How to solve systems of equations

By Taimur Hassan SiddiquiPublished 7 October 2026

A system of equations asks for values that satisfy several equations at once. This guide solves the same example three ways - substitution, elimination and Cramer's rule - and shows how to recognise a system with no solution or infinitely many.

The short version

Write each equation in standard form. ax + by = c, with the variables on the left.
Choose a method. Substitution, elimination or determinants.
Solve for one variable, then the other. Back-substitute to find the rest.
Check in both equations. The pair must satisfy every equation.

Skip the arithmetic. The free System of Equations Solver does this for you, shows the formula and the working with your own numbers, and runs in your browser - nothing you type is uploaded. Also useful: Quadratic Formula Calculator, Slope Calculator.

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The example

Solve 2x + 3y = 13 and x − y = −1.

Method 1: substitution

x − y = −1 → x = y − 1 2(y − 1) + 3y = 13 5y − 2 = 13 → y = 3 x = 3 − 1 = 2

Method 2: elimination

multiply the second equation by 3: 3x − 3y = −3 add it to the first: (2x + 3y) + (3x − 3y) = 13 + (−3) 5x = 10 → x = 2, then y = 3

Method 3: Cramer's rule

D = 2(−1) − 3(1) = −5 Dx = 13(−1) − 3(−1) = −10 x = −10 ÷ −5 = 2 Dy = 2(−1) − 13(1) = −15 y = −15 ÷ −5 = 3

Cramer's rule replaces one variable's column with the right-hand sides and divides determinants. It is easy to show as a list of steps and works the same for three equations, but is slower by hand for large systems.

Using it on a word problem

Tickets cost 12 for adults and 8 for children. 10 tickets were sold for a total of 104. How many of each?

a + c = 10 12a + 8c = 104 substitute c = 10 − a: 12a + 8(10 − a) = 104 4a + 80 = 104 → a = 6, c = 4 check: 12 × 6 + 8 × 4 = 72 + 32 = 104 ✓

Always define your variables in words first (here a = adult tickets, c = child tickets), turn each sentence of the problem into an equation, and check the answer against the original wording.

No solution and infinitely many solutions

Each equation in two variables is a line, and the solution is where they cross.

One solution: the lines cross once. The determinant D is not zero.
No solution: the lines are parallel. x + y = 2 and 2x + 2y = 5 contradict each other.
Infinitely many: the equations are the same line. x + y = 2 and 2x + 2y = 4 describe it twice.

When D = 0 you cannot divide by it, which is how you know a system is one of the two special cases. With three variables the equations are planes and the same three outcomes apply.

Frequently asked questions

Which method is best?

Substitution when one equation already isolates a variable, elimination when coefficients line up, and Cramer's rule when you want a mechanical procedure.

What if an equation is missing a variable?

Treat the coefficient as 0. For x = 4 and x + y = 9 write 1x + 0y = 4.

How do I solve three equations?

Eliminate one variable to get two equations in two unknowns, solve those, then back-substitute. Or use Cramer's rule with 3 × 3 determinants.

How do I know my answer is right?

Substitute the values into every original equation. They must all balance.

Can fractions or decimals appear in the solution?

Yes. Many systems have non-integer solutions; keeping them as fractions avoids rounding errors.