Solving Systems by Elimination

Learning goals

Add or subtract equations so one variable cancels
Scale one or both equations to match the coefficients
Justify the method from the properties of equality
Back-substitute for the second variable, then check both
Read $0 = k$ as no solution and $0 = 0$ as infinitely many
Prefer elimination in standard form, substitution when one is isolated

Adding equations to cancel a variable

A system of two linear equations asks for the pair $(x, y)$ that satisfies both equations at the same time. Substitution found that pair by isolating a variable and replacing it. Elimination finds it by combining the two equations so that one variable cancels outright.

Start with a system whose coefficients are already arranged for it:

x + y = 5, \qquad x - y = 1.

Look at the $y$ terms. The first equation has $+y$ , the second has $-y$ , and those are opposites. So if you add the equations column by column, the $y$ terms sum to zero and drop out. Add the left sides, add the right sides:

(x + y) + (x - y) = 5 + 1.

The $+y$ and $-y$ cancel, and what remains has only $x$ in it:

2x = 6, \qquad x = 3.

One variable is solved already. Recover the other by putting $x = 3$ back into either original equation, say $x + y = 5$ :

3 + y = 5, \qquad y = 2.

The candidate solution is $(3, 2)$ . Check it in both equations, exactly as you did with substitution: $3 + 2 = 5$ and $3 - 2 = 1$ , both true. So $(3, 2)$ solves the system.

The move that did the work was adding two equations to cancel a variable. It succeeds whenever a variable’s coefficients are opposites, because opposite numbers sum to zero. If instead the coefficients are equal, subtracting one equation from the other cancels the variable just as well. Adding and subtracting are the two faces of elimination. The only remaining question is what to do when the coefficients are neither opposite nor equal.

The elimination method step by step

Every system yields to the same routine.

Line up the equations in standard form $ax + by = c$ , so like terms sit in columns.
Match a variable’s coefficients. Pick the variable to eliminate, then multiply one or both equations by nonzero constants so its coefficients become opposites (to add) or equal (to subtract).
Add or subtract the equations to remove that variable, leaving one equation in one unknown.
Solve the one-variable equation.
Back-substitute the value into an original equation to find the other variable.
Check the pair in both original equations.

When a variable already has opposite coefficients, step 2 costs nothing and you go straight to adding.

Worked example 1 Solve $2x + y = 7$ and $3x - y = 8$

Scan the columns for a variable ready to cancel. The first equation has $+y$ and the second has $-y$ , opposites, so adding the equations eliminates $y$ with no setup:

(2x + y) + (3x - y) = 7 + 8.

The $y$ terms cancel and the $x$ terms combine:

5x = 15, \qquad x = 3.

Back-substitute $x = 3$ into an original equation. Using $2x + y = 7$ :

2(3) + y = 7, \qquad 6 + y = 7, \qquad y = 1.

So the candidate is $(3, 1)$ . Check it in both equations, not just the one you used:

2(3) + 1 = 7 \ \checkmark, \qquad 3(3) - 1 = 8 \ \checkmark.

Both hold, so the solution is $(3, 1)$ .

Check your understanding

To solve $3x + 4y = 10$ and $5x - 4y = 6$ by elimination, you add the two equations. Which variable is eliminated, and what equation results?

x

is eliminated, giving

8x = 16

y

is eliminated, giving

8x = 16

y

is eliminated, giving

8x = 4

x

is eliminated, giving

9y = 16

Why adding equations keeps every solution

Elimination keeps one of the two equations and replaces the other with a combined equation, then solves that smaller system. It is worth seeing why this is trustworthy: the new system is made of the equation you kept and the combined equation. That new system has exactly the same solution pairs as the original, so no common solution is lost and none is invented.

Why combining the equations preserves the solution set#

A system asks for every pair $(x, y)$ that makes both equations true at once, and that collection of pairs is its solution set. Two properties of equality do all the work. The addition property says that if $L_1 = R_1$ and $L_2 = R_2$ are both true, then adding them gives another true statement, $L_1 + L_2 = R_1 + R_2$ . The multiplication property says that if $L = R$ is true, then $kL = kR$ is true for any number $k$ . Scaling an equation and adding two equations are exactly the moves elimination uses.

First, no common solution is lost. Elimination keeps one original equation, say the first, and replaces the second with a combined equation built as $m$ times the first plus $n$ times the second. Suppose a pair $(x_0, y_0)$ solves the original system, so it makes both equations true. By the multiplication property, scaling the first equation by $m$ and the second by $n$ keeps each true for this pair. By the addition property, their sum is then true for this pair as well. That sum is exactly the combined equation, so the pair satisfies it. The pair also still satisfies the kept first equation, so it solves the new system made of the first equation and the combined equation. Every solution of the original system survives into the new one.

