System of Equations Calculator with Steps

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A

Equation Input Mode

3 (2-10 equations)
B

How to Use

Matrix Input Mode:

  • Enter matrix values in Matrix A
  • Click the RREF button to calculate reduced row echelon form

Equation Input Mode:

  • Use +/- buttons to adjust number of equations (2-10)
  • Enter coefficients for x, y, z variables
  • Enter constant after the equals sign
  • Click Calculate RREF button

Both modes will calculate the reduced row echelon form with detailed steps.

RREF Calculation Results

System of Equations Calculator

This article explains how a System of Equations Calculator works, including the mathematical definition, solution formulas, worked examples, common mistakes, and practice problems with solutions. It is designed as companion content for an online system of equations calculator.


What Is a System of Equations?

A system of equations is a collection of two or more equations that share the same variables. The goal is to find values of the variables that satisfy all equations simultaneously.

A general linear system can be written as:

$$ \begin{cases} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m \end{cases} $$

In matrix form, this system is written compactly as:

$$ A\mathbf{x} = \mathbf{b} $$

where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the vector of variables, and \(\mathbf{b}\) is the constant vector.


Worked Examples

Example 1: Unique Solution

Solve the system:

$$ \begin{cases} 5x + 2y = 10 \\ 7x + 2y = 14 \end{cases} $$
Show Solution
Subtract the first equation from the second: \[ (7x+2y) - (5x+2y) = 14-10 \Rightarrow 2x=4 \Rightarrow x=2 \] Substitute \(x=2\) into the first equation: \[ 5*2 + 2y = 10 \Rightarrow 10 + 2y = 10 \Rightarrow 2y = 0 \Rightarrow y=0 \] Solution: \[ (x,y)=(2,0) \]

Example 2: Unique Solution

Solve the system:

$$ \begin{cases} 3x + y = 7 \\ 5x + y = 11 \end{cases} $$
Show Solution
Subtract the first equation from the second: \[ (5x+y)-(3x+y)=11-7 \Rightarrow 2x=4 \Rightarrow x=2 \] Substitute \(x=2\) into the first equation: \[ 3*2 + y =7 \Rightarrow 6+y=7 \Rightarrow y=1 \] Solution: \[ (x,y)=(2,1) \]

Common Mistakes When Solving Systems of Equations

  • Arithmetic errors during elimination or substitution.
  • Forgetting to check whether equations are dependent or inconsistent.
  • Using the matrix inverse method when the determinant is zero.
  • Stopping before reaching reduced row echelon form.

Practice Problems

Try solving the following systems using your system of equations calculator. The answers are hidden below each problem.

Exercise 1

$$ \begin{cases} 4x + 3y = 8 \\ 2x + 3y = 4 \end{cases} $$
Show Answer
Subtract the second equation from the first: \[ (4x+3y)-(2x+3y)=8-4 \Rightarrow 2x=4 \Rightarrow x=2 \] Substitute \(x=2\) into the second equation: \[ 2*2 + 3y =4 \Rightarrow 4 + 3y=4 \Rightarrow 3y=0 \Rightarrow y=0 \] Solution: \[ (x,y)=(2,0) \]

Exercise 2

$$ \begin{cases} 6x + y = 13 \\ 2x + y = 5 \end{cases} $$
Show Answer
Subtract the second equation from the first: \[ (6x+y)-(2x+y)=13-5 \Rightarrow 4x=8 \Rightarrow x=2 \] Substitute \(x=2\) into the second equation: \[ 2*2 + y=5 \Rightarrow 4+y=5 \Rightarrow y=1 \] Solution: \[ (x,y)=(2,1) \]

Exercise 3

$$ \begin{cases} 3x + 2y = 10 \\ 5x + 2y = 14 \end{cases} $$
Show Answer
Subtract the first equation from the second: \[ (5x+2y)-(3x+2y)=14-10 \Rightarrow 2x=4 \Rightarrow x=2 \] Substitute \(x=2\) into the first equation: \[ 3*2 + 2y=10 \Rightarrow 6+2y=10 \Rightarrow 2y=4 \Rightarrow y=2 \] Solution: \[ (x,y)=(2,2) \]

Exercise 4

$$ \begin{cases} 7x + 2y = 14 \\ 5x + 2y = 10 \end{cases} $$
Show Answer
Subtract the second equation from the first: \[ (7x+2y)-(5x+2y)=14-10 \Rightarrow 2x=4 \Rightarrow x=2 \] Substitute \(x=2\) into the second equation: \[ 5*2 + 2y =10 \Rightarrow 10 + 2y=10 \Rightarrow 2y=0 \Rightarrow y=0 \] Solution: \[ (x,y)=(2,0) \]

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