How to Solve Systems of Equations: Substitution, Elimination, Examples and Practice Problems

To solve a system of two linear equations, find values of x and y that make both equations true at the same time. The most common methods are substitution, which replaces one variable with an equivalent expression, and elimination, which combines equations to cancel a variable. A third approach is to graph both equations and look for their point of intersection.

For example, the system y = 2x + 1 and 3x + y = 16 has the solution (3, 7), because 7 = 2(3) + 1 and 3(3) + 7 = 16. The steps below explain how to obtain—not simply guess—that answer. This lesson supports the U.S. Common Core Grade 8 standard 8.EE.C.8.b and later algebra courses.

What is a system of linear equations?

A system consists of two or more equations that share variables. In a typical two-variable linear system, each equation represents a straight line on the coordinate plane. If the lines intersect, their shared point is an ordered pair (x, y) that makes both statements true.

The three possibilities are:

What the lines do	Number of solutions	Algebraic result
Intersect at one point	Exactly one	A unique ordered pair
Are distinct and parallel	None	A contradiction such as 0 = 5
Are the same line	Infinitely many	An identity such as 0 = 0

The OpenStax Intermediate Algebra 2e lesson explains all three cases and the standard algebraic methods.

Method 1: solve by substitution

Substitution is often convenient if one equation already says x = … or y = …. Replace that variable in the second equation, solve the single-variable equation and return to find the other unknown.

Worked example: y = 2x + 1 and 3x + y = 16

1. Start with the two equations: y = 2x + 1 and 3x + y = 16.
2. Replace y in the second equation with its equivalent expression: 3x + (2x + 1) = 16.
3. Combine like terms: 5x + 1 = 16.
4. Subtract one: 5x = 15, giving x = 3.
5. Substitute x = 3 into y = 2x + 1: y = 2(3) + 1 = 7.
6. Write the solution as (3, 7).

Check: 7 = 2×3 + 1 and 3×3 + 7 = 16. Both equations are true. Do not stop after finding x; a complete two-variable answer normally needs both values.

Method 2: solve by elimination (addition)

Elimination works well when the two equations have matching or opposite coefficients for one variable. Add or subtract the equations to remove one variable. Multiplying both sides of an entire equation by the same nonzero constant is allowed, but multiplying only one term changes the equation and gives incorrect results.

Worked example: 2x + 3y = 13 and 4x − 3y = 17

1. Write the equations aligned: 2x + 3y = 13 and 4x − 3y = 17.
2. Add them. The terms +3y and −3y cancel.
3. That leaves 6x = 30, so x = 5.
4. Use the first equation: 2(5) + 3y = 13.
5. Subtract 10: 3y = 3, so y = 1.
6. The ordered-pair solution is (5, 1).

Check both: 2(5) + 3(1) = 13 and 4(5) − 3(1) = 17. Both work.

Elimination when you must multiply the equations first

Consider 2x + 3y = 16 and 3x + 2y = 14. Neither variable cancels if you simply add the original equations. To eliminate x, multiply the first whole equation by 3 and the second whole equation by −2:

1. First equation × 3: 6x + 9y = 48.
2. Second equation × −2: −6x − 4y = −28.
3. Add: 5y = 20, so y = 4.
4. Substitute into 2x + 3y = 16: 2x + 12 = 16; therefore x = 2.
5. Check the second equation: 3(2) + 2(4) = 14.

The answer is (2, 4). The essential idea is to create opposite coefficients, not to memorize a particular choice of variable. You could instead eliminate y and reach the same result.

Method 3: solve a system by graphing

Rewrite both equations as y = mx + b when possible, plot each line and locate the intersection. Graphing is useful for visual understanding and estimates, but a hand-drawn graph may be too imprecise for a fractional answer. Algebra provides exact solutions.

For y = 2x + 1 and 3x + y = 16, the second line becomes y = −3x + 16. Their intersection is at (3, 7). Since their slopes, 2 and −3, differ, the lines intersect at exactly one point. For a refresher, see our slope-intercept form guide.

