Solving Systems of Equations Using the Elimination Method

When solving systems of linear equations, the elimination method is one of the most popular and systematic approaches. This method involves adding or subtracting equations to eliminate one of the variables, allowing us to solve for the other. Once one variable is found, we substitute it back into one of the original equations to find the second variable.

Steps for Solving Using the Elimination Method

Here are the general steps for solving a system of equations using the elimination method:

Align the equations so that like terms (variables and constants) are in columns.
Eliminate one variable by adding or subtracting the equations. If necessary, multiply one or both equations by a constant to align coefficients.
Solve for the remaining variable.
Substitute the solution back into one of the original equations to find the other variable.
Verify the solution by substituting the values into both original equations.

Example 1: A Simple Elimination Scenario

Solve the following system of equations:

$\begin{aligned}[t] 2x + 3y &= 8 \\ 4x - 3y &= 10 \end{aligned}$

Step 1: Align the equations

The equations are already aligned:

$\begin{aligned}2x + 3y &= 8 \\ 4x - 3y &= 10 \end{aligned}$

Step 2: Eliminate one variable

Notice that the coefficients of $y$ are opposites: $+3y$ in the first equation and $−3y$ in the second equation. To eliminate $y$ , add the two equations:

$(2x+3y)+(4x -3y)=8+10$

Simplify:

$6x=18$

Step 3: Solve for $x$

Divide both sides by $6$ :

$x=3$

Step 4: Solve for $y$

Substitute $x=3$ into one of the original equations. Let’s use the first equation:

$2(3)+3y=8$

Simplify:

$6+3y=8$

Subtract $6$ from both sides:

$3y=2$

Divide by $3$ :

$y=\frac{2}{3}$

Step 5: Verify the solution

Substitute $x=3$ and $y=\frac{2}{3}$ into both original equations to confirm they work:

For $2x+3y=8$ : $2(3) + 3\left(\frac{2}{3}\right) = 8 \quad \text{✓}$
For $4x−3y=10$ : $4(3) - 3\left(\frac{2}{3}\right) = 10 \quad \text{✓}$

Thus, the solution is: $\boxed{(x, y) = (3, \frac{2}{3})}$

Visualization

In the plot below, you can see the two lines represented by the equations. The point of intersection, marked on the graph, represents the solution to the system of equations.

Example 2: Multiplying to Align Coefficients

Solve the following system of equations:

$\begin{aligned}[t] 2y + 3x&= 12 \\ 5x - 4y &= -2 \end{aligned}$

Step 1: Align the equations

The equations are aligned below:

$\begin{aligned}[t] 3x + 2y &= 12 \\ 5x - 4y &= -2 \end{aligned}$

Step 2: Eliminate one variable

We need to align the coefficients of either $x$ or $y$ . Let’s eliminate $y$ . To do this, multiply the first equation by $2$ and the second equation by $1$ (so that the coefficients of $y$ are opposites):

$\begin{aligned}[t] 2(3x + 2y) &= 2(12) \\ 1(5x - 4y) &= 1(-2) \end{aligned}$

This gives us:

$\begin{aligned}[t] 6x + 4y &= 24 \\ 5x - 4y &= -2 \end{aligned}$

Now, add the equations to eliminate $y$ :

$(6x + 4y) + (5x - 4y) = 24 + (-2)$

Simplify: $11x = 22$

Step 3: Solve for $x$

Divide both sides by $11$ : $x = 2$

Step 4: Solve for $y$

Substitute $x=2$ into one of the original equations. Let’s use the first equation:

$3(2) + 2y = 12$

Simplify:

$6 + 2y = 12$

Subtract $6$ from both sides:

$2y = 6$

Divide by $2$ :

$y = 3$

Step 5: Verify the solution

Substitute $x = 2$ and $y=3$ into both original equations to confirm they work:

For $3x+2y=12$ : $3(2) + 2(3) = 12 \quad \text{✓}$
For $5x−4y=−2$ : $5(2) - 4(3) = -2 \quad \text{✓}$

Thus, the solution is: $\boxed{(x, y) = (2, 3)}$

Visualization

The plot below illustrates the two lines corresponding to the equations. The intersection point indicates the solution to the system of equations.

