Solving Systems by Elimination
By the end of this lesson, you’ll be able to:
- Solve systems of linear equations using the elimination method.
- Eliminate a variable by adding or subtracting equations.
- Multiply one or both equations to create opposite coefficients.
- Decide when elimination is the quickest and cleanest method.
Key Ideas
Elimination is often the fastest approach when:
- one variable already has coefficients that are opposites (like \(y\) and \(-y\)), or
- you can make opposites with a small multiplier, or
- substitution would create fractions early on.
General plan:
- Align the equations vertically.
- Multiply one or both equations if needed to create opposite coefficients.
- Add or subtract to eliminate a variable.
- Solve the resulting one-variable equation.
- Substitute back to get the second variable.
- Check the solution in both equations.
Common Problem Types
Naturally Aligned Coefficients
The variable eliminates immediately when you add or subtract.
Coefficients That Become Opposites After Scaling
Multiply one or both equations to get opposite coefficients.
Choosing Between Addition and Subtraction
Pick whichever avoids a sign mistake or unnecessary negatives.
When Substitution Gets Messy
Elimination helps when isolating a variable first would create fractions.
Strategies
- Pick the variable that’s easiest to eliminate—look for matching or opposite coefficients.
- Use subtraction or addition based on which yields clean cancellation.
- When multiplying equations, keep numbers as small as possible.
- After eliminating, rewrite the new single-variable equation cleanly on its own line.
- Always substitute into the simpler original equation when solving for the second variable.
Worked Examples
Example 1 — Ready for Elimination
Solve:
\[ \begin{cases} 2x + y = 9 \\ 2x - y = 1 \end{cases} \]
Subtract the second equation from the first:
\[ \begin{split} (2x + y) - (2x - y) &= 9 - 1 \\ 2y &= 8 \\ y &= 4 \end{split} \]
Substitute back:
\[ \begin{split} 2x + 4 &= 9 \\ 2x &= 5 \\ x &= \frac{5}{2} \end{split} \]
Solution: \(\left(\frac{5}{2}, 4\right)\)
Example 2 — Multiply, Then Eliminate
Solve:
\[ \begin{cases} 3x + 2y = 16 \\ x - y = 1 \end{cases} \]
Multiply the second equation by 2:
\[ \begin{split} 2x - 2y &= 2 \end{split} \]
Add the equations:
\[ \begin{split} (3x + 2y) + (2x - 2y) &= 16 + 2 \\ 5x &= 18 \\ x &= \frac{18}{5} \end{split} \]
Substitute back to find \(y\):
\[ \begin{split} 3\left(\frac{18}{5}\right) + 2y &= 16 \\ \frac{54}{5} + 2y &= 16 \\ 2y &= 16 - \frac{54}{5} = \frac{26}{5} \\ y &= \frac{13}{5} \end{split} \]
Solution: \(\left(\frac{18}{5}, \frac{13}{5}\right)\)
- Multiplying an equation but forgetting to multiply the constant term.
- Adding when subtraction was needed for cancellation.
- Dropping negative signs when combining equations.
- Solving for one variable and forgetting to find the other.
Practice Problems
\[ \begin{cases} x + y = 8 \\ x - y = 2 \end{cases} \]
\[ \begin{cases} 2x - 3y = 5 \\ 4x + 3y = 19 \end{cases} \]
\[ \begin{cases} 3x + y = 10 \\ 2x - y = 1 \end{cases} \]
1.
Add equations:
\[ \begin{split} (x + y) + (x - y) &= 8 + 2 \\ 2x &= 10 \\ x &= 5 \end{split} \]
Then:
\[ y = 8 - 5 = 3 \]
Answer: \((5, 3)\)
2.
Add equations:
\[ \begin{split} (2x - 3y) + (4x + 3y) &= 5 + 19 \\ 6x &= 24 \\ x &= 4 \end{split} \]
Substitute:
\[ 2(4) - 3y = 5 \Rightarrow 8 - 3y = 5 \Rightarrow y = 1 \]
Answer: \((4, 1)\)
3.
Add equations:
\[ \begin{split} (3x + y) + (2x - y) &= 10 + 1 \\ 5x &= 11 \\ x &= \frac{11}{5} \end{split} \]
Substitute:
\[ y = 10 - 3\left(\frac{11}{5}\right) = \frac{-13}{5} \]
Answer: \(\left(\frac{11}{5}, \frac{-13}{5}\right)\)
Summary
- Elimination is ideal when coefficients already match or can be matched with small multipliers.
- Eliminate one variable, solve the new equation, then back-substitute.
- Multiply equations carefully—every term gets multiplied.
- A clean elimination step often prevents messy algebra later.
- Scan for the easiest variable to eliminate—often the one with matching or opposite coefficients.
- Use the smallest multipliers possible.
- Keep track of signs when adding or subtracting equations.
- Once a variable is found, substitute into the simplest equation available.