3 & 6 \end{array} \right]

Examples

The system $\begin{cases} 2x - 7y = 4 \\ x = 3y + 1 \end{cases}$ is first rewritten with the second equation in standard form as $x - 3y = 1$ . The augmented matrix is $\left[\begin{array}{cc|c} 2 & -7 & 4 \\ 1 & -3 & 1 \end{array}\right]$ .

The system $\begin{cases} x + 2y - z = 5 \\ 3x + y + 4z = 8 \\ 2x - y + 9z = -2 \end{cases}$ corresponds to the augmented matrix $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 5 \\ 3 & 1 & 4 & 8 \\ 2 & -1 & 9 & -2 \end{array}\right]$ .