How to Solve Systems of Equations? (+FREE Worksheet!)

Algebra 1 · Systems & Coordinate Algebra

How to Solve Systems of Equations

A system of equations is a set of equations that share variables, and its solution is the point that makes every equation true. You can solve a two-variable system by graphing (find where the lines cross), substitution (solve one equation for a variable and plug it into the other) or elimination (add or subtract the equations to cancel a variable). For example, y = 2x + 3 and y = −3x + 3 give 2x + 3 = −3x + 3, so x = 0 and y = 3.

Two equations, two unknowns, one neat answer that makes both true at the same time. By the end of this lesson you’ll be able to solve any system three different ways, and know exactly which one to reach for. Practice tools are one tap away.

Tutor-style math help

Solve Systems of Equations: what to notice and how to work it

Systems skill

A system asks for values that make every equation true at the same time. On a graph, the solution is where the graphs meet.

What to notice first

Choose the method based on the form you are given. Substitution is friendly when a variable is isolated; elimination is friendly when coefficients line up.

Common student mistake

Do not stop after finding one variable. A two-variable system usually needs an ordered pair, and that pair must check in every original equation.

Key formulas and cues

$\text{linear system solution}=(x,y)$

$\text{same slope, different intercepts}\Rightarrow\text{no solution}$

$\text{same line}\Rightarrow\text{infinitely many solutions}$

A reliable path

Choose a methodGraph, substitute, or eliminate depending on the form.
Solve one variableUse the cleanest equation to find one value.
Find and check the pairSubstitute back and verify both equations.

Worked examples

Substitution

Example: $y=x+2$ and $y=2x-1$

Set the right sides equal.
Solve x + 2 = 2x – 1 to get x = 3.
Substitute to find y.

Answer: $(3,5)$

Elimination

Example: $x+y=8$, $x-y=2$

Add the equations to eliminate y.
2x = 10, so x = 5.
Use x + y = 8 to find y = 3.

Answer: $(5,3)$

Open pop-up practice

Try one before moving on

Try: Solve $y=x+1$ and $y=3x-3$.

Answer: $(2,3)$.

Next step: do the matching worksheet or quiz while the method is still fresh, then come back and explain the first step in your own words.

Step-by-Step Solver Worksheet Maker ✏Practice Drills Flashcards Video

Let’s start with the big idea, because once it clicks, everything else is just careful arithmetic. When you see a single equation like $2x + 3 = 11$, there’s one unknown and you solve for it. But a lot of real problems have two unknowns at the same time, how many adult tickets and how many child tickets, the price of apples and the price of oranges. One equation isn’t enough to pin down two unknowns, so we use two equations together. That pair is what we call a system of equations, and your job is to find the one $(x, y)$ pair that satisfies both at once. Don’t worry if that feels abstract right now. We’ll build it up slowly, with plenty of worked examples.

The big idea

What Is a System of Equations?

A system of equations is just two (or more) equations that share the same variables and are meant to be true together. Think of each equation as a clue. One clue narrows things down, but two clues together can point to a single answer. The solution is the ordered pair $(x, y)$ that makes every equation in the system true at the same moment.

Here’s a friendly example: $\begin{cases} x – y = 1 \\ x + y = 5 \end{cases}$. If you test $x = 3$ and $y = 2$, the first equation gives $3 – 2 = 1$ ✓ and the second gives $3 + 2 = 5$ ✓. Both check out, so $(3, 2)$ is the solution. That’s the whole game: find the pair that keeps every equation happy.

Because each linear equation is really a straight line on a graph, a system is just two lines, and the solution is the point where they cross. That picture leads to three possible outcomes, and it helps to know them up front:

One solutionThe lines cross at a single point, the usual case.

No solutionThe lines are parallel and never meet.

Infinitely manyThe two equations are the same line in disguise.

Three Ways to Solve a System

There isn’t one “right” method. There are three reliable ones, and good students pick whichever fits the problem in front of them. Let’s walk through each, when to use it, and a fully worked example so you can see the moves.

