Calculator

Equation Solver with Steps

Solve linear, quadratic, rational, and radical equations with the operation applied to both sides named on every line, and extraneous roots tested rather than quietly kept.

Solve a equation problem now

First solution free

How to use it

Paste the equation exactly as written, including any fractions or radicals; there is no need to simplify it first. If more than one letter appears, say which one you are solving for. If your answer disagrees with the one shown, the labelled lines let you find the first step where the two diverge instead of re-solving from scratch. Word problems that need translating into an equation first are better sent to the word problem solver.

The method behind it

An equation is a claim that two expressions have the same value. Solving it means finding every value of the unknown that makes the claim true, and the only tool is doing identical things to both sides so the claim stays true.

A = B ⟹ A + c = B + c, A = B ⟹ c A = c B (c \neq = 0)

Those two moves are reversible, so they never gain or lose solutions. Undoing operations in the reverse of the order you would use to evaluate the expression is what isolates the variable — outermost operation first. If the priority order is not automatic for you, order of operations beyond PEMDAS is the prerequisite.

Once the equation is nonlinear, isolation stops working and structure takes over. Collect everything on one side against zero, factor, and use the fact that a product is zero only when a factor is.

A B = 0 ⟺ A = 0 or B = 0

Two operations break the guarantee that you neither gain nor lose solutions, and they are the source of nearly every wrong answer in this topic. Squaring is not reversible: $A = B$ implies $A^{2} = B^{2}$ , but the converse is false, because $A = - B$ squares to the same thing.

A = B ⟹ A^{2} = B^{2}, A^{2} = B^{2} \neq ⟹ A = B

Multiplying by an expression containing the variable is the other one. It clears denominators, which is exactly what you want, but if that expression can be zero, you have multiplied by zero for that value and invented a solution. Both operations are still worth using. They simply oblige you to substitute every candidate back into the original equation before writing it down as an answer.

Note down the excluded values before you clear denominators, not after. For a rational equation, any value that makes a denominator zero is outside the domain permanently — no later step can rescue it. When the equation reduces to a quadratic, the quadratic formula calculator takes it the rest of the way.

Worked examples

Worked example

\frac{2 x - 1}{3} - \frac{x + 2}{4} = 1

1
Clear both fractions at once by multiplying every term by the least common denominator, 12. The denominators are constants, so nothing is excluded.
$12 \cdot \frac{2 x - 1}{3} - 12 \cdot \frac{x + 2}{4} = 12 \cdot 1$
2
Each multiplication cancels one denominator, leaving brackets that must stay intact until they are expanded.
$4 (2 x - 1) - 3 (x + 2) = 12$
3
Expand, distributing the −3 across both terms of the second bracket.
$8 x - 4 - 3 x - 6 = 12$
4
Collect like terms on the left.
$5 x - 10 = 12$
5
Add 10, then divide by 5. Check: the two fractions become 13/5 and 8/5, which differ by 1.
$5 x = 22 \Rightarrow x = \frac{22}{5}$

Answer

x = \frac{22}{5}

Worked example

\frac{x}{x - 3} = \frac{3}{x - 3} + 2

1
Before anything else, note the excluded value. The denominator vanishes at x = 3, so 3 can never be a solution no matter what the algebra produces.
$x - 3 \neq = 0 \Rightarrow x \neq = 3$
2
Multiply every term by (x − 3) to clear the fractions.
$x = 3 + 2 (x - 3)$
3
Expand the right-hand side.
$x = 3 + 2 x - 6 = 2 x - 3$
4
Subtract 2x from both sides and divide by −1.
$- x = - 3 \Rightarrow x = 3$
5
The only candidate is the excluded value, so it is extraneous. Substituting 3 into the original equation divides by zero on both sides.
$x = 3 is excluded$

Answer

No solution

Work one through yourself

Radical equations are where the two dangerous operations meet: you have to square, and squaring can manufacture roots. Try each step before you reveal it.

Solve the square root of (x + 7) equals x − 5

0 of 7 steps

Try it yourself first — you'll remember it longer.

Common mistakes

Dividing both sides by the variable

From $x^{2} = 3 x$ , dividing by $x$ gives $x = 3$ and silently destroys the root $x = 0$ . Division is only safe by something known to be non-zero. Collect to $x^{2} - 3 x = 0$ , factor to $x (x - 3) = 0$ , and keep both.

Skipping the check

After squaring or clearing denominators, an unchecked candidate is not an answer. Substituting into the original equation takes one line and is the difference between full marks and a solution set containing a value that does not solve anything.

Distributing a minus sign to only the first term

$- 3 (x + 2)$ is $- 3 x - 6$ , not $- 3 x + 6$ . The sign belongs to the whole bracket. Writing the multiplier as $(- 3)$ in brackets makes the second product harder to get wrong.

Practice

Answers are checked here — nothing is sent anywhere.

1
$Solve 5 (x - 2) = 3 x + 4$
2
$Solve \frac{x}{4} + \frac{x}{6} = 5$
3
$Solve 2 x + 3 = x$

Frequently asked questions

Which kinds of equation does it handle?

Linear, quadratic, polynomial, rational, radical, absolute value, exponential, and logarithmic, plus literal equations where you rearrange a formula for one of its letters.

Why did it reject one of the values it found?

Squaring both sides and multiplying by an expression containing the variable can both create values that satisfy the transformed equation but not the original. Those are extraneous, and the check that eliminates them is part of the solution, not an optional extra.

Can it rearrange a formula rather than find a number?

Yes. Ask it to solve a formula for a named variable and the working treats every other letter as a constant, showing which operation moves each one across.

Does it show what happens to both sides?

Every line states the operation applied to both sides. That is the layout most markers expect, and it is what makes a mistake findable rather than just visible in the final answer.

What does it cost?

Your first solution is free without an account. Free accounts get three solutions a day, and Gauth Plus lifts that for $11.99 a month with a three-day free trial.

Keep going

Algebra solverEquations, systems, and inequalities with every step shown in one place.Order of operations beyond PEMDASRead an expression the way a grader does, so you undo it in the right order.The quadratic formula: when to use itWhat to do once an equation collapses into a quadratic.Inequality calculatorThe same balance rules, plus the one that reverses the sign.