Learn on PengiPengi Math (Grade 8)Chapter 4: Systems of Linear Equations

Lesson 3: Solving Systems by Elimination

In this Grade 8 lesson from Pengi Math Chapter 4, students learn to solve systems of linear equations using the elimination method by adding or subtracting equations to cancel out a variable. The lesson covers writing equations in standard form (Ax + By = C), identifying additive inverses, and multiplying equations to create opposite coefficients when needed. Students then practice the full elimination process, solving for one variable and back-substituting to find the complete solution.

Section 1

Standard Form of Linear Equations

Property

A linear equation in standard form is written as Ax+By=CAx + By = C, where AA, BB, and CC are real numbers, and AA and BB are not both zero.
This is one of the most useful forms for linear equations because it clearly shows the relationship between the variables and makes certain calculations easier.

Examples

Section 2

The Step-by-Step Method (Multiplying First)

Property

When solving systems by elimination, you must write both equations in standard form first and clear any fractions. Sometimes, you must multiply one or both equations by a constant to make the coefficients of one variable opposites before adding.

Examples

  • Multiplying One Equation: For the system 3x+4y=113x + 4y = 11 and 5x2y=135x - 2y = 13, multiply the second equation by 2 to get opposite yy-coefficients. This results in 3x+4y=113x + 4y = 11 and 10x4y=2610x - 4y = 26.
  • Multiplying Both Equations: For the system 2x+5y=82x + 5y = 8 and 3x+7y=123x + 7y = 12, multiply the first by 3 and the second by -2 to get opposite xx-coefficients. This results in 6x+15y=246x + 15y = 24 and 6x14y=24-6x - 14y = -24.
  • Full Process: Solve 2x+5y=342x + 5y = -34 and 3x+2y=44-3x + 2y = -44. Multiply to make the x-terms opposites: 3(2x+5y)=3(34)3(2x + 5y) = 3(-34) and 2(3x+2y)=2(44)2(-3x + 2y) = 2(-44). This simplifies to 6x+15y=1026x + 15y = -102 and 6x+4y=88-6x + 4y = -88. Adding them gives 19y=19019y = -190, so y=10y = -10. Substitute back to get x=8x = 8, making the solution (8,10)(8, -10).

Explanation

Strategic multiplication creates the opposite coefficients needed for elimination to work effectively. Look for the variable that will be easiest to eliminate by examining both coefficients and choosing appropriate multipliers. Once you add the equations and solve for the first variable, always substitute that answer back into an original equation to find the second variable, and check your final ordered pair.

Section 3

The Elimination Method: Step-by-Step

Property

Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
Step 2. Make the coefficients of one variable opposites.

  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.

Step 3. Add the equations resulting from Step 2 to eliminate one variable.
Step 4. Solve for the remaining variable.
Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Step 6. Write the solution as an ordered pair.
Step 7. Check that the ordered pair is a solution to both original equations.

Examples

  • Solve {4x+3y=1x3y=4\begin{cases} 4x + 3y = 1 \\ x - 3y = 4 \end{cases}. The equations are in standard form and the yy terms are opposites. Add them: 5x=55x=5, so x=1x=1. Substitute into the second equation: 13y=41-3y=4, so 3y=3-3y=3 and y=1y=-1. The solution is (1,1)(1, -1).
  • Solve {x+13y=212x+14y=2\begin{cases} x + \frac{1}{3}y = 2 \\ \frac{1}{2}x + \frac{1}{4}y = 2 \end{cases}. First, clear fractions by multiplying the first equation by 3 and the second by 4 to get {3x+y=62x+y=8\begin{cases} 3x+y=6 \\ 2x+y=8 \end{cases}. Multiply the second equation by 1-1 and add: (3x+y)+(2xy)=68(3x+y) + (-2x-y) = 6-8, which gives x=2x=-2. Then 3(2)+y=63(-2)+y=6, so y=12y=12. The solution is (2,12)(-2, 12).
  • Solve {y=52x3x2y=4\begin{cases} y = 5 - 2x \\ 3x - 2y = 4 \end{cases}. First, rewrite the first equation in standard form: 2x+y=52x+y=5. Now multiply it by 2 to get 4x+2y=104x+2y=10. Add this to 3x2y=43x-2y=4 to get 7x=147x=14, so x=2x=2. Substitute into y=52xy = 5 - 2x to get y=52(2)=1y=5-2(2)=1. The solution is (2,1)(2, 1).

Explanation

This is a step-by-step recipe for success. First, get your equations into Ax+By=CAx+By=C form. Next, multiply to create opposite terms. Then, add, solve for one variable, substitute back to find the other, and always check your answer.

