Learn on PengiReveal Math, Course 3Module 6: Systems of Linear Equations

Lesson 6-4: Solve Systems of Equations by Elimination

In this Grade 8 lesson from Reveal Math, Course 3 (Module 6), students learn how to solve systems of linear equations using elimination, an algebraic method that removes one variable by adding or subtracting equations with opposite or equal coefficients. Students practice both elimination by addition, when coefficients are opposites, and elimination by subtraction, when coefficients are the same, then substitute back to find the solution as an ordered pair. This lesson builds core algebra skills needed for working with systems of equations throughout Course 3.

Section 1

The Concept of Elimination (Add or Subtract)

Property

If two pairs of expressions are equal, you can add or subtract the corresponding sides to create a new, valid equation. When coefficients of a variable have opposite signs, add the equations to eliminate that variable. When coefficients of a variable have the exact same sign, subtract the equations to eliminate the variable.

Examples

  • Adding to Eliminate: Given the equations x4y=5x - 4y = 5 and 2x+4y=102x + 4y = 10, the coefficients of yy are 4-4 and 44 (opposite signs). Adding the equations eliminates the yy term completely.
  • Subtracting to Eliminate: Given the equations 3x+2y=83x + 2y = 8 and x+2y=4x + 2y = 4, the coefficients of yy are both 22 (same signs). Subtracting the second equation from the first gives (3x+2y)(x+2y)=84(3x + 2y) - (x + 2y) = 8 - 4, which simplifies to 2x=42x = 4.

Explanation

Think of it as teaming up your equations to knock out a variable! The key to elimination is recognizing coefficient signs to determine the correct operation. When coefficients have opposite signs, adding the equations causes them to cancel out because a positive plus a negative equals zero. When coefficients have the same sign, subtracting eliminates the variable. It’s a clean knockout that simplifies the whole problem instantly!

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

Special Cases and Graphical Verification

Property

When using elimination, if both variables are eliminated, the resulting mathematical statement determines the number of solutions:

  • If the resulting equation is a false statement (e.g., 0=c0 = c), the system has no solution.
  • If the resulting equation is a true statement (e.g., 0=00 = 0), the system has infinitely many solutions.

You can always verify your algebraic solution by graphing the two lines to confirm they intersect exactly at your calculated (x,y)(x, y) coordinate.

Examples

  • No Solution: For the system 2xy=52x - y = 5 and 4x+2y=12-4x + 2y = -12, multiply the first equation by 2 to get 4x2y=104x - 2y = 10. Adding this to the second equation results in 0=20 = -2, a false statement, meaning there is no solution.
  • Infinitely Many Solutions: For the system x+3y=6x + 3y = 6 and 2x+6y=122x + 6y = 12, multiply the first equation by -2 to get 2x6y=12-2x - 6y = -12. Adding this to the second equation results in 0=00 = 0, a true statement, meaning there are infinitely many solutions.
  • Graphical Check: If elimination gives (4,1)(4, -1) as a solution, both original equations should pass exactly through the point (4,1)(4, -1) when graphed.

Explanation

Sometimes when you perform elimination, both variables cancel out. If you are left with a false statement like 0=20 = -2, it means there is no pair of (x,y)(x, y) that can satisfy both equations because the lines are perfectly parallel. If you are left with a true statement like 0=00 = 0, it means the two equations describe the exact same line, resulting in infinitely many solutions.

Book overview

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Continue this chapter

Module 6: Systems of Linear Equations

  1. Lesson 1

    Lesson 6-1: Solve Systems of Equations by Graphing

  2. Lesson 2

    Lesson 6-2: Determine Number of Solutions

  3. Lesson 3

    Lesson 6-3: Solve Systems of Equations by Substitution

  4. Lesson 4Current

    Lesson 6-4: Solve Systems of Equations by Elimination

  5. Lesson 5

    Lesson 6-5: Write and Solve Systems of Equations

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

The Concept of Elimination (Add or Subtract)

Property

If two pairs of expressions are equal, you can add or subtract the corresponding sides to create a new, valid equation. When coefficients of a variable have opposite signs, add the equations to eliminate that variable. When coefficients of a variable have the exact same sign, subtract the equations to eliminate the variable.

Examples

  • Adding to Eliminate: Given the equations x4y=5x - 4y = 5 and 2x+4y=102x + 4y = 10, the coefficients of yy are 4-4 and 44 (opposite signs). Adding the equations eliminates the yy term completely.
  • Subtracting to Eliminate: Given the equations 3x+2y=83x + 2y = 8 and x+2y=4x + 2y = 4, the coefficients of yy are both 22 (same signs). Subtracting the second equation from the first gives (3x+2y)(x+2y)=84(3x + 2y) - (x + 2y) = 8 - 4, which simplifies to 2x=42x = 4.

Explanation

Think of it as teaming up your equations to knock out a variable! The key to elimination is recognizing coefficient signs to determine the correct operation. When coefficients have opposite signs, adding the equations causes them to cancel out because a positive plus a negative equals zero. When coefficients have the same sign, subtracting eliminates the variable. It’s a clean knockout that simplifies the whole problem instantly!

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

Special Cases and Graphical Verification

Property

When using elimination, if both variables are eliminated, the resulting mathematical statement determines the number of solutions:

  • If the resulting equation is a false statement (e.g., 0=c0 = c), the system has no solution.
  • If the resulting equation is a true statement (e.g., 0=00 = 0), the system has infinitely many solutions.

You can always verify your algebraic solution by graphing the two lines to confirm they intersect exactly at your calculated (x,y)(x, y) coordinate.

Examples

  • No Solution: For the system 2xy=52x - y = 5 and 4x+2y=12-4x + 2y = -12, multiply the first equation by 2 to get 4x2y=104x - 2y = 10. Adding this to the second equation results in 0=20 = -2, a false statement, meaning there is no solution.
  • Infinitely Many Solutions: For the system x+3y=6x + 3y = 6 and 2x+6y=122x + 6y = 12, multiply the first equation by -2 to get 2x6y=12-2x - 6y = -12. Adding this to the second equation results in 0=00 = 0, a true statement, meaning there are infinitely many solutions.
  • Graphical Check: If elimination gives (4,1)(4, -1) as a solution, both original equations should pass exactly through the point (4,1)(4, -1) when graphed.

Explanation

Sometimes when you perform elimination, both variables cancel out. If you are left with a false statement like 0=20 = -2, it means there is no pair of (x,y)(x, y) that can satisfy both equations because the lines are perfectly parallel. If you are left with a true statement like 0=00 = 0, it means the two equations describe the exact same line, resulting in infinitely many solutions.

Book overview

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

Continue this chapter

Module 6: Systems of Linear Equations

  1. Lesson 1

    Lesson 6-1: Solve Systems of Equations by Graphing

  2. Lesson 2

    Lesson 6-2: Determine Number of Solutions

  3. Lesson 3

    Lesson 6-3: Solve Systems of Equations by Substitution

  4. Lesson 4Current

    Lesson 6-4: Solve Systems of Equations by Elimination

  5. Lesson 5

    Lesson 6-5: Write and Solve Systems of Equations