Solving Systems by Elimination

The elimination method solves a system of equations by combining the equations so that one variable disappears.

Consider:

x + y = 7

x − y = 1

If we add the equations:

x + y + x − y = 7 + 1

The +y and −y cancel:

2x = 8

So:

x = 4

Substitute x = 4 into either original equation:

4 + y = 7

Therefore:

y = 3

The solution is:

(4, 3)

The central idea is:

Align → Eliminate → Solve → Substitute Back → Verify

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Why Elimination Works

Suppose:

x + y = 7

and:

x − y = 1

Both equations are true at the solution.

Adding equal quantities to equal quantities preserves equality:

(x + y) + (x − y) = 7 + 1

The y terms are opposites:

+y − y = 0

so they disappear.

This leaves:

2x = 8

The system has been reduced from two variables to one variable.

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What Does It Mean to Eliminate a Variable?

To eliminate a variable means to make its coefficient become zero.

For example:

3y + (−3y) = 0

or:

5x − 5x = 0

Once one variable disappears, the remaining equation can be solved normally.

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When Is Elimination Appropriate?

Elimination is especially efficient when:

• the equations are written in similar forms
• the x-coefficients are opposites
• the y-coefficients are opposites
• the x-coefficients are equal
• the y-coefficients are equal
• one or both equations can easily be multiplied to create matching coefficients

For example:

3x + 2y = 12

5x − 2y = 8

is ideal because:

+2y

and:

−2y

are already opposites.

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Equations Ready for Elimination

Consider:

2x + 3y = 13

4x − 3y = 11

The y terms are:

+3y

and:

−3y

Adding the equations eliminates y immediately.

No rearranging or multiplication is necessary.

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The Elimination Method

A reliable process is:

Align → Choose → Adjust → Eliminate → Solve → Substitute Back → Verify

Align

Write like terms underneath each other.

Choose

Decide whether x or y will be easiest to eliminate.

Adjust

If necessary, multiply one or both equations to create equal or opposite coefficients.

Eliminate

Add or subtract the equations.

Solve

Solve the resulting one-variable equation.

Substitute Back

Use the value found to determine the other variable.

Verify

Check the ordered pair in both original equations.

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Elimination by Addition

Solve:

2x + 3y = 13

4x − 3y = 11

Add the equations:

2x + 3y = 13

4x − 3y = 11

——————————

6x = 24

Therefore:

x = 4

Substitute into the first equation:

2(4) + 3y = 13

8 + 3y = 13

3y = 5

Therefore:

y = 5/3

Solution:

(4, 5/3)

Solutions do not always have to be whole numbers.

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Why Addition Worked

The y-coefficients were opposites:

+3

and:

−3

Therefore:

3y + (−3y) = 0

This eliminated y.

A useful rule is:

Opposite coefficients → Add the equations

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Elimination by Subtraction

Consider:

3x + 2y = 16

3x − y = 7

The x-coefficients are both:

3

If we subtract the second equation from the first:

(3x + 2y) − (3x − y) = 16 − 7

The x terms cancel:

3x − 3x = 0

Be careful with the second y term:

2y − (−y) = 3y

Therefore:

3y = 9

y = 3

Substitute into:

3x − y = 7

3x − 3 = 7

3x = 10

x = 10/3

Solution:

(10/3, 3)

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Equal Coefficients Usually Suggest Subtraction

If you see:

5x + ...

5x + ...

then subtracting the equations can eliminate x.

Similarly:

... + 7y

... + 7y

can be eliminated by subtraction.

A useful rule is:

Same coefficients → Subtract the equations

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Opposite Coefficients Usually Suggest Addition

If you see:

+4x

and:

−4x

add.

If you see:

+6y

and:

−6y

add.

This often provides the fastest route to a solution.

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Align the Equations Carefully

Write corresponding terms in columns.

For example:

3x + 4y = 18

2x − 4y = 7

Think of the equations as:

This makes the opposite y terms easy to recognize.

