Learn on PengiSaxon Algebra 1Chapter 7: Rational Expressions and Radicals

Lesson 63: Solving Systems of Linear Equations by Elimination

In this Grade 9 Saxon Algebra 1 lesson, students learn how to solve systems of linear equations using the elimination method by adding, subtracting, or multiplying one or both equations to cancel out a variable. The lesson covers all key cases, including eliminating a variable with opposite coefficients, matching coefficients, and scaling equations before combining. Students also apply the elimination method to real-world word problems, such as solving a coin collection scenario using a system of two equations.

Section 1

πŸ“˜ Solving Systems of Linear Equations by Elimination

New Concept

It is easier to eliminate one of the variables by combining the two equations using addition or subtraction.

What’s next

Next, you’ll apply this idea by adding, subtracting, and even multiplying entire equations to strategically remove variables and find the solution.

Section 2

Elimination by Addition

Property

When two linear equations have a variable with opposite coefficients (e.g., axax and βˆ’ax-ax), add the equations to eliminate that variable.

Explanation

Think of this as a mathematical disappearing act! When one variable in an equation is the exact opposite of its counterpart in the other equation, like a hero and its anti-hero (+5x+5x and βˆ’5x-5x), you can simply add the two equations. The opposite terms cancel each other out, leaving a much simpler problem to solve.

Examples

  • Given 3x+4y=103x + 4y = 10 and βˆ’3x+2y=8-3x + 2y = 8, adding them yields 6y=186y = 18, so y=3y = 3.
  • For the system x+y=5x + y = 5 and βˆ’x+y=1-x + y = 1, adding the equations results in 2y=62y = 6, so y=3y = 3.
  • In the system 8aβˆ’5b=118a - 5b = 11 and βˆ’8aβˆ’3b=5-8a - 3b = 5, adding them gives βˆ’8b=16-8b = 16, which simplifies to b=βˆ’2b = -2.

Section 3

Example Card: Elimination by Adding Equations

When two equations have perfectly opposite terms, adding them together can reveal the answer. This example uses the first key idea of this lesson, direct elimination.

Example Problem

Solve the system by elimination and check the answer: 4x+3y=54x + 3y = 5 and βˆ’4x+5y=11-4x + 5y = 11.

Step-by-Step

  1. The two equations have equal and opposite coefficients for the variable xx, so we can add the equations to eliminate it.
4x+3y=5βˆ’4x+5y=118y=16\begin{array}{r} 4x + 3y = 5 \\ -4x + 5y = 11 \\ \hline 8y = 16 \end{array}
  1. Now, we solve for yy.
y=2y = 2
  1. Substitute 22 for yy in the first original equation to solve for xx.
4x+3(2)=54x+6=54x=βˆ’1x=βˆ’14\begin{aligned} 4x + 3(2) &= 5 \\ 4x + 6 &= 5 \\ 4x &= -1 \\ x &= -\frac{1}{4} \end{aligned}
  1. The solution is (βˆ’14,2)(-\frac{1}{4}, 2).
  2. Check: Substitute (βˆ’14,2)(-\frac{1}{4}, 2) into the second equation to verify.
βˆ’4x+5y=11βˆ’4(βˆ’14)+5(2)=?111+10=?1111=11βœ“\begin{aligned} -4x + 5y &= 11 \\ -4(-\frac{1}{4}) + 5(2) &\stackrel{?}{=} 11 \\ 1 + 10 &\stackrel{?}{=} 11 \\ 11 &= 11 \quad βœ“ \end{aligned}

By adding the equations, the x-terms cancelled out, making the problem much simpler. This additive elimination works whenever you spot opposite coefficients.

Section 4

Elimination by Subtraction

Property

When two linear equations have a variable with identical coefficients (e.g., byby and byby), subtract one equation from the other to eliminate that variable.

Explanation

What happens when two terms are identical twins, like 3y3y and 3y3y? You can eliminate one by subtracting the entire equation from the other! This is like having two identical songs playing and hitting mute on one. The duplicate term vanishes, simplifying the system so you can easily find the value of the remaining variable.

