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System of Equations Solver

Solve two or three simultaneous equations by substitution, elimination, or graphing, with the reason one method beats the others on your particular system spelled out.

Solve a system of equations problem now

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How to use it

Enter the equations separated by “and” or a comma. They do not need to be rearranged into any particular form first. Ask for a specific method if your course requires one, and ask for the graph if you want to see where the lines meet. Systems that arrive as a story — two prices, two quantities, one total — are usually easier through the word problem solver, which shows the translation into equations before solving.

The method behind it

A linear system asks where two lines meet. That geometric picture predicts everything: two lines in a plane either cross once, are parallel and never cross, or are the same line drawn twice.

{a_{1} x + b_{1} y = c_{1} a_{2} x + b_{2} y = c_{2}

Which case you are in is decided by one number. If $a_{1} b_{2} - a_{2} b_{1} \neq = 0$ the lines have different slopes and there is exactly one solution. If it is zero the slopes match, and then the constants decide whether the lines coincide or sit apart.

Substitution solves one equation for one variable and puts that expression into the other, reducing two unknowns to one. It is fastest when some variable already has a coefficient of 1 or −1, because the rearrangement introduces no fractions.

Elimination scales one or both equations so that a variable has equal and opposite coefficients, then adds them so that the variable disappears. It is fastest when no coefficient is 1, and it is the method that scales to three or more variables — at which point it is called row reduction and gets written as a matrix.

Graphing converts both equations to slope-intercept form and reads off the intersection. It is the least accurate method for exact answers but the most informative when you want to know why a system has no solution, and it is the only one where parallel lines are obvious at a glance.

For a two-by-two system there is also a closed form. It is Cramer's rule, and the denominator is the same quantity that decides the number of solutions.

x = \frac{c _{1} b _{2} - c _{2} b _{1}}{a _{1} b _{2} - a _{2} b _{1}}, y = \frac{a _{1} c _{2} - a _{2} c _{1}}{a _{1} b _{2} - a _{2} b _{1}}

The denominator is the determinant of the coefficient matrix, which is why a zero determinant and a system without a unique solution are the same statement. That connection is developed in inverting a matrix two ways, and the choice between the three hand methods is argued in detail in substitution, elimination, or graphing.

Worked examples

Worked example

4 x + 3 y = 10, 3 x - 2 y = - 1

1
No coefficient is 1, so substitution would create fractions immediately. Use elimination and pick the variable to remove — y needs the smaller multipliers.
$eliminate y : lcm (3, 2) = 6$
2
Scale the first equation by 2 and the second by 3 so the y coefficients become +6 and −6.
$8 x + 6 y = 20, 9 x - 6 y = - 3$
3
Add the two equations. The y terms cancel because the coefficients are exact opposites.
$17 x = 17 \Rightarrow x = 1$
4
Back-substitute into an original equation, not a scaled one, so an arithmetic slip in the scaling would show up.
$4 (1) + 3 y = 10 \Rightarrow 3 y = 6 \Rightarrow y = 2$
5
Check in the equation you did not use for the back-substitution.
$3 (1) - 2 (2) = 3 - 4 = - 1 ✓$

Answer

(x, y) = (1, 2)

Worked example

2 x - 4 y = 6, - x + 2 y = 1

1
Scale the second equation by 2 so that the x coefficients are opposites.
$- 2 x + 4 y = 2$
2
Add it to the first equation. Both variables vanish at once, which never happens when a unique solution exists.
$(2 x - 4 y) + (- 2 x + 4 y) = 6 + 2$
3
What is left is a false statement. No pair of values can make it true, so nothing solves the system.
$0 = 8$
4
Confirm geometrically by putting both lines in slope-intercept form. Same slope, different intercept.
$y = \frac{1}{2} x - \frac{3}{2}, y = \frac{1}{2} x + \frac{1}{2}$
5
The lines are parallel and distinct, so they never meet. Had the result been 0 = 0 instead, the lines would coincide and every point on the line would be a solution.
$parallel, distinct$

Answer

No solution

One system, three methods

The three methods are not alternatives with different answers — they are three routes to the same point. Watching the same system go through all three shows exactly what each one costs.

Solve 2x + 3y = 12 and x − y = 1 three ways

0 of 9 steps

Try it yourself first — you'll remember it longer.

Common mistakes

Adding equations that are not aligned

Elimination works only when the coefficients of one variable are exact opposites. Adding $2 x + 3 y = 12$ to $x - y = 1$ gives $3 x + 2 y = 13$ , which is true but contains both unknowns and has achieved nothing. Scale first.

Back-substituting into the rearranged equation

If you rearranged the second equation to $x = y + 1$ , substitute your value of $y$ into an equation you have not manipulated. Feeding it back into the same rearrangement will confirm an error rather than catch it.

Confusing no solution with infinitely many

Both cases lose all the variables. Ending at $0 = 8$ is a false statement, so no solution exists. Ending at $0 = 0$ is always true, so the equations describe the same line and there are infinitely many. Read the surviving statement rather than the disappearance of the variables.

Practice

Answers are checked here — nothing is sent anywhere.

1
$Solve x + y = 7, x - y = 3$
2
$Solve 2 x + y = 11, x + 3 y = 18$
3
$How many solutions? 3 x - 6 y = 9, - x + 2 y = 4$

Frequently asked questions

Can it solve systems with three or more variables?

Yes. Beyond two variables the working switches to elimination arranged as row reduction, because tracking substitutions by hand across three equations is where most errors appear.

Which method will it use?

Whichever is shortest for your system, and it says why. If your class requires a particular method, name it in the question and the working follows that route instead.

What does an answer with infinitely many solutions look like?

As a parametric description, for example x = 2t + 1 and y = t, rather than the words on their own. The two equations describe the same line, so every point on it is a solution.

Does it handle non-linear systems?

Yes, such as a line meeting a circle or two parabolas. Substitution is almost always the method there, and the result can be zero, one, or two intersection points.

What does it cost?

The first solution is free with no account. Free accounts include three solutions a day, and Gauth Plus is unlimited at $11.99 a month with a three-day free trial.

Keep going

Algebra solverSystems alongside equations and inequalities, with every step shown.Substitution, elimination, or graphing?The same problem solved three ways, with a rule for picking the fastest.Inverting a matrix two waysWhy a zero determinant and a system without a unique solution are the same fact.Matrix calculatorWrite the system as a matrix and solve it by row reduction instead.