What does the substitution calculation do?

It starts from an original equation, isolates a variable as a constant plus a multiple of the other variable, and replaces that variable in the second equation. For the starting example it gets x = 3 - y, substitutes into x - y = 1, then solves -2y = -2 and substitutes y = 1 back to get x = 2.

What happens if substitution leaves 0 = 0?

The second equation supplies no new restriction, so one variable remains free and the first equation describes the solution family. If the remaining equation is 0 equal to a nonzero constant, there is no solution. If both original equations are 0 = 0, both variables are free.

Can this substitution view solve three equations?

This separate substitution calculation is for exactly two equations in two variables. The main calculator can solve 2 through 6 equations in 2 through 6 variables using Gaussian elimination and back-substitution.

System of Equations by Substitution

Solve two linear equations by isolating one variable and substituting it into the other. The page opens on x + y = 3 and x - y = 1, whose solution is x = 2, y = 1. Read the separate substitution working below the elimination check.

2 variables, 2 equationsEquation textEvery step shown

 x + y =  3 x - y =  1elimination         1 row operation, pivoting on the largest coefficientback-substitution   2 stepsx = 2        y = 1residual            0   0        unique solution

Enter as
Read from the text: x, y in 2 equations
Working

x: 2
y: 1
Case: Unique

Your equations

One equation per line, 2 to 6 of them, in 2 to 6 unknowns. Variables may sit on both sides, coefficients may be fractions or decimals, and x/2 means half of x.

Equations, one per line

The working

Forward elimination first, largest coefficient leading each column, then back-substitution from the bottom row up. Every multiplier is an exact fraction.

R2 -> R2 - (1)R1 clears x from row 2
```
[  1   1 |  3 ]
[  0  -2 | -2 ]
```

Back-substitution

From row 2: -2y = -2, so y = 1.
From row 1: x + y = 3 with y = 1, so x = 2.

The check

x + y = 3, residual 0
x - y = 1, residual 0

Each value is substituted back into the equations as you typed them: every residual is exactly zero, because the arithmetic is done in fractions rather than decimals.

Substitution for two equations

Isolate a variable, replace it in the other equation, then substitute back. These steps use the original equations and exact fractions.

Isolate x in equation 1: x = 3 - y.
Substitute into equation 2: (1)(3 - y) + (-1)y = 1.
Collect terms: (-2)y = -2.
y = 1.
Substitute back: x = 3 + (-1)(1) = 2.

Graph of the two-variable system

Approximate graph, axes from -5 to 5. 2 of 2 equation lines cross this view. Read the exact solution and consistency result above.

Line 1: x + y = 3
Line 2: x - y = 1

The three answers a linear system can have

Load any of them to see which one you are looking at and why.

x + y = 3, x - y = 1two lines that cross once
x = 2, y = 1
x + y = 1, 2x + 2y = 3same line, different constant
No solution: the equations contradict each other
x + y = 3, x + y = 3one equation repeated, rank below the variable count
x = 3 - y, with y free

Three answers, and the rank tells you which. Count the rows left with a leading coefficient after elimination: that is the rank. Rank equal to the number of unknowns gives one solution. A row that reduces to 0 = something other than zero is a contradiction, so there is no solution at all. Rank below the number of unknowns leaves free variables, and every value of them is a solution, which is why the answer comes back as a formula rather than a number. Telling those apart needs zero to mean zero, so every coefficient here is held as a fraction of two whole numbers and nothing is rounded on the way.

Common questions

What does the substitution calculation do?: It starts from an original equation, isolates a variable as a constant plus a multiple of the other variable, and replaces that variable in the second equation. For the starting example it gets x = 3 - y, substitutes into x - y = 1, then solves -2y = -2 and substitutes y = 1 back to get x = 2.
What happens if substitution leaves 0 = 0?: The second equation supplies no new restriction, so one variable remains free and the first equation describes the solution family. If the remaining equation is 0 equal to a nonzero constant, there is no solution. If both original equations are 0 = 0, both variables are free.
Can this substitution view solve three equations?: This separate substitution calculation is for exactly two equations in two variables. The main calculator can solve 2 through 6 equations in 2 through 6 variables using Gaussian elimination and back-substitution.

Gaussian elimination that pivots on the largest available coefficient for numerical stability, with every row operation printed so you can follow it. The solution is substituted back into the equations you typed and the residual is shown, so you can see the arithmetic close. A nonlinear term is refused by name rather than quietly linearised.