Add:

5x = 25

Therefore:

x = 5

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Substitute Back

Continue the previous example:

3x + 4y = 18

We found:

x = 5

Substitute:

3(5) + 4y = 18

15 + 4y = 18

4y = 3

Therefore:

y = 3/4

Solution:

(5, 3/4)

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Verification

Check:

(5, 3/4)

First equation:

3(5) + 4(3/4) = 18

15 + 3 = 18 ✓

Second equation:

2(5) − 4(3/4) = 7

10 − 3 = 7 ✓

The solution satisfies both equations.

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When the Equations Are Not Ready

Consider:

2x + 3y = 12

x + 2y = 7

No coefficients are equal or opposite.

But we can multiply one equation to create coefficients that will cancel.

For example, multiply the second equation by −2:

−2(x + 2y) = −2(7)

giving:

−2x − 4y = −14

Now the system is:

2x + 3y = 12

−2x − 4y = −14

Add:

−y = −2

Therefore:

y = 2

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Finish the Solution

Substitute y = 2 into:

x + 2y = 7

x + 2(2) = 7

x + 4 = 7

Therefore:

x = 3

Solution:

(3, 2)

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Multiplying an Entire Equation

If you multiply an equation by a number, you must multiply every term on both sides.

For example:

x + 2y = 7

Multiply by −2:

−2x − 4y = −14

Incorrect:

−2x + 2y = 7

Every term must be multiplied.

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Why Multiplying an Equation Is Allowed

Consider:

x + 2y = 7

Multiplying both sides by 3 gives:

3x + 6y = 21

These equations are equivalent.

Any ordered pair satisfying:

x + 2y = 7

also satisfies:

3x + 6y = 21

Multiplying an entire equation by the same nonzero number does not change its solution set.

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Choosing Which Variable to Eliminate

Consider:

3x + 2y = 11

5x + 4y = 19

To eliminate y, multiply the first equation by −2:

−6x − 4y = −22

Now add it to the second:

5x + 4y = 19

——————————

−x = −3

Therefore:

x = 3

This required only one multiplication.

Eliminating x would require larger multipliers.

So y was the more efficient choice.

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Finish the Example

Substitute:

x = 3

into:

3x + 2y = 11

3(3) + 2y = 11

9 + 2y = 11

2y = 2

Therefore:

y = 1

Solution:

(3, 1)

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Look for the Easiest Multiplication

Consider:

4x + 3y = 18

2x − 5y = −2

The x-coefficients are 4 and 2.

Multiply the second equation by −2:

−4x + 10y = 4

Now add:

4x + 3y = 18

−4x + 10y = 4

——————————

13y = 22

This is easier than trying to eliminate y using coefficients 3 and −5.

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Multiplying Both Equations

Sometimes both equations must be multiplied.

Consider:

2x + 3y = 7

5x + 2y = 8

To eliminate y, the least common multiple of 3 and 2 is 6.

Multiply the first equation by 2:

4x + 6y = 14

Multiply the second equation by −3:

−15x − 6y = −24

Add:

−11x = −10

Therefore:

x = 10/11

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Finish the Example

Use:

2x + 3y = 7

Substitute:

2(10/11) + 3y = 7

20/11 + 3y = 77/11

3y = 57/11

Therefore:

y = 19/11

Solution:

(10/11, 19/11)

Again, fractional solutions are perfectly valid.

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Using the Least Common Multiple

When coefficients do not match, look for their least common multiple (LCM).

For coefficients 3 and 4:

LCM = 12

For coefficients 4 and 6:

LCM = 12

For coefficients 5 and 3:

LCM = 15

Create equal or opposite coefficients using the smallest convenient multiples.

This usually keeps the arithmetic manageable.

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Example Using the LCM

Solve:

3x + 2y = 7

4x − 3y = 2

To eliminate y:

LCM of 2 and 3:

6

Multiply the first equation by 3:

9x + 6y = 21

Multiply the second by 2:

8x − 6y = 4

Add:

17x = 25

Therefore:

x = 25/17

Substitution can then be used to find y.

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Elimination with Negative Coefficients

Solve:

−2x + 3y = 4

2x + 5y = 12

The x-coefficients are already opposites.