How to recognize no solution

The system y = 2x + 1 and y = 2x − 4 has no solution: the lines have the same slope (2) but different y-intercepts (1 and −4), so they never meet. Setting the right sides equal gives 2x + 1 = 2x − 4, which simplifies to 1 = −4, a false statement.

In standard-form elimination, this can appear as 0 = 5 or another contradiction. It means the system is inconsistent, not that you made an arithmetic mistake—provided your transformations were valid.

How to recognize infinitely many solutions

Compare 2x + y = 6 and 4x + 2y = 12. Multiplying every term of the first equation by 2 gives the second equation. They describe exactly the same line, so every point satisfying 2x + y = 6 also satisfies the other equation. Subtracting one from twice the other yields 0 = 0, a true identity.

This system is consistent and dependent. It has infinitely many solutions, not one particular ordered pair. Do not label a contradiction as infinite solutions or a true identity as no solution.

Real-world example: ticket sales

A school event sells 20 tickets in total. Adult tickets cost $12 and student tickets cost $8. Total revenue is $208. How many of each type were sold?

1. Let a be adult tickets and s student tickets.
2. The number of tickets gives a + s = 20.
3. The revenue gives 12a + 8s = 208.
4. From the first equation, s = 20 − a.
5. Substitute: 12a + 8(20 − a) = 208, so 4a + 160 = 208.
6. That gives a = 12. Then s = 20 − 12 = 8.

Check: 12 + 8 = 20 tickets, and 12×$12 + 8×$8 = $144 + $64 = $208. The answer satisfies both the quantity and money constraints. This is the kind of modeling described by the Common Core Grade 8 mathematical applications standard.

Substitution vs. elimination: which method should you choose?

Choose	When it is often convenient	Watch out for
Substitution	One variable is already alone or has coefficient 1 or −1	Forgetting parentheses when replacing an expression
Elimination	Matching/opposite coefficients are present or easy to create	Multiplying only one side or one term
Graphing	You need to visualize the intersection or estimate a result	Reading an approximate graph coordinate as an exact answer

OpenStax’s College Algebra 2e chapter on two-variable systems provides additional method comparisons and advanced examples.

Six original practice problems

Try each independently. Write an ordered pair if there is exactly one solution, or describe why the system has no solution or infinitely many.

1. x + y = 7 and x − y = 1.
2. y = 3x − 2 and x + y = 14.
3. 2x + 3y = 12 and 4x + 6y = 24.
4. y = −x + 5 and x + y = 2.
5. 3x − 2y = 4 and x + y = 8.
6. 2x + y = 11 and 3x − y = 4.

Practice answer key with brief reasoning

1. (4, 3): add both equations to get 2x = 8; then y = 3.
2. (4, 10): 4x − 2 = 14; x = 4 and y = 10.
3. Infinitely many solutions: the second equation is exactly twice the first.
4. No solution: x + y = 2 is y = −x + 2, parallel to y = −x + 5.
5. (4, 4): substitute y = 8 − x into 3x − 2y = 4, giving 5x = 20.
6. (3, 5): add the equations to obtain 5x = 15 and then y = 5.

Five mistakes that change a correct answer

• Replacing a variable with an expression without parentheses, especially when a negative sign comes before it.
• Multiplying only one term rather than the entire equation when trying to eliminate a variable.
• Changing a plus sign to minus when adding or subtracting equations.
• Stopping after finding only x, even though the question asks for an ordered pair.
• Checking the answer in only one equation instead of both originals.

Official study references and related lessons

This explanation follows the methods in OpenStax Intermediate Algebra 2e, OpenStax College Algebra 2e and the Common Core Grade 8 expressions-and-equations standards. For additional foundations, review our slope-intercept form examples and reading graphs and slopes in a map context.

Original student tutorial prepared November 9, 2026. All worked examples and practice answers are algebraically checked. Featured image: real scientific calculator photographed by KneeHallHawk in December 2025, CC0 1.0 via Wikimedia Commons. The pictured calculator is illustrative; the equation solutions above are explained step by step and do not depend on a specific calculator model.