Elimination Method for Systems of Equations
The elimination method solves a system of equations by adding or subtracting the equations so that one variable cancels out. If nothing cancels right away, first multiply one or both equations so a pair of coefficients become opposites. Solve for the variable that’s left, then substitute to find the other one.
Before this lesson: You should know what a system of equations is and how to use the substitution method to find the second variable.
How the Elimination Method Works
If two things are equal, and you add the same amount to both, they stay equal. Elimination uses this idea with whole equations.
Look at this system:
\begin{cases}
x + y = 10 \\
x - y = 4
\end{cases}The y-terms are opposites: +y and -y. When you add the two equations, they cancel, or “eliminate” each other.
\begin{array}{rrcl}
& x + y & = & 10 \\
+ & x - y & = & 4 \\
\hline
& 2x & = & 14
\end{array}Now only x is left, so x = 7. Then 7 + y = 10, so y = 3. The solution is (7, 3).
Steps for the Elimination Method
- Line up the equations so the x-terms, y-terms, equal signs and numbers are in columns.
- Make opposites. If no pair of coefficients are opposites, multiply one or both equations by a number so they are. Multiply every term, on both sides.
- Add the equations to eliminate one variable.
- Solve for the variable that’s left.
- Substitute that value into either original equation to find the other variable.
- Check the pair in both original equations.
Add or subtract? If the coefficients are opposites, like 3y and −3y, add. If they are the same, like 2y and 2y, subtract, or multiply one equation by −1 and then add.
Worked Examples
Example 1: Add the equations
Solve the system 2x + y = 11 and 3x - y = 9.
Step 1. The y-terms are opposites, so add the equations.
\begin{array}{rrcl}
& 2x + y & = & 11 \\
+ & 3x - y & = & 9 \\
\hline
& 5x & = & 20 \\
& x & = & 4
\end{array}Step 2. Substitute into the first equation: 2(4) + y = 11, so y = 3.
Check: 2(4) + 3 = 11 and 3(4) - 3 = 9.
Answer: (4, 3)
Example 2: Subtract the equations
Solve the system 3x + 2y = 16 and x + 2y = 8.
Step 1. The y-terms are the same, 2y and 2y. Subtract the second equation from the first. Subtract every term.
\begin{array}{rrcl}
& 3x + 2y & = & 16 \\
- & (x + 2y) & = & 8 \\
\hline
& 2x & = & 8 \\
& x & = & 4
\end{array}Step 2. Substitute into the second equation: 4 + 2y = 8, so 2y = 4 and y = 2.
Check: 3(4) + 2(2) = 16 and 4 + 2(2) = 8.
Answer: (4, 2)
Example 3: Multiply one equation first
Solve the system 2x + 3y = 12 and x - y = 1.
Step 1. Nothing cancels yet. Multiply the second equation by 3, so its y-term becomes −3y.
3(x - y) = 3(1) \quad \Rightarrow \quad 3x - 3y = 3
Step 2. Add it to the first equation.
\begin{array}{rrcl}
& 2x + 3y & = & 12 \\
+ & 3x - 3y & = & 3 \\
\hline
& 5x & = & 15 \\
& x & = & 3
\end{array}Step 3. Substitute into x - y = 1: 3 - y = 1, so y = 2.
Check: 2(3) + 3(2) = 12 and 3 - 2 = 1.
Answer: (3, 2)
Example 4: Multiply both equations
Solve the system 3x + 4y = 10 and 2x - 3y = 1.
Step 1. To eliminate y, make the y-terms 12y and −12y. Multiply the first equation by 3 and the second by 4.
\begin{aligned}
3(3x + 4y) = 3(10) \quad &\Rightarrow \quad 9x + 12y = 30 \\
4(2x - 3y) = 4(1) \quad &\Rightarrow \quad 8x - 12y = 4
\end{aligned}Step 2. Add the new equations: 17x = 34, so x = 2.
Step 3. Substitute into 3x + 4y = 10: 6 + 4y = 10, so y = 1.
Check: 3(2) + 4(1) = 10 and 2(2) - 3(1) = 1.
Answer: (2, 1)
Example 5: Word problem
Three tacos and two drinks cost $13. Five tacos and two drinks cost $19. How much does one taco cost? How much does one drink cost?
Step 1. Let t be the price of a taco and d the price of a drink.
\begin{cases}
3t + 2d = 13 \\
5t + 2d = 19
\end{cases}Step 2. The d-terms are the same, so subtract the first equation from the second: 2t = 6, so t = 3.