Add:

8y = 16

Therefore:

y = 2

Substitute into:

2x + 5y = 12

2x + 10 = 12

2x = 2

x = 1

Solution:

(1, 2)

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Be Careful When Subtracting Equations

Suppose:

4x + 5y = 19

4x − 2y = 5

Subtract the second equation from the first:

(4x + 5y) − (4x − 2y) = 19 − 5

Distribute the subtraction:

4x + 5y − 4x + 2y = 14

Therefore:

7y = 14

y = 2

The −2y becomes +2y because the entire second equation is being subtracted.

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An Alternative to Subtraction

Instead of subtracting an equation, you can multiply it by −1 and then add.

For:

4x + 5y = 19

4x − 2y = 5

multiply the second equation by −1:

−4x + 2y = −5

Now add:

4x + 5y = 19

−4x + 2y = −5

——————————

7y = 14

This gives the same result and can sometimes reduce sign errors.

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Equations May Need Rearranging First

Consider:

y = 2x + 1

3x + y = 11

For elimination, rewrite the first equation in standard form:

−2x + y = 1

Now:

−2x + y = 1

3x + y = 11

The y-coefficients are equal.

Subtract the first equation from the second:

5x = 10

Therefore:

x = 2

Then:

y = 2(2) + 1 = 5

Solution:

(2, 5)

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Standard Form Helps

Elimination is often easiest when equations are written as:

Ax + By = C

This is called standard form.

Examples:

3x + 2y = 12

5x − y = 9

Writing equations in standard form aligns:

• x terms
• y terms
• constants

and makes elimination easier to see.

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Elimination and Graphs

A system such as:

x + y = 7

x − y = 1

represents two lines.

Elimination gives:

x = 4

and then:

y = 3

Therefore the lines intersect at:

(4, 3)

Elimination finds the same intersection that would appear on a graph, but algebraically.

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No Solution

Consider:

2x + 3y = 8

4x + 6y = 20

Multiply the first equation by −2:

−4x − 6y = −16

Add:

4x + 6y = 20

——————————

0 = 4

This statement is impossible.

Therefore:

No solution

The equations represent distinct parallel lines.

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Why 0 = 4 Means No Solution

Elimination removed both variables but produced a false statement.

There are no values of x and y that can make:

0 = 4

true.

Therefore no ordered pair satisfies both equations.

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Infinitely Many Solutions

Consider:

2x + 3y = 8

4x + 6y = 16

Multiply the first equation by −2:

−4x − 6y = −16

Add:

4x + 6y = 16

——————————

0 = 0

This is always true.

Therefore:

Infinitely many solutions

The two equations represent the same line.

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Three Possible Outcomes

This is the same set of possible outcomes seen with substitution.

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Elimination Versus Substitution

Consider:

3x + 2y = 12

5x − 2y = 8

Elimination is especially convenient because:

+2y

and:

−2y

cancel immediately.

But for:

y = 3x + 2

5x + 2y = 17

substitution may be more convenient because y is already isolated.

The best method depends on the structure of the system.

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Both Methods Give the Same Solution

Whether we use:

• graphing
• substitution
• elimination

we are solving the same system.

Therefore, if all methods are performed correctly, they must produce the same solution.

The methods differ only in how we find that solution.

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Verifying an Elimination Solution

Suppose elimination produces:

(2, 3)

for:

2x + y = 7

3x − y = 3

Check the first equation:

2(2) + 3 = 7

7 = 7 ✓

Check the second:

3(2) − 3 = 3

3 = 3 ✓

Therefore:

(2, 3)

is correct.

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Why Verification Matters

Elimination involves several opportunities for mistakes:

• sign errors
• multiplication errors
• subtracting equations incorrectly
• forgetting to multiply every term
• arithmetic mistakes
• writing an ordered pair backward

Checking both original equations can catch these errors.

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Real-World Application: Ticket Sales

Adult tickets cost $12 and student tickets cost $8.

A total of 50 tickets are sold for $480.

Let:

a = number of adult tickets

s = number of student tickets

Number equation:

a + s = 50

Revenue equation:

12a + 8s = 480

Multiply the first equation by −8:

−8a − 8s = −400

Add:

12a + 8s = 480

——————————

4a = 80

Therefore:

a = 20

Then:

s = 30

So:

20 adult tickets and 30 student tickets were sold.

