Lesson 3: Elimination

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.