System of Linear Equations Solver

What this solver answers

It finds the values of the unknowns that make every equation of a system true at once: x and y for two linear equations, or x, y and z for three. The answer is exact, as fractions, with decimals beside them. When there isn’t exactly one answer, it names the case and shows why: no solution, with the combination of equations that contradicts itself, or infinitely many, with a formula that gives all of them.

How to enter your system

Choose the System size, then how to enter it:

• Grid: one box per number, laid out like the equation, [2] x + [3] y = [7]. Each box takes a whole number, a decimal such as 0.25 or a fraction such as -3/4 or 2 1/3. Move the unknowns to the left side and the plain number to the right before you type. The line under each equation reads it back, so a slip shows right away.
• Equations: type each equation as written, such as 2x + 3 = y - 4. Terms may come in any order, on either side; the solver collects them into standard form (here 2x − y = −7) and shows that under the box. Any letters work as unknowns, taken in alphabetical order, and 3/4x means (3/4)x.

Switching between the two carries your system across. The Try buttons load the examples worked through on this page.

Missing terms and blank cells

If an equation doesn’t contain an unknown, its coefficient is 0: x + z = 4 is x + 0y + z = 4. In the grid, type that 0. A blank box is never read as 0, because it may be a number you haven’t typed yet. Blank boxes are dashed, and the result lists them with a Fill blank cells with 0 button for when every blank really is a missing term. Typed equations don’t need this: leaving y out of x + z = 4 already says its coefficient is 0.

Fractions and decimals stay exact

Every number is kept as a fraction, so 0.1 is exactly 1/10 and 1/3 is never cut to 0.333 partway through. Type 1/3 rather than 0.333: a rounded decimal is used exactly as typed, which gives the exact answer to a slightly different system, and a note under the box says so. The Fractions example solves (1/2)x + (1/3)y = 4 and 0.25x − y = −1; the answer is x = 44/7 and y = 18/7, shown beside their decimals, about 6.285714 and 2.571429.

Worked example: solving a 3 × 3 system step by step

A baker has 16 cups of sugar, 17 cups of butter and 38 cups of flour. A batch of cookies (x) takes 1 cup of sugar, 1 of butter and 2 of flour; a batch of muffins (y) takes 2, 1 and 3; a batch of scones (z) takes 1, 2 and 4. How many batches of each use everything up? One equation per ingredient:

$$\begin{aligned} x + 2y + z &= 16 \quad \text{(sugar)} \\ x + y + 2z &= 17 \quad \text{(butter)} \\ 2x + 3y + 4z &= 38 \quad \text{(flour)} \end{aligned}$$

1. Write the augmented matrix, one row per equation. Putting the sugar equation first gives a 1 in the top-left corner, which keeps the first steps free of fractions.

   $$\left[\begin{array}{rrr|r} 1 & 2 & 1 & 16 \\ 1 & 1 & 2 & 17 \\ 2 & 3 & 4 & 38 \end{array}\right]$$

2. Clear the x column below the pivot: R2 ← R2 − R1 gives 0, −1, 1 | 1, and R3 ← R3 − 2R1 gives 0, −1, 2 | 6.

3. Make the y pivot 1: R2 ← −R2 gives 0, 1, −1 | −1.

4. Clear the rest of the y column: R1 ← R1 − 2R2 gives 1, 0, 3 | 18, and R3 ← R3 + R2 gives 0, 0, 1 | 5.

5. The z pivot is already 1. Clear its column: R1 ← R1 − 3R3 gives 1, 0, 0 | 3, and R2 ← R2 + R3 gives 0, 1, 0 | 4.

6. Each row now reads unknown = number: x = 3, y = 4, z = 5, so 3 batches of cookies, 4 of muffins and 5 of scones.

7. Check in every original equation: sugar 3 + 2 × 4 + 5 = 16, butter 3 + 4 + 2 × 5 = 17 and flour 2 × 3 + 3 × 4 + 4 × 5 = 38.

The calculator shows these seven row operations with the matrix after each one, and the same answer by Cramer’s rule: D = −1, Dx = −3, Dy = −4 and Dz = −5, so x = −3 ÷ (−1) = 3, y = 4 and z = 5.

How to solve a system of linear equations by elimination (row reduction)

Elimination adds multiples of one equation to another until each equation is left with one unknown. Row reduction does the same on the augmented matrix [A | b]: one row per equation, one column per unknown and the right sides after the bar. Three operations never change the solutions:

• swap two rows (R1 ↔ R2);
• multiply a row by a number other than 0 (R2 ← −R2);
• add a multiple of one row to another (R3 ← R3 − 2R1).

Work one column at a time. Take as the pivot the first row, from the top, with a nonzero entry in that column; make the pivot 1; then clear every other entry in its column. At the end, in reduced row-echelon form, each pivot is 1 and the only nonzero entry in its column, so the rows read x = …, y = …, z = ….

If the pivot position holds a 0, swap in a row below it first. The Row swap example starts with 2y + z = 7, which has no x term, so its first step is R1 ↔ R2; it ends at x = 1, y = 2, z = 3.

Turning a word problem into a system

Give each quantity the question asks for its own letter, then write one equation for each fact that links them, keeping the units the same on both sides. In the Coffee blend example, a roaster mixes beans costing $9 and $14 a pound into 20 pounds that should be worth $12 a pound. With x pounds of the $9 beans and y pounds of the $14 beans:

• weight: x + y = 20 (pounds);
• value: 9x + 14y = 12 × 20 = 240 (dollars).

