Learn on PengiBig Ideas Math, Course 3Chapter 5: Systems of Linear Equations

Lesson 2: Solving Systems of Linear Equations by Substitution

In this Grade 8 lesson from Big Ideas Math, Course 3, students learn how to solve systems of linear equations using the substitution method. The lesson walks through a three-step process: solving one equation for a variable, substituting that expression into the second equation, and back-substituting to find both values. Students apply this skill to real-life word problems and practice verifying solutions by checking answers in both original equations.

Section 1

The Substitution Method: Step-by-Step

Property

The substitution method is an algebraic technique for solving a system of equations by replacing one variable with an equivalent expression.
To solve a system by substitution, follow these four steps:

  1. Isolate: Choose one equation and isolate one variable (make its coefficient 1 or -1 to avoid fractions).
  2. Substitute: Plug that isolated expression into the OTHER equation. This creates a new equation with only one variable.
  3. Solve: Solve this new one-variable equation.
  4. Back-Substitute: Plug the value you just found back into the isolated equation from Step 1 to find the second variable's value.

Examples

  • Isolating First: Solve 3x+y=103x + y = 10 and 2x+y=52x + y = 5.

The easiest variable to isolate is yy in the first equation. Subtract 3x3x to get y=103xy = 10 - 3x.

  • Full Substitution: Solve y=3xy = 3x and x+y=12x + y = 12.

Since yy is already isolated, substitute 3x3x for yy in the second equation: x+3x=12x + 3x = 12.
Solve: 4x=124x = 12, so x=3x = 3.
Back-substitute: y=3(3)=9y = 3(3) = 9. The solution is (3,9)(3, 9).

  • Multi-Step: Solve a2b=5a - 2b = 5 and 3a+b=83a + b = 8.

Isolate aa in the first equation: a=2b+5a = 2b + 5.
Substitute into the second: 3(2b+5)+b=83(2b + 5) + b = 8.
Solve: 6b+15+b=87b=7b=16b + 15 + b = 8 \rightarrow 7b = -7 \rightarrow b = -1.
Back-substitute: a=2(1)+5=3a = 2(-1) + 5 = 3. The solution is (3,1)(3, -1).

Explanation

Think of the substitution method as a perfectly legal mathematical swap. Because an equation tells you two things are perfectly equal (like y=3xy = 3x), you can completely remove the yy from the other equation and drop the 3x3x in its place. Choosing to isolate a variable that already has a coefficient of 1 or -1 is a pro-tip—it saves time and prevents you from having to do algebra with messy fractions!

Section 2

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 xx first, substitute that value into either original equation to find yy.
If you solved for yy first, substitute that value to find xx.

Examples

Section 3

Solving Real-World Systems by Substitution

Property

Many real-world situations can be modeled by a system of linear equations. The process involves:

  1. Defining variables to represent the unknown quantities.
  2. Writing two linear equations that describe the relationships between the variables.
  3. Solving the system using the substitution method to find the solution.

Examples

Book overview

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

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Chapter 5: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Solving Systems of Linear Equations by Graphing

  2. Lesson 2Current

    Lesson 2: Solving Systems of Linear Equations by Substitution

  3. Lesson 3

    Lesson 3: Solving Systems of Linear Equations by Elimination

  4. Lesson 4

    Lesson 4: Solving Special Systems of Linear Equations

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

The Substitution Method: Step-by-Step

Property

The substitution method is an algebraic technique for solving a system of equations by replacing one variable with an equivalent expression.
To solve a system by substitution, follow these four steps:

  1. Isolate: Choose one equation and isolate one variable (make its coefficient 1 or -1 to avoid fractions).
  2. Substitute: Plug that isolated expression into the OTHER equation. This creates a new equation with only one variable.
  3. Solve: Solve this new one-variable equation.
  4. Back-Substitute: Plug the value you just found back into the isolated equation from Step 1 to find the second variable's value.

Examples

  • Isolating First: Solve 3x+y=103x + y = 10 and 2x+y=52x + y = 5.

The easiest variable to isolate is yy in the first equation. Subtract 3x3x to get y=103xy = 10 - 3x.

  • Full Substitution: Solve y=3xy = 3x and x+y=12x + y = 12.

Since yy is already isolated, substitute 3x3x for yy in the second equation: x+3x=12x + 3x = 12.
Solve: 4x=124x = 12, so x=3x = 3.
Back-substitute: y=3(3)=9y = 3(3) = 9. The solution is (3,9)(3, 9).

  • Multi-Step: Solve a2b=5a - 2b = 5 and 3a+b=83a + b = 8.

Isolate aa in the first equation: a=2b+5a = 2b + 5.
Substitute into the second: 3(2b+5)+b=83(2b + 5) + b = 8.
Solve: 6b+15+b=87b=7b=16b + 15 + b = 8 \rightarrow 7b = -7 \rightarrow b = -1.
Back-substitute: a=2(1)+5=3a = 2(-1) + 5 = 3. The solution is (3,1)(3, -1).

Explanation

Think of the substitution method as a perfectly legal mathematical swap. Because an equation tells you two things are perfectly equal (like y=3xy = 3x), you can completely remove the yy from the other equation and drop the 3x3x in its place. Choosing to isolate a variable that already has a coefficient of 1 or -1 is a pro-tip—it saves time and prevents you from having to do algebra with messy fractions!

Section 2

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 xx first, substitute that value into either original equation to find yy.
If you solved for yy first, substitute that value to find xx.

Examples

Section 3

Solving Real-World Systems by Substitution

Property

Many real-world situations can be modeled by a system of linear equations. The process involves:

  1. Defining variables to represent the unknown quantities.
  2. Writing two linear equations that describe the relationships between the variables.
  3. Solving the system using the substitution method to find the solution.

Examples

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 1

    Lesson 1: Solving Systems of Linear Equations by Graphing

  2. Lesson 2Current

    Lesson 2: Solving Systems of Linear Equations by Substitution

  3. Lesson 3

    Lesson 3: Solving Systems of Linear Equations by Elimination

  4. Lesson 4

    Lesson 4: Solving Special Systems of Linear Equations