Lesson 3: Solve Systems of Equations by Elimination

New Concept

The elimination method simplifies a system of equations by strategically adding them. By making the coefficients of one variable opposites, adding the equations eliminates that variable, leaving a single, solvable equation. This is our third method for solving linear systems.

What’s next

This card is just the start. Next, you'll tackle interactive examples and practice problems to master solving systems of equations using the elimination method.

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.

Examples

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

Explanation

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

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 $\begin{cases} 4x + 3y = 1 \\ x - 3y = 4 \end{cases}$. The equations are in standard form and the $y$ terms are opposites. Add them: $5x=5$, so $x=1$. Substitute into the second equation: $1-3y=4$, so $-3y=3$ and $y=-1$. The solution is $(1, -1)$.
• Solve $\begin{cases} x + \frac{1}{3}y = 2 \\ \frac{1}{2}x + \frac{1}{4}y = 2 \end{cases}$. First, clear fractions by multiplying the first equation by 3 and the second by 4 to get $\begin{cases} 3x+y=6 \\ 2x+y=8 \end{cases}$. Multiply the second equation by $-1$ and add: $(3x+y) + (-2x-y) = 6-8$, which gives $x=-2$. Then $3(-2)+y=6$, so $y=12$. The solution is $(-2, 12)$.
• Solve $\begin{cases} y = 5 - 2x \\ 3x - 2y = 4 \end{cases}$. First, rewrite the first equation in standard form: $2x+y=5$. Now multiply it by 2 to get $4x+2y=10$. Add this to $3x-2y=4$ to get $7x=14$, so $x=2$. Substitute into $y = 5 - 2x$ to get $y=5-2(2)=1$. The solution is $(2, 1)$.

Explanation

This is a step-by-step recipe for success. First, get your equations into $Ax+By=C$ form. Next, multiply to create opposite terms. Then, add, solve for one variable, substitute back to find the other, and always check your answer.

Property

When solving a system of equations, if the elimination process results in a true statement with no variables (e.g., $0=0$), the system has infinitely many solutions because the equations represent the same line (coincident lines). If the process results in a false statement (e.g., $0=-5$), the system has no solution because the lines are parallel.

Examples

• The system $\begin{cases} 2x - y = 4 \\ -4x + 2y = -8 \end{cases}$ becomes $\begin{cases} 4x - 2y = 8 \\ -4x + 2y = -8 \end{cases}$ after multiplying the first equation by 2. Adding them gives $0=0$. This is a true statement, so there are infinitely many solutions.
• For the system $\begin{cases} 3x + 2y = 5 \\ 3x + 2y = 1 \end{cases}$, multiplying the second equation by $-1$ gives $-3x-2y=-1$. Adding this to the first equation results in $0=4$. This is a false statement, so there is no solution.
• Given $\begin{cases} y = 4x - 1 \\ -8x + 2y = 5 \end{cases}$, rewrite the first as $4x - y = 1$. Multiply by 2 to get $8x - 2y = 2$. Adding to $-8x + 2y = 5$ gives $0=7$. This is false, so there is no solution.

Explanation

If both variables vanish when you add the equations, look at what is left. A true statement like $0=0$ means infinite solutions (the lines are identical). A false statement like $0=9$ means no solution (the lines are parallel).

Property

Examples

• For $\begin{cases} y = 4x - 1 \\ 3x + 2y = 9 \end{cases}$, substitution is most convenient because the first equation is already solved for $y$.
• For $\begin{cases} 5x - 3y = 11 \\ 2x + 3y = 3 \end{cases}$, elimination is most convenient because the equations are in standard form and the coefficients for $y$ are already opposites.
• For $\begin{cases} x + y = 10 \\ 2x - y = 5 \end{cases}$, elimination is very convenient because the $y$ coefficients are opposites. Substitution would also work easily by solving the first equation for $x$ or $y$.

Explanation

Pick your method strategically. Substitution is perfect if an equation is already in $y=...$ or $x=...$ form. Elimination shines when both equations are lined up in standard form ($Ax+By=C$). Graphing is best for visualizing the answer.