How to Solve Systems of Equations? (+FREE Worksheet!)
How to Solve Systems of Equations
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.
Solve Systems of Equations: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
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
- Set the right sides equal.
- Solve x + 2 = 2x – 1 to get x = 3.
- Substitute to find y.
Elimination
- Add the equations to eliminate y.
- 2x = 10, so x = 5.
- Use x + y = 8 to find y = 3.
Try one before moving on
Solve Systems of Equations: pop-up practice
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.
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:
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.
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.
Swap \(y\) into eq. 2: \(3x+(2x+1)=11\Rightarrow x=2\), then \(y=5\) → (2, 5)
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.
Subtract to cancel \(x\): \(4y=8\Rightarrow y=2\), then \(x=3\) → (3, 2)
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.
The lines meet at (2, 3) — see the graph just below.
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 systemWorked 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
- \(\color{blue}{x=2,\ y=-\frac{5}{2}}\)
- \(\color{blue}{x=0,\ y=-3}\)
- \(\color{blue}{x=1,\ y=2}\)
- \(\color{blue}{x=20,\ y=2}\)
- \(\color{blue}{x=5,\ y=0}\)
- \(\color{blue}{x=0,\ y=2}\)
Make Your Own Worksheet
When you’re ready for more, generate a fresh worksheet — unlimited problems, never the same set twice, with a full answer key you can print or save as a PDF.
Frequently Asked Questions
What does it actually mean to “solve” a system?
It means finding the value of each variable that makes every equation true at the same time. For two lines, that’s simply the point where they cross. If someone hands you a candidate answer, you can confirm it by plugging it into both equations — if both come out true, you’ve got it.
Which method should I use — substitution, elimination, or graphing?
Use substitution when one variable is already by itself (like \(y = \dots\)). Use elimination when both equations are lined up in \(ax+by=c\) form. Use graphing when you want to see the answer or double-check. They all reach the same solution, so there’s no wrong choice — just easier and harder ones for a given problem.
How can I tell “no solution” from “infinitely many”?
Look at the lines. Same slope but different \(y\)-intercepts means they’re parallel — no solution. If both equations simplify to the exact same line, there are infinitely many solutions. In the algebra, a false statement like \(3=-1\) signals no solution, while an always-true statement like \(0=0\) signals infinitely many.
Can a system have more than two equations?
Absolutely. Systems can have three or more equations and variables — for example, three planes in three dimensions. The thinking is identical (find values that satisfy everything at once), though bigger systems are usually handled with elimination or matrices.
Do I always have to graph it?
No. Graphing is a great way to understand and check, but for exact answers — especially with fractions — substitution or elimination is faster and more precise. Many students graph just to confirm the answer they already found algebraically.
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