Examples

  • Given 4x+5y=154x + 5y = 15 and 2x+5y=92x + 5y = 9, subtracting the second from the first gives 2x=62x = 6, so x=3x = 3.
  • For 8aβˆ’2b=128a - 2b = 12 and 3aβˆ’2b=23a - 2b = 2, subtracting the second equation results in 5a=105a = 10, so a=2a = 2.
  • In the system 10x+3y=2010x + 3y = 20 and 2x+3y=42x + 3y = 4, subtraction yields 8x=168x = 16, which simplifies to x=2x = 2.

Section 5

Multiplying to Create Opposites

Property

If no variables have the same or opposite coefficients, multiply one or both equations by a non-zero constant to create a pair of opposite coefficients.

Explanation

Sometimes, your equations aren't ready to cooperate and don't have matching or opposite terms. This is where you become the director! You can multiply one or even both equations by a strategic number to create the perfect set of opposites. Once you've engineered the cancellation, you just add the equations together and solve away.

Examples

  • System: 3x+2y=73x + 2y = 7 and 6xβˆ’5y=46x - 5y = 4. Multiply the first equation by βˆ’2-2 to get βˆ’6xβˆ’4y=βˆ’14-6x - 4y = -14 and eliminate xx.
  • System: 3a+2b=83a + 2b = 8 and 2aβˆ’5b=βˆ’12a - 5b = -1. Multiply the first by 22 and the second by βˆ’3-3 to get 6a+4b=166a+4b=16 and βˆ’6a+15b=3-6a+15b=3.

Book overview

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

Continue this chapter

Chapter 7: Rational Expressions and Radicals

  1. Lesson 1

    Lesson 61: Simplifying Radical Expressions

  2. Lesson 2

    Lesson 62: Displaying Data in Stem-and-Leaf Plots and Histograms

  3. Lesson 3Current

    Lesson 63: Solving Systems of Linear Equations by Elimination

  4. Lesson 4

    Lesson 64: Identifying, Writing, and Graphing Inverse Variation

  5. Lesson 5

    Lesson 65: Writing Equations of Parallel and Perpendicular Lines

  6. Lesson 6

    Lesson 66: Solving Inequalities by Adding or Subtracting

  7. Lesson 7

    Lesson 67: Solving and Classifying Special Systems of Linear Equations

  8. Lesson 8

    Lesson 68: Mutually Exclusive and Inclusive Events

  9. Lesson 9

    Lesson 69: Adding and Subtracting Radical Expressions

  10. Lesson 10

    Lesson 70: Solving Inequalities by Multiplying or Dividing

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Solving Systems of Linear Equations by Elimination

New Concept

It is easier to eliminate one of the variables by combining the two equations using addition or subtraction.

What’s next

Next, you’ll apply this idea by adding, subtracting, and even multiplying entire equations to strategically remove variables and find the solution.

Section 2

Elimination by Addition

Property

When two linear equations have a variable with opposite coefficients (e.g., axax and βˆ’ax-ax), add the equations to eliminate that variable.

Explanation

Think of this as a mathematical disappearing act! When one variable in an equation is the exact opposite of its counterpart in the other equation, like a hero and its anti-hero (+5x+5x and βˆ’5x-5x), you can simply add the two equations. The opposite terms cancel each other out, leaving a much simpler problem to solve.

Examples

  • Given 3x+4y=103x + 4y = 10 and βˆ’3x+2y=8-3x + 2y = 8, adding them yields 6y=186y = 18, so y=3y = 3.
  • For the system x+y=5x + y = 5 and βˆ’x+y=1-x + y = 1, adding the equations results in 2y=62y = 6, so y=3y = 3.
  • In the system 8aβˆ’5b=118a - 5b = 11 and βˆ’8aβˆ’3b=5-8a - 3b = 5, adding them gives βˆ’8b=16-8b = 16, which simplifies to b=βˆ’2b = -2.

Section 3

Example Card: Elimination by Adding Equations

When two equations have perfectly opposite terms, adding them together can reveal the answer. This example uses the first key idea of this lesson, direct elimination.

Example Problem

Solve the system by elimination and check the answer: 4x+3y=54x + 3y = 5 and βˆ’4x+5y=11-4x + 5y = 11.