Step 3. Substitute: 3(3) + 2d = 13, so 2d = 4 and d = 2.
Check: 5(3) + 2(2) = 15 + 4 = 19.
Answer: A taco costs $3, and a drink costs $2.
Which Method Should You Use?
| Method | Best when | Example |
|---|---|---|
| Graphing | You want to see the solution, and it has whole numbers | y = x + 1 and y = -x + 5 |
| Substitution | A variable is already alone or has a coefficient of 1 | y = 2x - 6 and 4x - 3y = 6 |
| Elimination | Both equations are in the form Ax + By = C | 3x + 4y = 10 and 2x - 3y = 1 |
When Both Variables Cancel
Sometimes x and y both cancel. Look at what is left.
- A false statement, like 0 = 6, means no solution. The lines are parallel.
- A true statement, like 0 = 0, means infinitely many solutions. The equations describe the same line.
Learn more in Number of Solutions.
Common Mistakes
| Mistake | What it looks like | How to fix it |
|---|---|---|
| Multiplying only one side | 3(x - y) = 1 becomes 3x - 3y = 1 | Multiply both sides: 3x - 3y = 3. |
| Multiplying only one term | 3(x - y) becomes 3x - y | Multiply every term: 3x - 3y. |
| Adding when the terms match | Adding 2y + 2y gives 4y, which doesn’t cancel | Same terms: subtract. Opposite terms: add. |
| Subtracting only part | (3x + 2y) - (x + 2y) = 2x + 4y | Subtract every term: 2y - 2y = 0. |
| Stopping after one variable | Answering “x = 4“ | Substitute to find the other variable too. |
Practice Problems
Solve each system by elimination. Check your answers.
- x + y = 9 and x - y = 3
- 2x + y = 11 and 3x - y = 9
- 4x + 3y = 25 and 4x - y = 13
- x + 2y = 7 and 3x - 2y = 5
- 2x + 5y = 1 and x + 3y = 1
- 3x - 2y = 8 and 5x + 4y = 6
- 2x + 3y = 7 and 3x + 4y = 10
- 4x - 6y = 10 and 2x - 3y = 5
- Two adult tickets and three student tickets cost $36. One adult ticket and three student tickets cost $27. Find the price of each ticket.
Show answers
- Add: 2x = 12, so x = 6. Then y = 3. Answer: (6, 3).
- Add: 5x = 20, so x = 4. Then 2(4) + y = 11, so y = 3. Answer: (4, 3).
- Subtract: 4y = 12, so y = 3. Then 4x - 3 = 13, so x = 4. Answer: (4, 3).
- Add: 4x = 12, so x = 3. Then 3 + 2y = 7, so y = 2. Answer: (3, 2).
- Multiply the second equation by 2: 2x + 6y = 2. Subtract it from the first: -y = -1, so y = 1. Then x + 3 = 1, so x = -2. Answer: (−2, 1).
- Multiply the first equation by 2: 6x - 4y = 16. Add: 11x = 22, so x = 2. Then 6 - 2y = 8, so y = -1. Answer: (2, −1).
- Multiply the first by 3 and the second by 2: 6x + 9y = 21 and 6x + 8y = 20. Subtract: y = 1. Then 2x + 3 = 7, so x = 2. Answer: (2, 1).
- Multiply the second equation by 2: 4x - 6y = 10. Subtract: 0 = 0. The equations describe the same line, so there are infinitely many solutions.
- 2a + 3s = 36 and a + 3s = 27. Subtract: a = 9. Then 9 + 3s = 27, so s = 6. An adult ticket costs $9, and a student ticket costs $6.
Elimination Method FAQ
What is the elimination method?
The elimination method solves a system by adding or subtracting the equations so one variable cancels. You may need to multiply an equation first to make opposite coefficients.
When is elimination better than substitution?
Elimination is usually faster when both equations are in standard form, Ax + By = C, and no variable is already alone.
Should you eliminate x or y?
Either one works and gives the same answer. Choose the variable that needs the least multiplying, or one that already has opposite coefficients.
What does it mean if both variables cancel?
If you get a false statement like 0 = 6, there is no solution. If you get a true statement like 0 = 0, there are infinitely many solutions.
Summary
- Line up the equations by x-terms, y-terms and numbers.
- Make one pair of coefficients opposites, multiplying every term if you need to.
- Add the equations to eliminate a variable. Subtract if the terms are the same.
- Solve, substitute to find the other variable, and check in both equations.
- 0 = 0 means infinitely many solutions. A false statement means no solution.
Previous lesson: Substitution Method
Next lesson: Number of Solutions
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