How to Solve Systems of Equations (Substitution, Elimination & Graphing)

If you have ever stared at two equations sitting side by side and wondered how a single answer could possibly satisfy both at once, you are in the right place. A system of equations is simply a set of two or more equations that share the same variables, and your job is to find the values that make every equation true at the same time. It sounds tricky, but once you learn the three standard techniques, it becomes one of the most predictable topics in all of algebra.

In this guide you will learn how to solve systems of equations three different ways: by substitution, by elimination, and by graphing. We will work through several fully solved examples, show every single step, point out the mistakes students make most often, and help you decide which method to reach for on any given problem. If you ever get stuck or want to check your work, you can verify any answer in seconds with our free algebra solver.

By the end, you will be able to look at almost any pair of linear equations and know exactly what to do. Let’s start with what a “solution” really means.

What Is a System of Equations?

A system of equations is a group of equations that you want to solve together. The most common type you will meet is a system of two linear equations in two variables, usually called \(x\) and \(y\). For example:

$$\begin{cases} y = 2x + 1 \\ 3x + y = 11 \end{cases}$$

The solution is the ordered pair \((x, y)\) that satisfies both equations at the same time. Geometrically, each linear equation is a straight line, so the solution is the point where the two lines cross. That single idea explains everything that follows.

Not every system has exactly one answer. Because two lines can relate to each other in only three ways, a linear system has exactly three possible outcomes:

Situation	What the lines do	Number of solutions
Consistent & independent	Cross at one point	Exactly one solution
Inconsistent	Parallel, never touch	No solution
Dependent	Same line on top of itself	Infinitely many solutions

We will return to the second and third cases later. For now, know that most homework problems are designed to have one clean solution.

How to Solve Systems of Equations: The Three Main Methods

There are three reliable methods, and all three give the same correct answer. They differ only in how they get there. Here is a quick comparison so you can see the big picture before we dive into each one.

Method	Core idea	Works best when…
Substitution	Solve one equation for a variable, then plug it into the other	A variable is already alone (like \(y = \dots\)) or easy to isolate
Elimination	Add or subtract the equations to cancel a variable	The equations are lined up as \(ax + by = c\)
Graphing	Draw both lines and read off the crossing point	You want to see the solution or estimate quickly

Let’s master each method with worked examples.

Method 1: Substitution

The substitution method works by isolating one variable in one equation and then “substituting” that expression into the other equation. This leaves you with a single equation in a single variable, which you already know how to solve.

Steps for the substitution method

1. Isolate one variable in one of the equations (pick the easiest one).
2. Substitute that expression into the other equation.
3. Solve the resulting one-variable equation.
4. Back-substitute to find the second variable.
5. Check your pair in both original equations.

Worked Example 1

Solve the system:

$$\begin{cases} y = 2x + 1 \\ 3x + y = 11 \end{cases}$$

Step 1 — Isolate a variable. The first equation is already solved for \(y\):

$$y = 2x + 1$$

Step 2 — Substitute this expression for \(y\) into the second equation:

$$3x + (2x + 1) = 11$$

Step 3 — Solve for \(x\). Combine like terms:

$$5x + 1 = 11$$ $$5x = 10$$ $$x = 2$$

Step 4 — Back-substitute \(x = 2\) into \(y = 2x + 1\):

$$y = 2(2) + 1 = 4 + 1 = 5$$

Step 5 — Check the pair \((2, 5)\) in the second equation: \(3(2) + 5 = 6 + 5 = 11\). It works.

If you can comfortably isolate and substitute a single variable, you have already done the hardest part of solving any linear equation. For a refresher on those one-variable skills, see our guide on How to Solve Linear Equations Step by Step.

Method 2: Elimination

The elimination method (also called the addition method) adds or subtracts the two equations so that one variable cancels out completely. It is often the fastest method when both equations are written in standard form \(ax + by = c\).