Second, no new common solution is invented. The replacement is reversible. The combined equation is $m$ times the first plus $n$ times the second. Subtracting $m$ times the first equation from that combined equation, and then dividing by $n$ , gives back the second equation exactly. This undoing is valid because the multiplier $n$ on the second equation is not zero. That is guaranteed, since you never eliminate by multiplying an equation by zero, which would only erase it. So from the new system, the first equation together with the combined equation, you can rebuild the original second equation. Any pair that solves the new system therefore also satisfies the second equation, and so it solves the original system too.

The new system is the first equation paired with the combined equation. The two directions together show that this new system has exactly the same solution set as the original. Elimination therefore trades the system for an equivalent one in which a variable is already gone. This also explains the unusual outcomes. When the combined equation still carries a variable, it pins that variable to a single value and the system has one solution. When both variables cancel and leave a false statement such as $0 = 8$ , no pair can satisfy the equations, so the lines are parallel and there is no solution. When both cancel and leave a statement that is always true such as $0 = 0$ , the second equation was a multiple of the first all along. The two equations therefore describe the same line, and every point on that line solves the system.

Matching the coefficients: multiply, then eliminate

Most systems do not hand you opposite coefficients. You create them. Multiplying an entire equation by a nonzero constant produces an equation with the same solutions, by the multiplication property of equality. The new equation carries rescaled coefficients, and you can aim those at a match.

Two situations arise. Sometimes one variable’s coefficient in one equation is a multiple of its coefficient in the other, so multiplying just one equation lines them up. When neither coefficient is a multiple of the other, multiply both equations, scaling the chosen variable’s coefficients up to their least common multiple. Make one of those coefficients positive and the other negative, so they cancel on adding.

Worked example 2 Multiply one equation: solve $x + 2y = 11$ and $3x - y = 5$

Aim to eliminate $y$ . The first equation has $+2y$ and the second has $-y$ . Doubling the second turns its $-y$ into $-2y$ , the opposite of the $+2y$ above it. Multiply every term of $3x - y = 5$ by $2$ :

6x - 2y = 10.

Now add this to the first equation, $x + 2y = 11$ :

(x + 2y) + (6x - 2y) = 11 + 10.

The $y$ terms cancel:

7x = 21, \qquad x = 3.

Back-substitute $x = 3$ into $x + 2y = 11$ :

3 + 2y = 11, \qquad 2y = 8, \qquad y = 4.

The solution is $(3, 4)$ . Check the second equation too: $3(3) - 4 = 9 - 4 = 5$ , correct. Notice that only one equation needed multiplying, because the coefficient $2$ was already a multiple of the $1$ on the $-y$ .

Check your understanding

You want to solve $3x + 2y = 7$ and $x + y = 3$ by elimination. If you multiply the second equation by $-2$ and add it to the first, which variable is eliminated?

A both variables B

x

y

D neither variable

When neither coefficient divides the other, you scale both equations. Aim the variable you are eliminating at the least common multiple of its two coefficients.

Worked example 3 Multiply both equations: solve $2x + 3y = 7$ and $3x + 2y = 8$

Eliminate $x$ . Its coefficients are $2$ and $3$ , and neither divides the other, so scale both up to their least common multiple, $6$ . Multiply the first equation by $3$ and the second by $2$ :

6x + 9y = 21, \qquad 6x + 4y = 16.

Both now carry $6x$ . Since the $x$ coefficients are equal rather than opposite, subtract the second equation from the first so the $6x$ terms cancel. Subtracting flips every sign in the second equation:

(6x + 9y) - (6x + 4y) = 21 - 16, \qquad 5y = 5, \qquad y = 1.

Back-substitute $y = 1$ into $2x + 3y = 7$ :

2x + 3(1) = 7, \qquad 2x = 4, \qquad x = 2.

The solution is $(2, 1)$ , and $3(2) + 2(1) = 8$ confirms the second equation. You could instead have multiplied the second equation by $-2$ to get $-6x - 4y = -16$ and then added; the result is identical. Making coefficients equal and subtracting, or opposite and adding, are two routes to the same cancellation.

When a variable cancels to a number: no solution and infinitely many

Usually elimination leaves a one-variable equation that pins a value down. Once in a while both variables cancel together and you are left with a bare numerical statement. As with substitution, that is not a dead end; it is the system telling you how the two lines lie.

Worked example 4 A system with no solution

Solve

2x + y = 5, \qquad 4x + 2y = 3.

