Equation, inequality and system solver with steps

This tool solves an equation, an inequality or a system of two equations and lays out the working the way it is written by hand: moving terms, the discriminant, Vieta's formulas, domain restrictions, the sign chart, substitution and elimination. You can verify the answer right on the page — each root is substituted back, and each answer interval is tested with a point.

How to solve an equation

1

Choose the problem type and type it in — for example, x^2 - 5x + 6 = 0 or 2x + 3y = 13; x - y = 1. A photo works too.

2

The answer and the substitution check appear right under the field.

3

Next to it is the working: switch the method, copy the solution, save it as PDF or open the graph.

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Enter a problem and follow every step of the solution

Equation

Examples:
Solve from a photo

A photo gives a draft: what was read goes into the field above and has to be checked before solving. A printed line reads noticeably better than handwriting.

Drop image here
JPG, PNG, HEIC, WebP ·
Practice and worksheets

A problem of the same kind with different numbers: solve it on paper and check your answer.

Class worksheet

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Up to 30 problems of this kind per version, several versions and an answer key — a printable PDF.

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Features

A quadratic is solved with the method you pick: the discriminant, Vieta's formulas, k = b/2, completing the square, or as an incomplete quadratic.
Inequalities are solved with a sign chart: zeros, excluded points of the denominator, signs on each interval and an answer with round and square brackets.
A system of two equations is solved by substitution, elimination or Cramer's rule; a nonlinear one (xy = 6, x² + y² = 25) by substitution from the linear equation, with both curves on a graph.
Exact answers: √2, (1 + √5)/2 and fractions stay as they are, with the decimal shown beside them.
Every root is substituted into the original equation, and an inequality answer is tested with a point from each interval.
Practice mode gives a similar problem with new numbers and checks your answer; the solution copies as text, saves as PDF and travels as a link.

What it solves

  • Linear equations: 2x + 5 = 13, 3(x - 1) = x + 7
  • Quadratics, including incomplete ones: x^2 - 5x + 6 = 0, 3x^2 - 12x = 0, x^2 = 2
  • Biquadratics by substitution and higher degrees by synthetic division: x^4 - 5x^2 + 4 = 0, x^3 - 6x^2 + 11x - 6 = 0
  • Rational equations with domain restrictions: (x+1)/(x-1) = 2, 1/x + 1/(x+1) = 1
  • Inequalities, including rational and compound ones: x^2 - 5x + 6 < 0, (x-1)/(x+2) >= 0, -3 < 2x + 1 <= 5
  • Systems of two equations: 2x + 3y = 13; x - y = 1 and x + y = 5; xy = 6
  • Everything else — numerically, with the search interval stated: sin(x) = 0.5

What the input understands

Implicit multiplication (2x, 3(x+1), (x+1)(x-1)), powers with ^ or ², brackets, fractions, absolute value with vertical bars, decimal point or comma. Inequality signs: <, >, <=, >=, ≤, ≥, including compound inequalities. A system goes into two fields or one line separated by a semicolon, and xy written together is read as a product. Functions: sin, cos, tan, cot, ln, lg, log(base, number), sqrt, abs, exp; constants pi and e. Any letter can be the variable — 2y + 1 = 7 is solved for y.

How this differs from a root calculator

A calculator answers with a number, but homework needs a solution — and in the method taught in class. Here every step is labeled, and the method can be switched: discriminant or Vieta, substitution or elimination. The second difference is precision: where an ordinary calculator prints 1.414214, this keeps √2, and a fractional coefficient is written as x/2 rather than 1/2x. The third is honest limits: when there is no exact school technique, the tool does not invent a formula — it says plainly that the roots were found numerically and on which interval.

Frequently asked questions

Why is the answer written as √2 and not 1.41?

Because √2 is the exact value and 1.41 is already rounded. If that root is used in the next step of a problem, the error carries on and the answer drifts away from the textbook. The decimal is still shown next to it with ≈ — for a rough idea, not instead of the answer. For the same reason fractions stay fractions: 2/3, not 0.666666.

What are domain restrictions, and why was one of the roots dropped?

The domain is the set of values for which the equation makes sense at all. A denominator cannot be zero, so those values are excluded before solving. When the common denominator is cleared by multiplication, roots can appear that the original equation does not have: x²/(x−2) = 4/(x−2) formally gives ±2, but at x = 2 both fractions do not exist. That root is named extraneous and dropped rather than quietly listed in the answer.

Why does an inequality answer have one round bracket and one square bracket?

A square bracket means the endpoint belongs to the answer, a round one means it does not. Zeros of the expression belong only to a non-strict inequality (≤ or ≥). Points where the denominator is zero never belong: the expression does not exist there. So for (x − 1)/(x + 2) ≥ 0 the answer is (−∞, −2) ∪ [1, +∞), and writing [−2 would be a mistake. When the answer is a single point, it is written in curly braces: {1}.

Which systems of equations are solved, and how?

A system of two equations in two variables. A linear one can be worked three ways — substitution, elimination and Cramer's rule — by switching the method above the steps; all three give the same answer. A nonlinear one (x + y = 5, xy = 6 or x² + y² = 25, x − y = 1) is solved by substitution: one variable is expressed from the linear equation, and the rest is an ordinary equation with its own steps. If the system has no linear equation, the tool says so rather than guessing an answer.

Why does a cubic show synthetic division instead of a formula?

Cardano's formula exists, but it is not taught in school and that step cannot be copied into homework. The school technique is different: a root is guessed among the divisors of the constant term, and the polynomial is divided by (x − root), leaving a quadratic that is solved with the discriminant. That keeps the answer exact. If there is no rational root at all, the tool says so and the roots are then found numerically.

Where does the interval from −50 to 50 in an approximate answer come from?

Those are the limits of the numerical search, and they are stated on purpose. When no exact technique applies (sin x = 0.5, |x − 2| < 3, or a cubic without rational roots), roots and interval ends are found by scanning, and a scan always runs over a finite interval. Saying "three roots were found" without the interval would be untrue: there may be more beyond it. To look further, open the problem in the graph plotter, where the view can be moved and zoomed.

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🎓 Students 📚 Educators & Teachers
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