Should I use substitution or elimination?

Substitution when a variable is already isolated or has a coefficient of 1. Elimination when both equations are in standard form, especially if a variable already has matching or opposite coefficients. Both always give the same answer — one is just less work.

What does it mean when the variables all cancel?

Check what statement is left. If it is true, like 0 = 0, the equations describe the same line and there are infinitely many solutions. If it is false, like 0 = 3, the lines are parallel and there is no solution.

Why does a system need two equations?

One equation with two unknowns has infinitely many solutions — every point on a line. A second independent equation pins it down to the single point where the lines cross. In general you need one equation per unknown.

Can I solve a system with three variables?

Yes, and you need three equations. The usual approach is to eliminate one variable from two different pairs of equations, reducing it to a two-variable system, then back-substitute twice.

How do I check my answer?

Substitute the pair into both original equations. It must satisfy each one — a pair that works in only one equation is not a solution, and this check catches almost every arithmetic slip.

Can Photo Math Solver read both equations at once?

Yes — screenshot the whole system, including a brace if there is one, and the solution picks the method that suits how the equations are written, then shows every step and the final check.

You can start solving systems of equations for free by adding the Chrome extension — try it on your own worksheet before deciding if you need anything more.

AI-Powered Algebra Solver

Solve Systems of Equations Step by Step

Screenshot both equations and get the method choice, the elimination or substitution, and the checked pair.

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Photo Math Solver Solved

\begin{cases} y = 2x + 1 \\ 3x + y = 16 \end{cases}

1 $y$ is already isolated, so substitute into the second equation: $3x + (2x + 1) = 16$ .
2Combine and solve: $5x + 1 = 16$ , so $5x = 15$ and $x = 3$ .
3Back-substitute: $y = 2(3) + 1 = 7$ . Solution $(3, 7)$ .

On this page01Definition 02Formulas 03Methods 04Examples 05Practice 06Mistakes 07FAQ

01Definition

What is a system of equations?

A system of equations is two or more equations that must hold at the same time, sharing the same unknowns. Solving it means finding the values that satisfy every equation at once — not just one of them.

With two variables, each linear equation is a line on a graph, and the solution is where those lines cross. That picture explains all three possible outcomes: lines that cross once give one solution, parallel lines never meet and give no solution, and identical lines overlap everywhere and give infinitely many.

One equation with two unknowns has endlessly many answers, which is why a second equation is needed. As a rule you need as many independent equations as you have unknowns — two for $x$ and $y$ , three for $x$ , $y$ and $z$ .

Where you'll actually use this

Break-even analysis: where a cost line and a revenue line cross is a two-equation system.
Mixture problems in chemistry — combining two concentrations to hit a target.
Ticket, coin, and inventory totals, where one equation counts items and the other counts value.
Supply and demand equilibrium in economics is literally the intersection of two lines.
Comparing two phone or subscription plans to find the usage level where costs match.

System of equations: Two or more equations in the same unknowns that must all hold at once.
Substitution: Isolating one variable and putting its expression into the other equation.
Elimination: Adding or subtracting scaled equations so one variable cancels out.
Consistent system: A system with at least one solution — the lines cross or coincide.
Dependent system: Two equations describing the same line, giving infinitely many solutions.

02Formulas

The formulas you need

Keep these on hand — every method below builds on them.

Standard form

\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

Variables lined up in columns — the layout elimination needs.

The solution

(x, y)

An ordered pair, not a single number — it must satisfy both equations.

One solution

\frac{a_1}{a_2} \ne \frac{b_1}{b_2}

Different slopes, so the lines cross exactly once.

No solution

0 = 5 \;\text{(a false statement)}

Parallel lines — same slope, different intercepts.

Infinitely many

0 = 0 \;\text{(always true)}

The same line written twice; every point on it works.

03Methods

How to solve systems of equations

Pick the method that fits the problem in front of you.

When to use it: One equation already has a variable isolated, or has a coefficient of

1

that makes isolating it easy — like

y = 2x + 1

1Isolate one variable in whichever equation makes it cheapest.
2Substitute that whole expression into the other equation — never back into the one it came from.
3Solve the resulting single-variable equation, e.g. $3x + 2(2x + 1) = 16$ gives $x = 2$ .
4Back-substitute to find the second variable, then check the pair in both originals.

04Worked examples

Systems of Equations, solved step by step

From easy to hard — pick a problem to see its full solution.

\begin{cases} y = 2x + 1 \\ 3x + y = 16 \end{cases}

1 $y$ is already isolated, so substitute into the second equation: $3x + (2x + 1) = 16$ .
2Combine and solve: $5x + 1 = 16$ , so $5x = 15$ and $x = 3$ .
3Back-substitute: $y = 2(3) + 1 = 7$ . Solution $(3, 7)$ .

Try it on your own systems of equations homework

Photo Math Solver reads both equations of a system straight from a screenshot — a worksheet pair, a word problem, a graph question — and shows the full working: which method suits the way the equations are written, the substitution or cancellation, and the check of the answer in both original equations.

05Practice

Now you try

Work each one on paper first, then check your answer.

Problem 1

\begin{cases} y = x - 2 \\ 2x + y = 10 \end{cases}

Problem 2

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

Problem 3

\begin{cases} 2x + 3y = 7 \\ 4x - y = 7 \end{cases}

Problem 4

\begin{cases} x + y = 6 \\ 2x + 2y = 15 \end{cases}

Problem 5

\text{Two numbers sum to } 25 \text{ and differ by } 7.

Stuck on your systems of equations homework? Screenshot it

Systems are two equations that have to be read together, often stacked under a brace and easy to mistranscribe one sign at a time. Screenshot the whole system with Photo Math Solver and the solution keeps both equations aligned, shows the scaling factors, and checks the answer in each original.

06Common mistakes

Where points get lost

Each of these shows up on real graded work — and each has a simple fix.

The mistake	Why it happens	The fix
Substituting an expression back into the equation it was derived from.	That equation is the one in front of you, and the substitution feels like the next move.	Always substitute into the other equation, not the source one. Going back into the source collapses to $0 = 0$ and tells you nothing, which is a sign you have looped rather than a sign of infinite solutions.
Multiplying only one term of an equation when scaling for elimination.	Attention goes to the coefficient being matched, and the constant on the right gets left behind.	Multiply every term on both sides. Scaling $2x + 3y = 12$ by $2$ gives $4x + 6y = 24$ — the $12$ must double too.
Finding one variable and stopping.	Solving for $x$ feels like completing the problem, since that is where every previous chapter ended.	The solution is an ordered pair. Back-substitute to find the second variable and write the answer as $(x, y)$ .
Adding when the coefficients match in sign, so nothing cancels.	Elimination is remembered as "add the equations" without checking the signs first.	Add when the coefficients are opposites ( $+3y$ and $-3y$ ); subtract when they match ( $+3y$ and $+3y$ ). Or scale one equation by $-1$ so adding always works.
Reading $0 = 0$ as no solution and $0 = 5$ as infinitely many.	Both outcomes end with the variables gone, and the two cases are easily swapped.	Ask whether the leftover statement is true. $0 = 0$ is always true, so every point works — infinitely many. $0 = 5$ is never true, so nothing works — no solution.

Photo Math Solver shows every intermediate step, so slips like these are easy to catch before they cost you marks.

07FAQ