Method 1

Substitution

Reach for this when one variable is already alone (or easy to get alone).

Solve one equation for a single variable.
Plug that expression into the other equation.
Solve the one-variable equation you get.
Put that value back to find the second variable.

$\begin{cases} y = 2x+1 \\ 3x+y=11 \end{cases}$
Swap $y$ into eq. 2: $3x+(2x+1)=11\Rightarrow x=2$, then $y=5$ → (2, 5)

Method 2

Elimination

Reach for this when both equations are tidy, in $ax+by=c$ form.

Multiply so one variable has opposite coefficients.
Add the equations to cancel that variable.
Solve for what’s left.
Back-substitute for the other variable.

$\begin{cases} 2x+3y=12 \\ 2x-y=4 \end{cases}$
Subtract to cancel $x$: $4y=8\Rightarrow y=2$, then $x=3$ → (3, 2)

Method 3

Graphing

Reach for this when you want to see the answer or check your work.

Rewrite each line as $y=mx+b$.
Graph both lines.
The point where they cross is your solution.

$\begin{cases} y=x+1 \\ y=-x+5 \end{cases}$
The lines meet at (2, 3), see the graph just below.

Tutor tip: All three methods always give the same answer, so if you’re ever unsure, solve it one way and check it a second way. Substitution and elimination are usually faster and more exact; graphing is wonderful for understanding and for catching the “no solution” and “infinitely many” cases at a glance.

See it graphically

The solution is where the lines meet

Below are $y=x+1$ and $y=-x+5$ drawn together. Notice they cross at exactly one point, $(2,3)$, and that single point is the only place where both equations are true. Slide your own system into the solver and watch its graph appear instantly.

Graph my system

Worked Examples, Let’s Solve Some Together

Read these slowly and try to predict the next step before you read it. That little bit of effort is what makes the method stick.

Example 1, Elimination

Find $x+y$ for $\begin{cases} 2x+5y=11 \\ 4x-2y=-26 \end{cases}$.

The $x$-terms don’t match yet, so multiply eq. 1 by $-2$: $-4x-10y=-22$. Add it to eq. 2 and the $x$’s cancel: $-12y=-48\Rightarrow y=4$. Now back-substitute: $2x+5(4)=11\Rightarrow x=-4.5$.

So $x+y=$ −0.5. Check: $4(-4.5)-2(4)=-26$ ✓

Example 2, Substitution

Solve $\begin{cases} 3x-4y=-20 \\ -x+2y=10 \end{cases}$.

The second equation is friendly, solve it for $x$: $x=2y-10$. Substitute into eq. 1: $3(2y-10)-4y=-20\Rightarrow 2y=10\Rightarrow y=5$. Then $x=2(5)-10=0$.

Solution: (0, 5).

Example 3, When there’s no solution

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

Both lines have slope $2$ but different starting points, so they run side by side forever and never touch.

Result: no solution. (If you push the algebra, you get $3=-1$, a false statement, which is the tell-tale sign.)

Example 4, When there are infinitely many

Solve $\begin{cases} y=3x-2 \\ 2y=6x-4 \end{cases}$.

Divide the second equation by 2 and you get $y=3x-2$, exactly the first equation. They’re the same line.

Result: infinitely many solutions (every point on the line works).

Example 5, A real word problem

A school play sells 20 tickets for a total of $115. Adult tickets cost $8 and student tickets cost $5. How many of each were sold?

First, name the unknowns: let $a$ = adult tickets and $s$ = student tickets. Now turn the sentences into equations: “20 tickets” gives $a+s=20$, and “$115 total” gives $8a+5s=115$. Solve by substitution: $s=20-a$, so $8a+5(20-a)=115\Rightarrow 3a=15\Rightarrow a=5$, which means $s=15$.

Answer: 5 adult and 15 student tickets. Check: $8(5)+5(15)=115$ ✓ The hardest part of a word problem is usually the setup, not the solving, so always start by naming your variables.

