Linear Equation Solver

Written byOleh Soloninupdated 5 October 2026

a (coefficient of x)

b (constant on the left)

c (right-hand side)

Result

x5
Working3x = 20 − 5 = 15, so x = 15 ÷ 3
Check3 × 5 + 5 = 20
As a fraction15/3

The answer and how it was reached

Solves ax + b = c, shows the working step by step and substitutes the answer back as a check.

A linear equation takes two moves: shift the constant to the right, then divide by the coefficient. Both steps are printed, because for schoolwork the write-up matters as much as the answer.

The degenerate cases are handled properly too. When the coefficient is zero the equation either has no solution or holds for every x — and that is a correct answer, not an error.

How to use

Enter the coefficient of x, the constant term and the right-hand side.
Read the root, the working and the check by substitution.
Use the substitution check: it catches the sign slip that causes most errors here.

Two steps and a check

A linear equation ax + b = c takes one move and one division: 3x + 5 = 20 becomes 3x = 15, so x = 5. Substituting the answer back takes a second and catches the sign slip that causes most errors here.

The degenerate cases are worth recognising. With a = 0 and b = c any x works; with a = 0 and b ≠ c none does. The calculator names both instead of dividing by zero.

Worked out with this calculator.
Equation	Root	Check
3x + 5 = 20	x = 5	3 × 5 + 5 = 20
2x − 7 = 1	x = 4	2 × 4 − 7 = 1
0x + 4 = 4	any x	an identity
0x + 4 = 9	no root	a contradiction

What this calculator shows

The value of x as a decimal and as a fraction
The working with your numbers substituted in
A check that puts the answer back into the original equation

What to keep in mind

The equation must be in the form ax + b = c; collect terms in x on the left first.
With a = 0 and b = c any number works; with a = 0 and b ≠ c nothing does — both are reported explicitly.

FAQs

What if x appears on both sides?

Collect the x terms on the left and the numbers on the right. 5x + 3 = 2x + 12 becomes 3x + 3 = 12, so a = 3, b = 3, c = 12.

Why can a = 0 give infinitely many solutions?

Because x disappears from the equation. If what remains is true, every number works; if false, none does.

How do I handle brackets?

Expand and collect like terms first, then rearrange into ax + b = c. This solver works on the tidied form.

How is this different from a quadratic?

A linear equation has x only to the first power and, when a ≠ 0, exactly one root; when a = 0 there is either no solution or infinitely many. A quadratic has x² and can have two roots, one or none — there is a separate solver for that.

How do I solve 3x + 5 = 20?

Move the five across to get 3x = 15, then divide by three for x = 5. Substituting back confirms it: 3 × 5 + 5 = 20.

What if the coefficient of x is zero?

Then the equation is either always true or never true. 0x + 4 = 4 holds for any x, while 0x + 4 = 9 holds for none.

Worked example

3x + 5 = 20

Input: a=3, b=5, c=20

Output: x = 5

Note: Working: 3x = 20 − 5 = 15, so x = 15 ÷ 3. Check: 3 × 5 + 5 = 20.

Useful links

All tools in Math & Statistics

Linear Equation Solver

Result

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