Example 3: Infinite Solutions

Sometimes, a system of equations can have infinite solutions (the equations represent the same line). Let’s look at an example.

Solve the following system:

$\begin{aligned}[t] 2x + 4y &= 8 \\ x + 2y &= 4 \end{aligned}$

Step 1: Align the equations

The equations are already aligned:

$\begin{aligned} 2x + 4y &= 8 \\ x + 2y &= 4 \end{aligned}$

Step 2: Eliminate one variable

Notice that the second equation is just half of the first. Multiply the second equation by $2$ :

$\begin{aligned} 2x + 4y &= 8 \\ 2x + 4y &= 8 \end{aligned}$

Subtract the second equation from the first:

$(2x + 4y) - (2x + 4y) = 8 - 8$

Simplify:

$0 = 0$

This is a true statement, meaning the two equations represent the same line. Therefore, there are infinite solutions.

Visualization

The plot below shows the two equations as overlapping lines, indicating that every point on the line is a solution to the system.

Example 4: No Solution

Also, a system of equations can have no solution (the lines are parallel). Here is an example.

Solve the following system of equations:

$\begin{aligned}[t] 2x + 3y &= 6 \\ 6y + 4x &= 15 \end{aligned}$

Step 1: Align the equations

The equations are aligned below:

$\begin{aligned}[t] 2x + 3y &= 6 \ 4x + 6y &= 15 \end{aligned}$

Step 2: Eliminate one variable

To eliminate one variable, we need to make the coefficients of either $x$ or $y$ the same. Let’s eliminate $x$ . Multiply the first equation by $2$ so that the coefficients of $x$ : match:

$\begin{aligned}[t] 2(2x + 3y) &= 2(6) \\ 4x + 6y &= 15 \end{aligned}$

This gives us:

$\begin{aligned}[t] 4x + 6y &= 12 \\ 4x + 6y &= 15 \end{aligned}$

Now subtract the first equation from the second:

$(4x + 6y) - (4x + 6y) = 15 - 12$

Simplify:

$0 = 3$

This is a false statement, meaning the system of equations has no solution. The two lines are parallel and never intersect.

Visualization

The plot below illustrates the two parallel lines, confirming that there is no point of intersection. This graphical representation helps to visualize why the system has no solution.

Example 5: Solving a 3×3 System of Equations

Solve the following system of equations:

$\begin{cases}x + 2y + z &= 8 \\2x + y - z &= 3 \\3x - y + 2z &= 7\end{cases}$

Step 1: Align the equations
The equations are already aligned:

$\begin{cases}x + 2y + z &= 8 \\2x + y - z &= 3 \\3x - y + 2z &= 7\end{cases}$

Step 2: Eliminate one variable
We will eliminate ( z ) first. To do this, we can use the first two equations. We can add the first equation to the second equation after adjusting the coefficients of ( z ).

First, we can rewrite the second equation to align ( z ):

$2x + y - z = 3 \quad \text{(no change needed)}$

Now, we can add the first equation to the second equation:

$\begin{aligned}(x + 2y + z) + (2x + y - z) &= 8 + 3 \\3x + 3y &= 11\end{aligned}$

Simplify:

(Eq 4) $x + y = \frac{11}{3} \quad$

Next, we will eliminate ( z ) using the first and third equations. We can rewrite the first equation to align ( z ):

$2x + 4y + 2z = 16 \quad \text{(multiplied by two)}$

Now, we can add the first equation to the third equation after adjusting the coefficients of ( z ):

$(3x - y + 2z) - (2x + 4y + 2z) &= 7 - 16$

Simplify:

(Eq 5) $x - 5y = -9 \quad$

Step 3: Solve for one variable
Now we have two new equations (Equation 4 and Equation 5):

$x + y = \frac{11}{3}$ (Equation 4)
$x - 5y &= -9$ (Equation 5)

We can solve Equation 4 for $x$ :

$x = \frac{11}{3} - y$

Substituting $x$ into Equation 5:

$\left(\frac{11}{3} - y\right) - 5y = -9$

Simplify:

