System of Equations Calculator

How to use the system of equations calculator

1. Choose 2 or 3 equations. Two equations use x and y; three use x, y and z.
2. Write each equation in standard form, with the variables on the left and the constant on the right, such as 2x + y = 7. Enter the coefficient of each variable and the constant. Use 0 for a missing variable and a minus sign for negative numbers. Fractions such as 3/2 are fine.
3. Read the solution. The line above each row repeats the equation as you typed it, so you can spot sign errors. The result gives each variable exactly, with a decimal when it is a fraction.

The calculator opens on the system used throughout section 4.1 of OpenStax’s Intermediate Algebra 2e: 2x + y = 7 and x − 2y = 6. Its solution is x = 4, y = −1.

What a solution means

A solution of a system is a set of values that makes every equation true at once. Check (4, −1): 2(4) + (−1) = 7 and 4 − 2(−1) = 6. Both hold.

Each linear equation in two variables is a line, so solving a system of two is finding where two lines meet. That gives three possibilities, which the textbook names:

With three variables, each equation is a plane in space, and the same three outcomes apply: the planes meet at one point, share no common point, or share a whole line or plane.

Substitution and elimination

The textbook teaches three methods for two equations: graphing, substitution and elimination.

Substitution solves one equation for one variable and substitutes it into the other. From 2x + y = 7, y = 7 − 2x. Then x − 2(7 − 2x) = 6 gives 5x − 14 = 6, so x = 4 and y = 7 − 8 = −1. It is convenient when a variable already has a coefficient of 1.

Elimination adds multiples of the equations so that one variable cancels. Multiplying 2x + y = 7 by 2 gives 4x + 2y = 14; adding x − 2y = 6 gives 5x = 20, so x = 4. In Example 4.10, 4x − 3y = 9 and 7x + 2y = −6, neither equation can be multiplied alone to cancel a variable, so both are scaled: 2 times the first plus 3 times the second gives 29x = 0, so x = 0 and y = −3.

The calculator uses elimination in a systematic form, Gauss–Jordan elimination, carried out with exact fractions so no rounding creeps in.

Equations with fractions and decimals

Example 4.11 solves x + ½y = 6 and (3/2)x + (2/3)y = 17/2. By hand, the first step is to clear the fractions by multiplying each equation by its least common denominator. In the calculator, type the fractions directly, as 1/2, 3/2, 2/3 and 17/2; the answer is (3, 6). Decimals such as 0.25 are read exactly as 1/4, so 0.1 + 0.2 issues of floating-point arithmetic don’t arise.

Three equations in three variables

Section 4.4 extends elimination to three variables: use two pairs of equations to eliminate the same variable, which leaves two equations in two variables, then solve those and substitute back. Example 4.32 solves

and finds (4, −1, −3). Load it with the “3 × 3” example.

Examples 4.34 and 4.35 show the other two outcomes. In 4.34, elimination ends with a false statement such as 0 = 2, so the system is inconsistent and has no solution. In 4.35 it ends with 0 = 0, a dependent system with infinitely many solutions of the form x = 5z − 5, y = −2z + 3, with z any real number. The “Dependent” example reproduces that general solution.

The determinant shortcut

The determinant of the coefficients tells you at a glance whether there is exactly one solution. For two equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂, it is a₁b₂ − a₂b₁. For the default system, 2 × (−2) − 1 × 1 = −5.

• Not zero: exactly one solution.
• Zero: either no solution or infinitely many; the elimination result tells you which.

When the determinant isn’t zero, Cramer’s rule writes each variable as a ratio of determinants, for example x = (c₁b₂ − c₂b₁) ÷ (a₁b₂ − a₂b₁). It is handy for two equations but slower than elimination for larger systems.

Setting up word problems

Many applied problems turn into a system. Section 4.4’s theater example sells adult, student and child tickets at different prices; knowing the total tickets, total revenue and one more relationship gives three equations in three unknowns. Name a variable for each unknown, write one equation for each fact you are given, put each in standard form, and enter the coefficients. If the calculator reports no solution or infinitely many, recheck the setup: a missing fact or a repeated one is the usual cause.

Common mistakes

• Not in standard form. An equation like y = 3 − (3/4)x must first be rewritten as (3/4)x + y = 3, as the textbook does in Example 4.12, before its coefficients go into the boxes.
• Lost signs. x − 2y = 6 has a y coefficient of −2, not 2. The equation preview above each row helps catch this.
• Missing variables. In 3x − 4z = 0, the y coefficient is 0, not blank.
• Stopping at one variable. After finding x, substitute it back to find the others, then check the full solution in every original equation.

Seeing it on a graph

For two equations, solving by graphing means drawing both lines and reading off where they cross; Example 4.2 does this for 2x + y = 7 and x − 2y = 6, and the lines meet at (4, −1). Graphing shows the picture behind the three outcomes, but reading a crossing point accurately is hard when the answer is a fraction such as (13/7, −2/7). That is where the algebraic methods, and an exact calculator, help. To graph a line yourself, the slope calculator turns any two points into the slope-intercept form y = mx + b. Lines with equal slopes and different intercepts are parallel and don’t cross, the inconsistent case; equal slopes and equal intercepts mean the same line, the dependent case.

Sources & assumptions

How this calculator gets its numbers. Full method: our methodology.

• Linear equations in standard form; coefficients may be integers, decimals or fractions.
• Solved exactly by elimination (Gauss–Jordan) with fractions, so answers are exact.
• A zero determinant means no single solution: the system is either inconsistent or dependent.

Sources

1. OpenStax (Rice University), Intermediate Algebra 2e, 4.1 Solve Systems of Linear Equations with Two Variables. Graphing, substitution and elimination; consistent, inconsistent and dependent systems (Examples 4.9–4.11).
2. OpenStax (Rice University), Intermediate Algebra 2e, 4.4 Solve Systems of Equations with Three Variables. Elimination with three variables; Example 4.32 → (4, −1, −3), Example 4.34 no solution, Example 4.35 dependent.

Last reviewed November 2, 2026. For one equation in x², use the quadratic formula calculator.