Section 5.3: Solving Systems of Linear Equations by 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

When solving systems by elimination, you must write both equations in standard form first and clear any fractions. Sometimes, you must multiply one or both equations by a constant to make the coefficients of one variable opposites before adding.

Examples

• Multiplying One Equation: For the system $3x + 4y = 11$ and $5x - 2y = 13$, multiply the second equation by 2 to get opposite $y$-coefficients. This results in $3x + 4y = 11$ and $10x - 4y = 26$.
• Multiplying Both Equations: For the system $2x + 5y = 8$ and $3x + 7y = 12$, multiply the first by 3 and the second by -2 to get opposite $x$-coefficients. This results in $6x + 15y = 24$ and $-6x - 14y = -24$.
• Full Process: Solve $2x + 5y = -34$ and $-3x + 2y = -44$. Multiply to make the x-terms opposites: $3(2x + 5y) = 3(-34)$ and $2(-3x + 2y) = 2(-44)$. This simplifies to $6x + 15y = -102$ and $-6x + 4y = -88$. Adding them gives $19y = -190$, so $y = -10$. Substitute back to get $x = 8$, making the solution $(8, -10)$.

Explanation

Strategic multiplication creates the opposite coefficients needed for elimination to work effectively. Look for the variable that will be easiest to eliminate by examining both coefficients and choosing appropriate multipliers. Once you add the equations and solve for the first variable, always substitute that answer back into an original equation to find the second variable, and check your final ordered pair.

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.