Book overview

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Chapter 4: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Introduction to Systems and Solving by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems by Substitution

  3. Lesson 3Current

    Lesson 3: Solving Systems by Elimination

  4. Lesson 4

    Lesson 4: Number of Solutions and Special Cases

  5. Lesson 5

    Lesson 5: Real-World Applications of Linear Systems

Lesson overview

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Section 1

Standard Form of Linear Equations

Property

A linear equation in standard form is written as Ax+By=CAx + By = C, where AA, BB, and CC are real numbers, and AA and BB are not both zero.
This is one of the most useful forms for linear equations because it clearly shows the relationship between the variables and makes certain calculations easier.

Examples

Section 2

The Step-by-Step Method (Multiplying First)

Property

When solving systems by elimination, you must write both equations in standard form first and clear any fractions. Sometimes, you must multiply one or both equations by a constant to make the coefficients of one variable opposites before adding.

Examples

  • Multiplying One Equation: For the system 3x+4y=113x + 4y = 11 and 5x2y=135x - 2y = 13, multiply the second equation by 2 to get opposite yy-coefficients. This results in 3x+4y=113x + 4y = 11 and 10x4y=2610x - 4y = 26.
  • Multiplying Both Equations: For the system 2x+5y=82x + 5y = 8 and 3x+7y=123x + 7y = 12, multiply the first by 3 and the second by -2 to get opposite xx-coefficients. This results in 6x+15y=246x + 15y = 24 and 6x14y=24-6x - 14y = -24.
  • Full Process: Solve 2x+5y=342x + 5y = -34 and 3x+2y=44-3x + 2y = -44. Multiply to make the x-terms opposites: 3(2x+5y)=3(34)3(2x + 5y) = 3(-34) and 2(3x+2y)=2(44)2(-3x + 2y) = 2(-44). This simplifies to 6x+15y=1026x + 15y = -102 and 6x+4y=88-6x + 4y = -88. Adding them gives 19y=19019y = -190, so y=10y = -10. Substitute back to get x=8x = 8, making the solution (8,10)(8, -10).

Explanation

Strategic multiplication creates the opposite coefficients needed for elimination to work effectively. Look for the variable that will be easiest to eliminate by examining both coefficients and choosing appropriate multipliers. Once you add the equations and solve for the first variable, always substitute that answer back into an original equation to find the second variable, and check your final ordered pair.

Section 3

The Elimination Method: Step-by-Step

Property

Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
Step 2. Make the coefficients of one variable opposites.

  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.

Step 3. Add the equations resulting from Step 2 to eliminate one variable.
Step 4. Solve for the remaining variable.
Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Step 6. Write the solution as an ordered pair.
Step 7. Check that the ordered pair is a solution to both original equations.

Examples

  • Solve {4x+3y=1x3y=4\begin{cases} 4x + 3y = 1 \\ x - 3y = 4 \end{cases}. The equations are in standard form and the yy terms are opposites. Add them: 5x=55x=5, so x=1x=1. Substitute into the second equation: 13y=41-3y=4, so 3y=3-3y=3 and y=1y=-1. The solution is (1,1)(1, -1).
  • Solve {x+13y=212x+14y=2\begin{cases} x + \frac{1}{3}y = 2 \\ \frac{1}{2}x + \frac{1}{4}y = 2 \end{cases}. First, clear fractions by multiplying the first equation by 3 and the second by 4 to get {3x+y=62x+y=8\begin{cases} 3x+y=6 \\ 2x+y=8 \end{cases}. Multiply the second equation by 1-1 and add: (3x+y)+(2xy)=68(3x+y) + (-2x-y) = 6-8, which gives x=2x=-2. Then 3(2)+y=63(-2)+y=6, so y=12y=12. The solution is (2,12)(-2, 12).
  • Solve {y=52x3x2y=4\begin{cases} y = 5 - 2x \\ 3x - 2y = 4 \end{cases}. First, rewrite the first equation in standard form: 2x+y=52x+y=5. Now multiply it by 2 to get 4x+2y=104x+2y=10. Add this to 3x2y=43x-2y=4 to get 7x=147x=14, so x=2x=2. Substitute into y=52xy = 5 - 2x to get y=52(2)=1y=5-2(2)=1. The solution is (2,1)(2, 1).

Explanation

This is a step-by-step recipe for success. First, get your equations into Ax+By=CAx+By=C form. Next, multiply to create opposite terms. Then, add, solve for one variable, substitute back to find the other, and always check your answer.

Book overview

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

Continue this chapter

Chapter 4: Systems of Linear Equations

  1. Lesson 1

    Lesson 1: Introduction to Systems and Solving by Graphing

  2. Lesson 2

    Lesson 2: Solving Systems by Substitution

  3. Lesson 3Current

    Lesson 3: Solving Systems by Elimination

  4. Lesson 4

    Lesson 4: Number of Solutions and Special Cases

  5. Lesson 5

    Lesson 5: Real-World Applications of Linear Systems