How to Solve an Equation with Elimination: A Step-by-Step Guide
The elimination method is a powerful algebraic technique for solving systems of linear equations. By strategically adding or subtracting equations, this method eliminates one variable, allowing you to solve for the remaining variables systematically. Whether you’re dealing with two or three variables, mastering elimination can simplify complex problems and save time. This guide will walk you through the process, provide clear examples, and address common pitfalls to ensure you can confidently apply the elimination method.
Steps to Solve a System of Equations Using Elimination
Follow these steps to solve a system of equations using the elimination method:
- Align the Equations: Write the equations vertically, ensuring like terms (variables and constants) are aligned in columns.
- Multiply Equations if Necessary: If the coefficients of one variable are not already additive inverses (e.g., 2 and -2), multiply one or both equations by constants to create such pairs.
- Add or Subtract Equations: Add or subtract the equations to eliminate one variable. Choose the operation that cancels the terms.
- Solve for the Remaining Variable: After elimination, solve the resulting single-variable equation.
- Substitute Back: Plug the value of the solved variable into one of the original equations to find the other variable(s).
- Verify the Solution: Substitute both values into all original equations to ensure they satisfy the system.
Example 1: Simple Elimination
Problem: Solve the system:
$
\begin{align*}
2x + 3y &= 12 \quad \text{(Equation 1)} \
4x - 3y &= 6 \quad \text{(Equation 2)}
\end{align*}
$
Solution:
- Step 1: The equations are already aligned. Notice that the coefficients of $ y $ (3 and -3) are additive inverses.
- Step 2: Add Equation 1 and Equation 2 to eliminate $ y $:
$ (2x + 3y) + (4x - 3y) = 12 + 6 \
6x = 18
$ - Step 3: Solve for $ x $:
$ x = \frac{18}{6} = 3
$ - Step 4: Substitute $ x = 3 $ into Equation 1 to find $ y $:
$ 2(3) + 3y = 12 \
6 + 3y = 12 \
3y = 6 \implies y = 2
$ - Step 5: Verify the solution $ (3, 2) $ in both equations:
- Equation 1: $ 2(3) + 3(2) = 6 + 6 = 12 $ ✔️
- Equation 2: $ 4(3) - 3(2) = 12 - 6 = 6 $ ✔️
Answer: $ x = 3 $, $ y = 2 $ Worth keeping that in mind..
Example 2: Elimination with Multiplication
Problem: Solve the system:
$
\begin{align*}
3x + 2y &= 8 \quad \text{(Equation 1)} \
5x - 4y &= 2 \quad \text{(Equation 2)}
\end{align*}
$
Solution:
- Step 1: Align the equations. The coefficients of $ y $ (2 and -4) are not additive inverses.
- Step 2: Multiply Equation 1 by 2 to make the coefficients of $ y $ opposites:
$ 2(3x + 2y) = 2(8) \implies 6x + 4y = 16 \quad \text{(New Equation 3)}
$ - Step 3: Add Equation 3 and Equation 2 to eliminate $ y $:
$ (6x + 4y) + (5x - 4y) = 16 + 2 \
11x = 18
$ - Step 4: Solve for $ x $:
$ x = \frac{18}{11}
$ - Step 5: Substitute $ x = \frac{18}{11} $ into Equation 1 to find $ y $:
$ 3\left(\frac{18}{11}\right) + 2y = 8 \
\frac{54}{11} + 2y = 8 \
2y = 8 - \frac{54}{11} = \frac{88 - 54}{11} = \frac{34}{11} \
y = \frac{17}{11}
$ - Step 6: Verify the solution $ \left(\frac{18}{11}, \frac{17}{11}\right) $ in both equations.
Answer: $ x = \frac{18}{11} $, $ y = \frac{17}{11} $.
When to Use the Elimination Method
The elimination method is most effective when:
- The coefficients of one variable are already opposites (e.g., $ 5x $ and $ -5x $).
