Learn on PengiBig Ideas Math, Algebra 1Chapter 5: Solving Systems of Linear Equations

Lesson 3: Solving Systems of Linear Equations by Elimination

Property The Elimination Method is based on the Addition Property of Equality. When you add equal quantities to both sides of an equation, the results are equal. For any expressions $a, b, c$, and $d$, if $a = b$ and $c = d$, then $a+c = b+d$. To solve a system of equations by elimination, we start with both equations in standard form. We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Section 1

Solve by Elimination

Property

The Elimination Method is based on the Addition Property of Equality. When you add equal quantities to both sides of an equation, the results are equal. For any expressions a,b,ca, b, c, and dd, if a=ba = b and c=dc = d, then a+c=b+da+c = b+d. To solve a system of equations by elimination, we start with both equations in standard form. We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Examples

  • To solve the system {x+y=8xy=4\begin{cases} x + y = 8 \\ x - y = 4 \end{cases}, we add the equations. The yy terms are opposites and eliminate, giving 2x=122x=12, so x=6x=6. Substituting back, 6+y=86+y=8, so y=2y=2. The solution is (6,2)(6, 2).
  • To solve {2x+y=73x2y=0\begin{cases} 2x + y = 7 \\ 3x - 2y = 0 \end{cases}, multiply the first equation by 2 to make the yy coefficients opposites: 4x+2y=144x + 2y = 14. Adding this to 3x2y=03x - 2y = 0 gives 7x=147x=14, so x=2x=2. Then 2(2)+y=72(2)+y=7, so y=3y=3. The solution is (2,3)(2, 3).
  • To solve {3x+2y=82x+5y=9\begin{cases} 3x + 2y = 8 \\ 2x + 5y = 9 \end{cases}, multiply the first equation by 2 and the second by 3-3 to get 6x6x and 6x-6x. This gives {6x+4y=166x15y=27\begin{cases} 6x + 4y = 16 \\ -6x - 15y = -27 \end{cases}. Adding them yields 11y=11-11y = -11, so y=1y=1. Then 3x+2(1)=83x+2(1)=8, so x=2x=2. The solution is (2,1)(2, 1).

Explanation

This method adds two equations together. The goal is to make the coefficients of one variable opposites (like 5x5x and 5x-5x). When you add the equations, that variable cancels out, leaving a simple, one-variable equation to solve.

Section 2

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 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.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solve by Elimination

Property

The Elimination Method is based on the Addition Property of Equality. When you add equal quantities to both sides of an equation, the results are equal. For any expressions a,b,ca, b, c, and dd, if a=ba = b and c=dc = d, then a+c=b+da+c = b+d. To solve a system of equations by elimination, we start with both equations in standard form. We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Examples

  • To solve the system {x+y=8xy=4\begin{cases} x + y = 8 \\ x - y = 4 \end{cases}, we add the equations. The yy terms are opposites and eliminate, giving 2x=122x=12, so x=6x=6. Substituting back, 6+y=86+y=8, so y=2y=2. The solution is (6,2)(6, 2).
  • To solve {2x+y=73x2y=0\begin{cases} 2x + y = 7 \\ 3x - 2y = 0 \end{cases}, multiply the first equation by 2 to make the yy coefficients opposites: 4x+2y=144x + 2y = 14. Adding this to 3x2y=03x - 2y = 0 gives 7x=147x=14, so x=2x=2. Then 2(2)+y=72(2)+y=7, so y=3y=3. The solution is (2,3)(2, 3).
  • To solve {3x+2y=82x+5y=9\begin{cases} 3x + 2y = 8 \\ 2x + 5y = 9 \end{cases}, multiply the first equation by 2 and the second by 3-3 to get 6x6x and 6x-6x. This gives {6x+4y=166x15y=27\begin{cases} 6x + 4y = 16 \\ -6x - 15y = -27 \end{cases}. Adding them yields 11y=11-11y = -11, so y=1y=1. Then 3x+2(1)=83x+2(1)=8, so x=2x=2. The solution is (2,1)(2, 1).

Explanation

This method adds two equations together. The goal is to make the coefficients of one variable opposites (like 5x5x and 5x-5x). When you add the equations, that variable cancels out, leaving a simple, one-variable equation to solve.

Section 2

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 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.