Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 5: Multi-Variable Linear Equations

Lesson 2: Substitution

In this Grade 4 AMC Math lesson from AoPS: Introduction to Algebra, students learn the substitution method for solving systems of two-variable linear equations. They practice isolating one variable in one equation and substituting that expression into the second equation to produce a solvable one-variable linear equation. The lesson covers applying substitution to systems involving fractions and decimals, and verifying solutions by checking them in both original equations.

Section 1

Introduction to 2x2 Linear Systems and Substitution Method

Property

A solution to a 2×22 \times 2 linear system is an ordered pair that satisfies both equations.

The substitution method involves solving one equation for a variable, then substituting that expression into the other equation.

Section 2

The Substitution Method

Property

To solve a system by substitution, follow these steps:

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

Examples

  • Solve the system y=x+3y = x + 3 and 3x+2y=193x + 2y = 19. Substitute x+3x+3 for yy in the second equation: 3x+2(x+3)=193x + 2(x+3) = 19. This simplifies to 5x+6=195x+6=19, so 5x=135x=13 and x=135x=\frac{13}{5}. Then y=135+3=285y = \frac{13}{5} + 3 = \frac{28}{5}. The solution is (135,285)(\frac{13}{5}, \frac{28}{5}).
  • Solve the system 2xy=82x - y = 8 and x+3y=11x + 3y = 11. From the first equation, solve for yy: y=2x8y = 2x - 8. Substitute this into the second equation: x+3(2x8)=11x + 3(2x-8) = 11. This gives 7x24=117x-24=11, so 7x=357x=35 and x=5x=5. Then y=2(5)8=2y=2(5)-8=2, making the solution (5,2)(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.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Introduction to 2x2 Linear Systems and Substitution Method

Property

A solution to a 2×22 \times 2 linear system is an ordered pair that satisfies both equations.

The substitution method involves solving one equation for a variable, then substituting that expression into the other equation.

Section 2

The Substitution Method

Property

To solve a system by substitution, follow these steps:

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

Examples

  • Solve the system y=x+3y = x + 3 and 3x+2y=193x + 2y = 19. Substitute x+3x+3 for yy in the second equation: 3x+2(x+3)=193x + 2(x+3) = 19. This simplifies to 5x+6=195x+6=19, so 5x=135x=13 and x=135x=\frac{13}{5}. Then y=135+3=285y = \frac{13}{5} + 3 = \frac{28}{5}. The solution is (135,285)(\frac{13}{5}, \frac{28}{5}).
  • Solve the system 2xy=82x - y = 8 and x+3y=11x + 3y = 11. From the first equation, solve for yy: y=2x8y = 2x - 8. Substitute this into the second equation: x+3(2x8)=11x + 3(2x-8) = 11. This gives 7x24=117x-24=11, so 7x=357x=35 and x=5x=5. Then y=2(5)8=2y=2(5)-8=2, making the solution (5,2)(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.