Eliminate $x$ . Multiply the first equation by $-2$ so its $2x$ becomes $-4x$ , opposite the $4x$ below:

-4x - 2y = -10.

Add this to $4x + 2y = 3$ :

(-4x - 2y) + (4x + 2y) = -10 + 3.

Both variables cancel at once, and what is left is

0 = -7.

This is false: no values of $x$ and $y$ can make $0$ equal $-7$ . So the system has no solution. The two lines have the same steepness but different heights, so they are parallel and never meet.

Worked example 5 A system with infinitely many solutions

Solve

x - 3y = 2, \qquad 2x - 6y = 4.

Eliminate $x$ by doubling the first equation so its $x$ becomes $2x$ :

2x - 6y = 4.

This is now identical to the second equation, so subtracting them gives

0 = 0.

Every term has cancelled and the statement left over is always true. That means every pair solving the first equation already solves the second, so the system has infinitely many solutions. The second equation is just the first multiplied by $2$ , so the two describe the same line. Each point on that line, written $(x, y)$ with $x - 3y = 2$ , is a solution.

Check your understanding

Eliminating a variable from a system, you reach $0 = 6$ . What does this tell you?

A The system has no solution. B The system has infinitely many solutions. C The system has exactly one solution,

x = 0

. D You must have made an arithmetic error, since every system has a solution.

Substitution or elimination?

Both methods return the same pair, so the choice is about convenience. Reach for elimination when both equations are already in standard form $ax + by = c$ . Elimination fits especially well when a variable shares a coefficient or an easy multiple across the two equations, so a single add or subtract clears it. Reach for substitution when one equation already has a variable isolated, like $y = 3x - 2$ , because you can drop that expression straight into the other equation. Many systems accept either method comfortably; with a little practice you will glance at a system and pick the route that needs fewer steps.

Seeing the cancellation

The whole method comes down to lining up like terms and adding so that a column cancels. Here is the first worked example laid out that way. The $y$ column holds $+y$ directly above $-y$ , opposite terms that sum to zero, so adding the two equations erases $y$ and leaves a single equation in $x$ .

Elimination as column addition. The first worked example, 2x + y = 7 and 3x - y = 8, is stacked so like terms align. The y column holds +y above -y, opposites that sum to zero, so adding the equations cancels y and leaves 5x = 15.

Every elimination problem is this picture. When the canceling column is not already in place, you multiply one or both equations first to force it there, then add.

Common mistakes

Adding when you should subtract, or the reverse. Adding cancels a variable only when its coefficients are opposites; subtracting cancels it only when they are equal. Check the signs before you combine, or you will keep the variable you meant to remove.
Forgetting to multiply every term. Scaling an equation multiplies the whole thing, both variable terms and the constant. Multiplying $x + 2y = 11$ by $2$ gives $2x + 4y = 22$ , not $2x + 4y = 11$ . A forgotten constant quietly breaks the answer.
Multiplying only one side. Multiplying an equation means multiplying both sides. Scale only the left side and you have changed the equation into a different one.
Sign slips when subtracting. Subtracting an equation flips every sign inside it. So $(6x + 9y) - (6x + 4y)$ is $5y$ , and $(3x - y) - (x - y)$ is $2x$ , not $2x - 2y$ . Distribute the minus to every term.
Misreading a vanished variable. If both variables cancel and leave $0 = 8$ , the answer is no solution; if they leave $0 = 0$ , it is infinitely many solutions. Neither one means $x = 0$ .
Stopping at one variable. Elimination gives you one coordinate. Back-substitute for the other, state the full pair $(x, y)$ , and check it in both equations before calling the system solved.

Key takeaways

Elimination solves a system by combining the two equations so that one variable cancels, turning the system into a single one-variable equation.
Add when a variable’s coefficients are opposites; subtract when they are equal. If they are neither, multiply one or both equations by nonzero constants first to force a match, scaling to the least common multiple of the coefficients.
After eliminating, solve for the surviving variable, back-substitute into an original equation, and check the pair in both.
Elimination is valid because of the addition and multiplication properties of equality: any pair solving both equations also solves any scaled sum of them. The step is also reversible, so no solution is lost or invented.
A leftover $0 = k$ with $k$ not zero means no solution (parallel lines); a leftover $0 = 0$ means infinitely many (the same line).
Prefer elimination when both equations are in standard form $ax + by = c$ , and substitution when a variable is already isolated.

Solving Systems by Elimination

Learning goals

Adding equations to cancel a variable

The elimination method step by step

Why adding equations keeps every solution

Matching the coefficients: multiply, then eliminate

When a variable cancels to a number: no solution and infinitely many

Substitution or elimination?

Seeing the cancellation

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)