Solve Systems of Equations by Elimination

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the coefficients are small and the solution has integer values. Substitution works well when one equation is already solved for a variable, or can easily be solved for one.

The third method for solving systems of linear equations is called elimination. It is based on the Addition Property of Equality, which says that adding the same quantity to both sides of an equation preserves equality. We extend that idea: if $a=b$ and $c=d$, then $a+c=b+d$ — adding equal quantities to both sides of an equation gives equal results.

To solve a system by elimination, we start with both equations in standard form, then decide which variable will be easiest to eliminate. We want the coefficients of that variable to be opposites, so that adding the equations eliminates it. Notice how that works when we add these two equations together:

The $y$’s add to zero, leaving one equation with one variable. Now try this system:

This time no variable is ready to eliminate right away. But multiplying the first equation by $-2$ makes the coefficients of $x$ opposites: $-2(x+4y) = -2(2)$ gives $-2x - 8y = -4$, while the second equation stays the same. Adding the new first equation to the second eliminates $x$:

Sometimes we can’t make one variable’s coefficients opposite by multiplying only one equation — unless we’re willing to use fractions. Instead, we multiply both equations by (different) constants. For $4x - 3y = 10$ and $3x + 5y = -7$, multiplying the first equation by $3$ and the second by $-4$ turns the $x$-coefficients into $12$ and $-12$:

Adding these eliminates the $x$’s, leaving $-29y = 58$.

Once we have an equation with just one variable, we solve it, then substitute that value into one of the original equations to solve for the remaining variable — and, as always, check the answer in both original equations.

Example. Solve the system by elimination.

Now let’s look at a system where multiplying just one equation makes a pair of coefficients opposite.

Example. Solve the system by elimination.

Neither variable’s coefficients are opposites yet, but multiplying the first equation by $-3$ makes the $y$-coefficients opposite ($6$ and $-6$): $-3(3x-2y) = -3(-2)$ gives $-9x + 6y = 6$. Adding that to the second equation:

So $x = -4$. Substituting $x=-4$ into $3x-2y=-2$ gives $3(-4) - 2y = -2$, so $-12 - 2y = -2$, and $-2y = 10$, so $y = -5$. The solution is $(-4,-5)$. Check: $3(-4)-2(-5) = -12+10=-2$ ✓ and $5(-4)-6(-5)=-20+30=10$ ✓.

Sometimes we must multiply both equations by constants to make a pair of coefficients opposite.

Example. Solve the system by elimination.

Neither variable can be eliminated by multiplying just one equation. To make the $y$-coefficients opposite, multiply the first equation by $2$ and the second by $3$:

Adding these equations eliminates $y$: $29x = 0$, so $x=0$. Substituting $x=0$ into $7x+2y=-6$ gives $2y=-6$, so $y=-3$. The solution is $(0,-3)$.

When a system has fractions, clear them first by multiplying each equation by its LCD — then eliminate as usual. For $x + \tfrac{1}{2}y = 6$ and $\tfrac{3}{2}x + \tfrac{2}{3}y = \tfrac{17}{2}$, multiplying the first equation by $2$ and the second by $6$ clears the fractions: $2x+y=12$ and $9x+4y=51$. Multiplying the first of those by $-4$ gives $-8x-4y=-48$; adding it to $9x+4y=51$ leaves $x=3$, and substituting back gives $y=6$. The solution is $(3,6)$.

Not every system has exactly one solution. Recall from graphing that two equations describing the same line have infinitely many solutions (a consistent, dependent system), and two equations describing parallel lines have no solution at all (an inconsistent system). Elimination reveals both cases too.

Example. Solve the system by elimination.

Writing the second equation in standard form gives $\tfrac{3}{4}x + y = 3$; multiplying by $4$ clears the fraction: $3x+4y=12$ — the very same equation as the first! Multiplying this equation by $-1$ and adding it to the first equation gives $0=0$, a true statement. The system is consistent but dependent: the two equations describe the same (coincident) line, so the system has infinitely many solutions.

Example. Solve the system by elimination.

Multiplying the second equation by $3$ gives $6x - 15y = -15$. Adding this to the first equation, the variables cancel and we’re left with $0 = -5$ — a false statement. The system is inconsistent: the two equations describe parallel lines, so the system has no solution.

Solve Applications of Systems of Equations by Elimination

Many application problems translate directly into a system of two equations in standard form, which makes elimination a natural choice. As before, we follow the problem-solving strategy: read, identify, name, translate, solve, check, and answer.

Example. The sum of two numbers is 39. Their difference is 9. Find the numbers.

Let $n =$ the first number and $m =$ the second number. “The sum of two numbers is 39” translates to $n+m=39$; “their difference is 9” translates to $n - m = 9$. The system is:

The coefficients of $m$ are already opposite, so add the equations: $2n = 48$, giving $n = 24$. Substituting into $n+m=39$ gives $24+m=39$, so $m=15$. Check: $24+15=39$ and $24-15=9$. The numbers are 24 and 15.

Example. Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, for a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are in one order of medium fries? How many calories are in one small soda?

Let $f =$ the number of calories in one order of medium fries, and $s =$ the number of calories in one small soda. “One order of fries and two sodas had 620 calories” translates to $f+2s=620$; “two orders of fries and one soda had 820 calories” translates to $2f+s=820$. The system is:

Multiplying the first equation by $-2$ gives $-2f-4s=-1240$. Adding this to the second equation eliminates $f$: $-3s = -420$, so $s = 140$. Substituting $s=140$ into $f+2s=620$ gives $f + 280 = 620$, so $f = 340$. The fries have 340 calories and the soda has 140 calories.

Choose the Most Convenient Method to Solve a System of Linear Equations

You now know three ways to solve a system of linear equations: graphing, substitution, and elimination. In a later math class you’ll usually have to decide for yourself which method to use, so it helps to recognize which method is easiest for a given system.

Example. For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination.

(a) $3x+8y=40$ and $7x-4y=-32$ — both equations are already in standard form, so elimination will be most convenient.

(b) $5x+6y=12$ and $y=\tfrac{2}{3}x-1$ — the second equation is already solved for $y$, so substitution will be most convenient.

Key terms

elimination — a method for solving a system of linear equations in which the equations are added together, after being multiplied by constants if necessary, so that one variable cancels out. consistent, dependent system — a system whose equations describe the same line, so it has infinitely many solutions; eliminating a variable leaves a true statement like $0=0$. inconsistent system — a system whose equations describe parallel lines, so it has no solution; eliminating a variable leaves a false statement like $0=-5$.

This section is adapted from Elementary Algebra 2e, Section 5.3: Solve Systems of Equations by Elimination by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recast the seven-step worked example and the Choose-the-Method comparison as markdown tables; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback — including turning the consistent/inconsistent Try Its into multiple-choice questions about the number of solutions, since a word answer like “no solution” can’t be graded by the math checker.