Common Mistakes (and How to Dodge Them)

Forgetting to find both variables. A system’s answer is a pair $(x, y)$. Solving for $x$ is only half the job, always go back and find $y$ too.
Sign slips during elimination. When you multiply an equation by a negative, multiply every term. Writing the new equation on its own line before adding saves a lot of lost points.
Substituting into the same equation you started with. After you solve one equation for a variable, plug it into the other equation, otherwise you’ll just get a true statement like $0=0$ and learn nothing.
Misreading the special cases. A false statement (like $3=-1$) means no solution; an always-true statement (like $0=0$) means infinitely many. They’re easy to mix up, so say them out loud as you go.

Now You Try, Practice Exercises

Work these with paper and pencil, then reveal the answers to check yourself. If one trips you up, drop it into the step-by-step solver and watch how it’s done.

$\begin{cases} -4x-6y=7 \\ x-2y=7 \end{cases}$
$\begin{cases} -5x+y=-3 \\ 3x-7y=21 \end{cases}$
$\begin{cases} 3y=-6x+12 \\ 8x-9y=-10 \end{cases}$
$\begin{cases} x+15y=50 \\ x+10y=40 \end{cases}$
$\begin{cases} 3x-2y=15 \\ 3x-5y=15 \end{cases}$
$\begin{cases} 3x-6y=-12 \\ -x-3y=-6 \end{cases}$

Show answers

$\color{blue}{x=2,\ y=-\frac{5}{2}}$
$\color{blue}{x=0,\ y=-3}$
$\color{blue}{x=1,\ y=2}$
$\color{blue}{x=20,\ y=2}$
$\color{blue}{x=5,\ y=0}$
$\color{blue}{x=0,\ y=2}$

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Frequently Asked Questions

What is a system of equations?

A system of equations is a group of two or more equations that use the same variables, and its solution is the set of values that makes every equation true at once. For example, x + y = 10 and x − y = 2 form a system whose solution is (6, 4), because 6 + 4 = 10 and 6 − 4 = 2. On a graph, that solution is where the lines cross.

How do you solve systems of equations by substitution?

Solve one equation for one variable, then substitute that expression into the other equation. For y = x + 2 and 3x + y = 14, replace y to get 3x + x + 2 = 14, so 4x = 12 and x = 3. Then y = 3 + 2 = 5. Substitution works best when a variable is already isolated or has a coefficient of 1.

How do you solve a system of equations by elimination?

Add or subtract the equations so one variable cancels out. For 2x + y = 11 and x − y = 1, adding the equations gives 3x = 12, so x = 4. Substitute back: 4 − y = 1, so y = 3. If no variable cancels right away, multiply one or both equations by a number first so the coefficients match.

How do you solve a system of equations by graphing?

Graph both equations on the same coordinate plane and find the point where the lines intersect. For y = x + 1 and y = −x + 5, the lines cross at (2, 3). Parallel lines (same slope, different intercepts) mean no solution, and two equations that draw the same line have infinitely many solutions. Graphing is great for seeing the answer but less precise with fractions.

Is systems of equations Algebra 1 or Algebra 2?

Both. Students first learn to solve two-variable linear systems by graphing, substitution and elimination in Algebra 1, often around 8th or 9th grade. Algebra 2 builds on that with three-variable systems, systems that include a quadratic, and sometimes matrices. If you are in Algebra 1, focus on mastering the three methods with two equations and two unknowns.

How do you solve a system of equations with 3 variables?

Use elimination twice to reduce three equations to two, then solve that smaller system. Pair equations to cancel the same variable, say z, which leaves two equations in x and y. Solve those, then substitute back to find z. For x + y + z = 6, x − y = 0 and y + z = 4, you get x = 2, y = 2 and z = 2.

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How to Solve Systems of Equations? (+FREE Worksheet!)

How to Solve Systems of Equations