Learn on PengienVision, Algebra 1Chapter 4: Systems of Linear Equations and Inequalities

Lesson 2: Solving Systems of Equations by Substitution

In this Grade 11 enVision Algebra 1 lesson, students learn how to solve systems of linear equations using the substitution method by isolating one variable and substituting its expression into the other equation. The lesson covers finding unique solutions, identifying systems with infinitely many solutions or no solution, and compares substitution to the graphing method. Real-world applications, such as mixing saline solutions and calculating activity costs, help students see how substitution produces exact answers more efficiently than graphing.

Section 1

The Substitution Method

Property

To solve a system by substitution, follow these steps:

Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into one of the original equations to find the other variable.
Write the solution as an ordered pair and check that it is a solution to both original equations.

Examples

Solve the system $y = x + 3$ and $3x + 2y = 19$ . Substitute $x+3$ for $y$ in the second equation: $3x + 2(x+3) = 19$ . This simplifies to $5x+6=19$ , so $5x=13$ and $x=\frac{13}{5}$ . Then $y = \frac{13}{5} + 3 = \frac{28}{5}$ . The solution is $(\frac{13}{5}, \frac{28}{5})$ .
Solve the system $2x - y = 8$ and $x + 3y = 11$ . From the first equation, solve for $y$ : $y = 2x - 8$ . Substitute this into the second equation: $x + 3(2x-8) = 11$ . This gives $7x-24=11$ , so $7x=35$ and $x=5$ . Then $y=2(5)-8=2$ , making the solution $(5, 2)$ .

Explanation

This method simplifies a two-variable system into a single-variable equation. By isolating a variable in one equation and plugging its expression into the other, you can solve for one variable and then use that value to find the second.

Section 2

Proper Use of Parentheses in Substitution

Property

When substituting an expression for a variable, always enclose the entire expression in parentheses: if $y = mx + b$ , then substituting into $ax + cy = d$ gives $ax + c(mx + b) = d$ .

Examples

Section 3

Back-Substitution in Systems

Property

Back-substitution is the process of finding the first variable's value after solving for the second variable.
If you solved for $x$ first, substitute that value into either original equation to find $y$ .
If you solved for $y$ first, substitute that value to find $x$ .

Examples

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

The Substitution Method

Property

To solve a system by substitution, follow these steps:

Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into one of the original equations to find the other variable.
Write the solution as an ordered pair and check that it is a solution to both original equations.

Examples

Solve the system $y = x + 3$ and $3x + 2y = 19$ . Substitute $x+3$ for $y$ in the second equation: $3x + 2(x+3) = 19$ . This simplifies to $5x+6=19$ , so $5x=13$ and $x=\frac{13}{5}$ . Then $y = \frac{13}{5} + 3 = \frac{28}{5}$ . The solution is $(\frac{13}{5}, \frac{28}{5})$ .
Solve the system $2x - y = 8$ and $x + 3y = 11$ . From the first equation, solve for $y$ : $y = 2x - 8$ . Substitute this into the second equation: $x + 3(2x-8) = 11$ . This gives $7x-24=11$ , so $7x=35$ and $x=5$ . Then $y=2(5)-8=2$ , making the solution $(5, 2)$ .

Explanation

Section 2

Proper Use of Parentheses in Substitution

Property

When substituting an expression for a variable, always enclose the entire expression in parentheses: if $y = mx + b$ , then substituting into $ax + cy = d$ gives $ax + c(mx + b) = d$ .

Examples

Section 3

Back-Substitution in Systems

Property

Examples

Overview

Lesson overview

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

Overview

Section 1

The Substitution Method

Property

To solve a system by substitution, follow these steps:

Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into one of the original equations to find the other variable.
Write the solution as an ordered pair and check that it is a solution to both original equations.

Examples

Solve the system $y = x + 3$ and $3x + 2y = 19$ . Substitute $x+3$ for $y$ in the second equation: $3x + 2(x+3) = 19$ . This simplifies to $5x+6=19$ , so $5x=13$ and $x=\frac{13}{5}$ . Then $y = \frac{13}{5} + 3 = \frac{28}{5}$ . The solution is $(\frac{13}{5}, \frac{28}{5})$ .
Solve the system $2x - y = 8$ and $x + 3y = 11$ . From the first equation, solve for $y$ : $y = 2x - 8$ . Substitute this into the second equation: $x + 3(2x-8) = 11$ . This gives $7x-24=11$ , so $7x=35$ and $x=5$ . Then $y=2(5)-8=2$ , making the solution $(5, 2)$ .

Explanation

Section 2

Proper Use of Parentheses in Substitution

Property

When substituting an expression for a variable, always enclose the entire expression in parentheses: if $y = mx + b$ , then substituting into $ax + cy = d$ gives $ax + c(mx + b) = d$ .