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Real-World Application: Two Purchases

Suppose:

3 notebooks and 2 pens cost $13.

2 notebooks and 2 pens cost $10.

Let:

n = price of one notebook

p = price of one pen

System:

3n + 2p = 13

2n + 2p = 10

Subtract the second equation from the first:

n = 3

So one notebook costs:

$3

Substitute:

2(3) + 2p = 10

6 + 2p = 10

2p = 4

p = 2

Therefore:

• notebook = $3
• pen = $2

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Real-World Application: Coins

Suppose there are 20 coins consisting of $1 and $2 coins. Their total value is $32.

Let:

x = number of $1 coins

y = number of $2 coins

Number of coins:

x + y = 20

Value:

x + 2y = 32

Subtract the first equation from the second:

y = 12

Then:

x = 8

Therefore there are:

8 one-dollar coins and 12 two-dollar coins

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Real-World Application: Comparing Plans

Plan A:

C = 20 + 3x

Plan B:

C = 8 + 5x

To use elimination, write both equations in standard form:

C − 3x = 20

C − 5x = 8

Subtract the second equation from the first:

2x = 12

Therefore:

x = 6

Substitute:

C = 20 + 3(6)

C = 38

The plans cost the same when:

x = 6

at a cost of:

$38

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Real-World Application: Science

Suppose experimental measurements lead to:

2F + 3m = 19

2F − m = 7

Subtract the second equation from the first:

4m = 12

Therefore:

m = 3

Substitute:

2F − 3 = 7

2F = 10

F = 5

So the system gives:

F = 5

and:

m = 3

The variables and units would then be interpreted according to the experiment.

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A More Challenging Example

Solve:

4x + 7y = 29

6x − 5y = 11

Neither variable eliminates immediately.

To eliminate x, the LCM of 4 and 6 is 12.

Multiply the first equation by 3:

12x + 21y = 87

Multiply the second equation by −2:

−12x + 10y = −22

Add:

31y = 65

Therefore:

y = 65/31

Now substitute into either original equation to find x.

This example shows why elimination may sometimes produce fractional answers.

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Another Multi-Step Example

Solve:

5x − 3y = 7

2x + 4y = 18

To eliminate y, the LCM of 3 and 4 is 12.

Multiply the first equation by 4:

20x − 12y = 28

Multiply the second equation by 3:

6x + 12y = 54

Add:

26x = 82

Therefore:

x = 41/13

Substitute back to determine y.

The procedure remains exactly the same even when the numbers are less convenient.

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Explaining Why Elimination Works

A strong explanation might be:

Elimination works because we can add, subtract, or multiply equations without changing the values that satisfy the system, provided the operations are applied correctly to the entire equations. We choose operations that create opposite terms, causing one variable to become zero. This leaves an equation containing only one variable, which can then be solved.

The method is based on the properties of equality.

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Writing Clear Elimination Work

Keep equations aligned:

3x + 4y = 18

2x − 4y = 7

——————————

5x = 25

This layout makes it clear that:

4y − 4y = 0

and helps prevent mistakes.

Clear organization is particularly important when equations must first be multiplied.

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Common Mistakes

Mistake: Adding equations when equal coefficients have the same sign.

For:

3x + ...

3x + ...

addition gives 6x rather than eliminating x.

Subtract instead.

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Mistake: Subtracting opposite coefficients.

For:

+4y

and:

−4y

subtraction gives:

4y − (−4y) = 8y

Add instead.

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Mistake: Multiplying only one term in an equation.

If multiplying an equation by 3, multiply every term on both sides by 3.

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Mistake: Forgetting to change signs when subtracting an equation.

Treat the entire second equation as being inside parentheses.

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Mistake: Choosing unnecessarily large multipliers.

Use the LCM of coefficients when possible.

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Mistake: Stopping after finding one variable.

Substitute back to find the second variable.

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Mistake: Writing the ordered pair incorrectly.

Write:

(x, y)

in that order.

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Mistake: Assuming the solution must be a whole number.

Fractions and decimals are valid solutions.

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Mistake: Treating 0 = 5 as a solution.