Steps for the elimination method

1. Write both equations in the form \(ax + by = c\) and line them up.
2. If needed, multiply one or both equations so a pair of coefficients are opposites (or equal).
3. Add the equations (or subtract) to eliminate one variable.
4. Solve for the remaining variable.
5. Back-substitute and check.

Worked Example 2: when a variable cancels immediately

Solve the system:

$$\begin{cases} 2x + 3y = 12 \\ 4x - 3y = 6 \end{cases}$$

Notice the \(+3y\) and \(-3y\). They are already opposites, so if we add the equations, the \(y\) terms disappear:

$$(2x + 3y) + (4x - 3y) = 12 + 6$$ $$6x = 18$$ $$x = 3$$

Now back-substitute \(x = 3\) into the first equation:

$$2(3) + 3y = 12$$ $$6 + 3y = 12$$ $$3y = 6$$ $$y = 2$$

Check in the second equation: \(4(3) - 3(2) = 12 - 6 = 6\). Correct.

Worked Example 3: when you must multiply first

Most real systems are not so polite — the coefficients rarely line up on their own. When that happens, multiply one or both equations by a constant first.

Solve the system:

$$\begin{cases} 3x + 2y = 16 \\ 2x + 5y = 18 \end{cases}$$

Step 1 — Make the \(x\) coefficients match. The least common multiple of \(3\) and \(2\) is \(6\). Multiply the first equation by \(2\) and the second by \(3\):

$$2 \times (3x + 2y = 16) \;\Rightarrow\; 6x + 4y = 32$$ $$3 \times (2x + 5y = 18) \;\Rightarrow\; 6x + 15y = 54$$

Step 2 — Subtract the first new equation from the second to eliminate \(x\):

$$(6x + 15y) - (6x + 4y) = 54 - 32$$ $$11y = 22$$ $$y = 2$$

Step 3 — Back-substitute \(y = 2\) into the original \(3x + 2y = 16\):

$$3x + 2(2) = 16$$ $$3x + 4 = 16$$ $$3x = 12$$ $$x = 4$$

Check in the other original equation: \(2(4) + 5(2) = 8 + 10 = 18\). Perfect.

Method 3: Graphing

The graphing method turns the algebra into a picture. Since each equation is a line, you graph both lines on the same coordinate plane and find the point where they intersect. That intersection point is your solution.

Steps for the graphing method

1. Rewrite each equation in slope-intercept form, \(y = mx + b\).
2. Plot each line using its \(y\)-intercept \(b\) and slope \(m\).
3. Find the point of intersection.
4. Read off the coordinates and check them algebraically.

Worked Example 4

Solve the system by graphing:

$$\begin{cases} y = x + 1 \\ y = -x + 5 \end{cases}$$

Both equations are already in slope-intercept form, which makes graphing easy.

• The first line, \(y = x + 1\), has slope \(m = 1\) and \(y\)-intercept \(b = 1\). It passes through \((0, 1)\) and rises one unit for every step right.
• The second line, \(y = -x + 5\), has slope \(m = -1\) and \(y\)-intercept \(b = 5\). It passes through \((0, 5)\) and falls one unit for every step right.

When you draw these lines, they cross at a single point. To find that point exactly, set the two expressions for \(y\) equal to each other (this is actually substitution in disguise):

$$x + 1 = -x + 5$$ $$2x = 4$$ $$x = 2$$

Then \(y = x + 1 = 2 + 1 = 3\). The lines intersect at \((2, 3)\).

Check: In the second equation, \(y = -x + 5 = -2 + 5 = 3\). Both give \(y = 3\), so the point is confirmed.

Special Cases: No Solution and Infinitely Many Solutions

Not every system gives a neat ordered pair. Sometimes the algebra produces a strange-looking statement, and that statement is itself the answer.