The solution, x = 8 and y = 12, means 8 pounds of the cheaper beans and 12 of the pricier ones.

When a system has no solution

Row reduction then ends with a row that reads 0 = c, where c isn’t 0, and no values can make that true. Such a system is called inconsistent. With two unknowns the equations are parallel lines: the same slope and different intercepts. The No solution example, 2x − 3y = 6 and −4x + 6y = 5, has two lines of slope 2/3, and 2 × equation 1 + equation 2 gives 0x + 0y = 17.

With three unknowns the contradiction can take all three equations. In x + y + z = 2, x − y + 2z = 1 and 2x + 3z = 5, no two equations clash, but adding the first two gives 2x + 3z = 3 while the third says 2x + 3z = 5. The solver finds that combination and shows it: equation 1 + equation 2 − equation 3 gives 0x + 0y + 0z = −2. In the language of rank, the coefficients A have rank 2 but [A | b] has rank 3.

When a system has infinitely many solutions

Row reduction then leaves at least one unknown with no pivot and no contradiction. That unknown is free: give it any value t, and the other rows give the rest. In the Infinitely many example,

$$\begin{aligned} x + 2y - z &= 1 \\ 2x + 5y + z &= 8 \\ 3x + 7y &= 9 \end{aligned}$$

equation 3 is equation 1 + equation 2, so it adds nothing new: A and [A | b] both have rank 2, one less than the 3 unknowns. The reduced matrix leaves z free, and every solution, for any number t, is

$$\begin{aligned} x &= -11 + 7t \\ y &= 6 - 3t \\ z &= t \end{aligned}$$

t = 0 gives (−11, 6, 0) and t = 1 gives (−4, 3, 1), and both satisfy all three equations. With two unknowns this case means both equations describe the same line. If your textbook lets a different unknown be free, its formulas look different but describe the same points.

The determinant and Cramer’s rule

For a square system, the determinant D of the coefficients decides whether there is exactly one solution: D ≠ 0 means exactly one. Cramer’s rule then gives each unknown as a ratio of determinants,

$$x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D}, \qquad z = \frac{D_z}{D}$$

• $D$ is the determinant of the coefficients; for two equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ it is $a_1b_2 - b_1a_2$.
• $D_x$ is the same determinant with the x column replaced by the right sides $c_1, c_2$; likewise $D_y$ and $D_z$.

For the coffee blend, D = 1 × 14 − 1 × 9 = 5, Dx = 20 × 14 − 1 × 240 = 40 and Dy = 1 × 240 − 20 × 9 = 60, so x = 40 ÷ 5 = 8 and y = 60 ÷ 5 = 12. For a 3 × 3 system the steps expand each determinant along its first row, where $a_{ij}$ is the coefficient in row i and column j:

$$\begin{aligned} D &= a_{11}(a_{22}a_{33} - a_{23}a_{32}) \\ &\quad - a_{12}(a_{21}a_{33} - a_{23}a_{31}) \\ &\quad + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \end{aligned}$$

When D = 0, Cramer’s rule would divide by zero, and D = 0 alone doesn’t tell you which of the other two cases you have: the no-solution and the infinitely-many examples above both have D = 0. The ranks decide, which is why the solver always row-reduces as well.

Checking the answer by substitution

Put the values back into every original equation; each left side must come out exactly equal to its right side. The results do this for every equation, with the products written out, such as 1 × 3 + 2 × 4 + 1 × 5 = 16 for the sugar equation above. Because the arithmetic is exact, a check that holds really holds. Rounded decimals such as 6.285714 miss by a little, so substitute the fractions when you check by hand.

Reading the result

• The headline gives the solution, or “Infinitely many” with the general solution, or “None” with the contradiction that rules it out.
• The tiles give the determinant, the ranks of A and [A | b] that decide the case, and a check count, the free unknowns or the contradiction.
• The tables hold the solution as fractions and decimals, the check by substitution, the general solution at t = 0 and t = 1, or the combination of equations that reads 0 = c. Each downloads as CSV; fractions are written with a fraction slash (44⁄7) so that a spreadsheet keeps them as text instead of reading them as dates.
• The graph (2 × 2 only) draws both equations as lines, with the crossing point, or shows them parallel or on top of each other, plus each line’s slope-intercept form.
• How this was solved lists the reasoning, and the tabs underneath show every row operation with its matrix, and Cramer’s rule. Save for comparison keeps up to three systems side by side.

Assumptions and limitations

• Linear equations only: every unknown to the first power, with no products such as xy and no unknown in a denominator.
• Square systems of 2 or 3 equations in as many unknowns.
• Each number may have up to 15 digits above and below the fraction bar. Within that, everything is exact; decimals are rounded for display only, to about 7 significant digits, and marked ≈.
• Row reduction uses the hand method, pivoting on the first nonzero entry from the top. With exact fractions the order of the row operations can’t add rounding error, and every valid order leads to the same answer.

Common mistakes

• Leaving a missing term blank instead of typing 0, or typing a coefficient in the wrong column when the terms are in another order: for 3y + 2x = 7, the x box gets 2 and the y box 3.
• Using the grid before rearranging. 2x = 7 − 3y must become 2x + 3y = 7; a term changes sign when it moves across the = sign.
• Reading D = 0 as “no solution”. It means “not exactly one solution”; the system may have infinitely many.
• Rounding partway through. Writing 1/3 as 0.33 in a middle step can turn an exact answer into a nearly right one, or hide a contradiction.
• Checking only one equation. A solution has to satisfy every equation; a value that fits one line may miss the other.