Step-by-Step

  1. The two equations have equal and opposite coefficients for the variable xx, so we can add the equations to eliminate it.
4x+3y=5βˆ’4x+5y=118y=16\begin{array}{r} 4x + 3y = 5 \\ -4x + 5y = 11 \\ \hline 8y = 16 \end{array}
  1. Now, we solve for yy.
y=2y = 2
  1. Substitute 22 for yy in the first original equation to solve for xx.
4x+3(2)=54x+6=54x=βˆ’1x=βˆ’14\begin{aligned} 4x + 3(2) &= 5 \\ 4x + 6 &= 5 \\ 4x &= -1 \\ x &= -\frac{1}{4} \end{aligned}
  1. The solution is (βˆ’14,2)(-\frac{1}{4}, 2).
  2. Check: Substitute (βˆ’14,2)(-\frac{1}{4}, 2) into the second equation to verify.
βˆ’4x+5y=11βˆ’4(βˆ’14)+5(2)=?111+10=?1111=11βœ“\begin{aligned} -4x + 5y &= 11 \\ -4(-\frac{1}{4}) + 5(2) &\stackrel{?}{=} 11 \\ 1 + 10 &\stackrel{?}{=} 11 \\ 11 &= 11 \quad βœ“ \end{aligned}

By adding the equations, the x-terms cancelled out, making the problem much simpler. This additive elimination works whenever you spot opposite coefficients.

Section 4

Elimination by Subtraction

Property

When two linear equations have a variable with identical coefficients (e.g., byby and byby), subtract one equation from the other to eliminate that variable.

Explanation

What happens when two terms are identical twins, like 3y3y and 3y3y? You can eliminate one by subtracting the entire equation from the other! This is like having two identical songs playing and hitting mute on one. The duplicate term vanishes, simplifying the system so you can easily find the value of the remaining variable.

Examples

  • Given 4x+5y=154x + 5y = 15 and 2x+5y=92x + 5y = 9, subtracting the second from the first gives 2x=62x = 6, so x=3x = 3.
  • For 8aβˆ’2b=128a - 2b = 12 and 3aβˆ’2b=23a - 2b = 2, subtracting the second equation results in 5a=105a = 10, so a=2a = 2.
  • In the system 10x+3y=2010x + 3y = 20 and 2x+3y=42x + 3y = 4, subtraction yields 8x=168x = 16, which simplifies to x=2x = 2.

Section 5

Multiplying to Create Opposites

Property

If no variables have the same or opposite coefficients, multiply one or both equations by a non-zero constant to create a pair of opposite coefficients.

Explanation

Sometimes, your equations aren't ready to cooperate and don't have matching or opposite terms. This is where you become the director! You can multiply one or even both equations by a strategic number to create the perfect set of opposites. Once you've engineered the cancellation, you just add the equations together and solve away.

Examples

  • System: 3x+2y=73x + 2y = 7 and 6xβˆ’5y=46x - 5y = 4. Multiply the first equation by βˆ’2-2 to get βˆ’6xβˆ’4y=βˆ’14-6x - 4y = -14 and eliminate xx.
  • System: 3a+2b=83a + 2b = 8 and 2aβˆ’5b=βˆ’12a - 5b = -1. Multiply the first by 22 and the second by βˆ’3-3 to get 6a+4b=166a+4b=16 and βˆ’6a+15b=3-6a+15b=3.

Book overview

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

Continue this chapter

Chapter 7: Rational Expressions and Radicals

  1. Lesson 1

    Lesson 61: Simplifying Radical Expressions

  2. Lesson 2

    Lesson 62: Displaying Data in Stem-and-Leaf Plots and Histograms

  3. Lesson 3Current

    Lesson 63: Solving Systems of Linear Equations by Elimination

  4. Lesson 4

    Lesson 64: Identifying, Writing, and Graphing Inverse Variation

  5. Lesson 5

    Lesson 65: Writing Equations of Parallel and Perpendicular Lines

  6. Lesson 6

    Lesson 66: Solving Inequalities by Adding or Subtracting

  7. Lesson 7

    Lesson 67: Solving and Classifying Special Systems of Linear Equations

  8. Lesson 8

    Lesson 68: Mutually Exclusive and Inclusive Events

  9. Lesson 9

    Lesson 69: Adding and Subtracting Radical Expressions

  10. Lesson 10

    Lesson 70: Solving Inequalities by Multiplying or Dividing