A false statement means:

No solution

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Mistake: Treating 0 = 0 as one solution.

An identity such as 0 = 0 means:

Infinitely many solutions

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Mistake: Failing to verify.

Always check the final ordered pair in both original equations.

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Key Terms

System of equations — set of equations involving the same variables that must be satisfied simultaneously.

Elimination — method of solving a system by combining equations so that one variable disappears.

Eliminate — cause a variable's coefficient to become zero.

Coefficient — numerical factor multiplying a variable.

Opposite coefficients — coefficients with equal magnitude but opposite signs, such as +4 and −4.

Equal coefficients — coefficients with the same value and sign.

Standard form — linear equation written as Ax + By = C.

Least common multiple (LCM) — smallest positive multiple shared by two or more numbers.

Equivalent equation — equation with the same solution set as another equation.

Ordered pair — pair (x, y) representing values of two variables.

Solution to a system — values satisfying every equation in the system.

Verification — checking a solution in the original equations.

Consistent system — system having at least one solution.

Inconsistent system — system having no solution.

Dependent system — system in which the equations represent the same line.

Independent system — system in which two different lines intersect at one point.

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Key Takeaways

• Elimination solves systems by making one variable disappear.
• The equations should be aligned so corresponding variables are in the same columns.
• Opposite coefficients can usually be eliminated by addition.
• Equal coefficients can usually be eliminated by subtraction.
• If coefficients do not match, multiply one or both equations first.
• Every term in an equation must be multiplied by the chosen multiplier.
• Using the LCM often gives the simplest multipliers.
• Choose the variable that can be eliminated most efficiently.
• Once one variable is eliminated, solve the resulting one-variable equation.
• Substitute the known value into an original equation to find the second variable.
• Write the solution as an ordered pair (x, y).
• Verify the ordered pair in both original equations.
• Elimination works because valid operations on equations preserve equality.
• Adding equations with opposite terms causes those terms to sum to zero.
• Subtracting equations with equal terms causes those terms to become zero.
• Equations may need to be rearranged into standard form before elimination.
• Fractional and negative solutions are perfectly valid.
• A false statement such as 0 = 5 indicates no solution.
• A true statement such as 0 = 0 indicates infinitely many solutions.
• Graphically, one solution represents an intersection of two lines.
• No solution corresponds to distinct parallel lines.
• Infinitely many solutions correspond to the same line.
• Elimination is often preferable when coefficients already match or are easy to make match.
• Substitution is often preferable when a variable is already isolated.
• Graphing, substitution, and elimination are different methods for finding the same system solution.

Check Your Understanding

1. What does it mean to eliminate a variable?
2. Explain why +5y and −5y are ready for elimination by addition.
3. Explain why two +3x terms can be eliminated by subtraction.
4. Solve: x + y = 9 and x − y = 3.
5. Solve: 2x + 3y = 13 and 4x − 3y = 11.
6. Solve: 3x + 2y = 14 and 3x − y = 5.
7. Solve: 2x + y = 8 and 3x − y = 7.
8. Solve: x + 2y = 9 and 2x + 3y = 16 by multiplying one equation first.
9. Solve: 3x + 2y = 11 and 5x + 4y = 19.
10. Solve: 2x + 3y = 7 and 5x + 2y = 8 by multiplying both equations.
11. Explain why every term must be multiplied when preparing an equation for elimination.
12. Explain how the LCM can help determine useful multipliers.
13. Verify whether (2, 3) solves 2x + y = 7 and 3x − y = 3.
14. What does 0 = 7 tell you about a system?
15. What does 0 = 0 tell you about a system?
16. Solve and classify: x + 2y = 5 and 2x + 4y = 10.
17. Solve and classify: x + 2y = 5 and 2x + 4y = 13.
18. Three notebooks and two pens cost $17. Two notebooks and two pens cost $14. Create a system and use elimination to determine the price of each item.
19. A school event sells 60 tickets. Adult tickets cost $15 and student tickets cost $10. Total revenue is $750. Create a system and solve it using elimination.
20. Explain the complete Align → Choose → Adjust → Eliminate → Solve → Substitute Back → Verify process and describe why each step is necessary.