No solution (parallel lines)

Consider:

$$\begin{cases} y = 2x + 3 \\ y = 2x - 1 \end{cases}$$

Set the expressions equal: \(2x + 3 = 2x - 1\). Subtract \(2x\) from both sides:

$$3 = -1$$

This is false — \(3\) can never equal \(-1\). Whenever the variables vanish and you are left with a false statement, the system has no solution. The two lines have the same slope but different intercepts, so they are parallel and never meet.

Infinitely many solutions (the same line)

Now consider:

$$\begin{cases} x + y = 4 \\ 2x + 2y = 8 \end{cases}$$

Multiply the first equation by \(2\) and you get \(2x + 2y = 8\), which is exactly the second equation. The two equations describe the same line. Every point on that line satisfies both equations, so there are infinitely many solutions. If your algebra reduces to a true statement like \(0 = 0\) or \(8 = 8\), this is what is happening.

A Real-World Word Problem

Systems of equations are not just abstract puzzles — they model real situations whenever two unknown quantities are linked by two conditions. Here is a classic example.

Problem: A movie theater sells adult tickets for 12 dollars and child tickets for 8 dollars. One evening it sold 200 tickets and collected 2,040 dollars. How many adult tickets and how many child tickets were sold?

Step 1 — Define variables. Let \(a\) be the number of adult tickets and \(c\) be the number of child tickets.

Step 2 — Translate into equations. The total number of tickets and the total money give us two equations:

$$\begin{cases} a + c = 200 \\ 12a + 8c = 2040 \end{cases}$$

Step 3 — Solve by substitution. From the first equation, \(c = 200 - a\). Substitute into the second:

$$12a + 8(200 - a) = 2040$$ $$12a + 1600 - 8a = 2040$$ $$4a + 1600 = 2040$$ $$4a = 440$$ $$a = 110$$

Step 4 — Find \(c\): \(c = 200 - 110 = 90\).

Check: \(12(110) + 8(90) = 1320 + 720 = 2040\). The totals match.

The trick with word problems is always the same: name your unknowns, write one equation per condition, then solve the system with whichever method fits best.

Which Method Should You Use?

All three methods are correct, so the “best” one depends on how the problem is written. Use this quick decision guide:

• Choose substitution when one variable is already alone or has a coefficient of \(1\) — for example, \(y = 3x - 2\) or \(x = 5 - y\). Isolating is painless.
• Choose elimination when both equations are in standard form \(ax + by = c\), especially if a pair of coefficients are already opposites or easy to match with one multiplication.
• Choose graphing when you want a visual understanding, when the numbers are friendly, or when you only need an estimate.

Common Mistakes to Avoid

Most lost points on systems problems come from a handful of avoidable errors. Watch for these:

• Forgetting to find the second variable. A full solution is an ordered pair \((x, y)\), not just one number. Always back-substitute.
• Sign errors during elimination. Subtracting equations flips every sign on the line you subtract. Take it one term at a time.
• Multiplying only part of an equation. Every term — left and right of the equals sign — must be multiplied by the same factor.
• Skipping the check. Plugging your pair back into both originals takes ten seconds and catches almost every mistake.
• Misreading special cases. Remember: false statement means no solution; true statement means infinitely many.

If you are still building confidence with the individual steps, it helps to strengthen related skills too. Our walkthroughs on How to Solve Quadratic Equations (4 Methods) and How to Solve Inequalities Step by Step use many of the same moves — isolating variables, combining like terms, and checking solutions.

Putting It All Together

Solving a system of equations always comes down to the same goal: reduce two unknowns to one, solve that single equation, then work backward for the other unknown. Substitution swaps an expression in; elimination cancels a variable by adding or subtracting; graphing finds the crossing point of two lines. Pick the tool that matches the problem, and finish every time by checking your pair in both original equations.

With the four worked examples above — substitution, simple elimination, elimination with multiplication, and graphing — plus the word problem and the two special cases, you now have a complete toolkit. The only thing left is practice. Try a few problems on your own, then compare your work against a trusted answer.