$\frac{11}{3} - y - 5y = -9$

Combine like terms:

$\frac{11}{3} - 6y = -9$

Multiply through by 3 to eliminate the fraction:

$11 - 18y = -27$

Subtract 11 from both sides:

$-18y = -38$

Divide by -18:

$y = \frac{19}{9}$

Step 4: Solve for $x$ and $z$
Substituting $y = \frac{19}{9}$ back into Equation 4:

$x + \frac{19}{9} = \frac{11}{3}$

Subtract $\frac{19}{9}$ from both sides:

$x = \frac{11}{3} - \frac{19}{9} = \frac{33-19}{9} = \frac{14}{9}$

Now, substitute $x = \frac{14}{9}$ and $y = \frac{19}{9}$ back into the first original equation to find $z$ :

$\frac{14}{9} + 2\left( \frac{19}{9}\right) + z = 8$

Simplify:

$\frac{14}{9} + \frac{38}{9} + z = 8$

Combine the fractions:

$\frac{52}{9} + z = 8$

Subtract $\frac{52}{9}$ from both sides:

$z = 8 - \frac{52}{9} = \frac{72}{9} - \frac{52}{9} = \frac{20}{9}$

Step 5: Verify the solution
Substitute $x = \frac{14}{9}$ , $y = \frac{19}{9}$ , and $z = \frac{20}{9}$ into all original equations to confirm they work:

For $x + 2y + z = 8$ :
$\begin{aligned}&\frac{14}{9} + 2\left(\frac{19}{9} \right) + \frac{20}{9} \\&= \frac{14}{9} + \frac{38}{9} + \frac{20}{9} \\&= \frac{14+38+20}{9} \\&= 8 \quad \text{\checkmark}\end{aligned}$
For $2x + y - z = 3$ :
$\begin{aligned}&2(\frac{14}{9}) + \frac{19}{9} - \frac{20}{9} \\&= \frac{28}{9} + \frac{19}{9} - \frac{20}{9} \\&= - \frac{28+19-20}{9} \\&= 3 \quad \text{\checkmark}\end{aligned}$
For $3x - y + 2z = 7$ :
$\begin{aligned}&3(\frac{14}{9}) - \frac{19}{9} + 2\left(\frac{20}{9} \right) \\&= \frac{42}{9} - \frac{19}{9} + \frac{40}{9} \\&= \frac{42-19+40}{3} \\&= 7 \quad \text{\checkmark}\end{aligned}$

Thus, the solution is: $\boxed{(x, y, z) = \left(\frac{14}{9}, \frac{19}{9}, \frac{20}{9}\right)}$

Visualization

The plot below illustrates the three planes represented by the equations. The point of intersection, marked on the graph, represents the unique solution to the system of equations.

Key Takeaways

The elimination method is a powerful way to solve systems of equations by eliminating one variable at a time.
Always align your equations properly and, if necessary, multiply through by constants to align coefficients.
Verify your solution by substituting the values into both original equations.
Keep an eye out for special cases like infinite solutions or no solution.

Solving Systems of Equations Using the Elimination Method

Steps for Solving Using the Elimination Method

Example 1: A Simple Elimination Scenario

Step 1: Align the equations

Step 2: Eliminate one variable

Step 3: Solve for $x$

Step 4: Solve for $y$

Step 5: Verify the solution

Visualization

Example 2: Multiplying to Align Coefficients

Step 1: Align the equations

Step 2: Eliminate one variable

Step 3: Solve for $x$

Step 4: Solve for $y$

Step 5: Verify the solution

Visualization

Example 3: Infinite Solutions

Step 1: Align the equations

Step 2: Eliminate one variable

Visualization

Example 4: No Solution

Step 1: Align the equations

Step 2: Eliminate one variable

Visualization

Example 5: Solving a 3×3 System of Equations

Visualization

Key Takeaways

Comments

Leave a Reply Cancel reply

More posts

Stability Analysis of a First-Order System with Proportional Control – Bode Plot

Nodal Analysis – Dependent Current Source

Nodal Analysis – Dependent Voltage Source

Nodal Analysis – Circuit with Dependent Voltage Source