Algebra

Algebra Solver with Step-by-Step Answers

Type an equation, a system, or an inequality and get the full working — every operation applied to both sides, named as it happens.

Solve a algebra problem now

First solution free

Algebra fails in predictable places. A minus sign that only reaches the first term inside a bracket. A solution thrown away by dividing both sides by $x$ . A root that satisfies the rearranged equation but not the one you started with. Because the solver writes out each operation as it applies it, you can compare line for line against your own page and find where the two diverged, which is more useful than learning that the final number was wrong.

Write powers with ^, fractions with /, and roots as sqrt(x) if the symbol is awkward to type. You can also photograph a question straight from a textbook. If a line needs justifying, ask a follow-up on that line and the rest of the solution stays where it is.

What this solver handles

Linear equations with fractions, decimals, or brackets on both sides — the routine behind the step-by-step equation solver.
Quadratics by factoring, completing the square, or the formula, with the discriminant read first so you know how many real roots to expect.
Factoring: greatest common factor, difference of squares, trinomials, and four-term grouping, using the pattern-matching approach.
Systems of two or three linear equations by substitution or elimination, plus nonlinear systems where a line meets a curve.
Inequalities, including compound and absolute value, with the sign flip justified rather than just performed.
Rational equations, with excluded values listed before the algebra starts.
Radical equations, including ones needing two rounds of squaring.
Exponent and radical simplification, negative and fractional powers included.
Function notation: evaluating, composing, and inverting, and reading a solution off a graph.

Two things are worth knowing before you ask. It will not guess which method your class wants — if you spent this week completing the square, ask for that and the working follows it rather than reaching for the formula. And if your answer differs from the one shown, paste your own lines and ask which step diverges. That comparison finds the misunderstanding faster than reading a correct solution and trying to spot where yours stopped matching.

Three problems, worked

These are the three shapes most algebra homework reduces to: something linear hidden under fractions, something quadratic that factors if you look at the product $a c$ , and something rational that quietly excludes one of its own answers.

Worked example

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

1
The denominators are 3 and 2, so the lowest common denominator is 6. Multiply every term — including the lone 1 — by 6.
$6 \cdot \frac{2 x}{3} - 6 \cdot 1 = 6 \cdot \frac{x + 4}{2}$
2
Each fraction cancels down. On the right, 6 divided by 2 leaves 3 multiplying the whole bracket, so the bracket must stay.
$4 x - 6 = 3 (x + 4)$
3
Distribute the 3 across both terms.
$4 x - 6 = 3 x + 12$
4
Subtract 3x from both sides and add 6 to both sides.
$x = 18$
5
Check in the original equation, not the cleared one: the left side is 12 minus 1, and the right side is 22 over 2. Both give 11.
$\frac{2 ( 18 )}{3} - 1 = 11, \frac{18 + 4}{2} = 11$

Answer

x = 18

Worked example

6 x^{2} + 7 x - 20 = 0

1
The leading coefficient is not 1, so use the ac method. Multiply a by c to get the target product, and keep b as the target sum.
$a c = 6 \times (- 20) = - 120, b = 7$
2
Find two numbers multiplying to −120 and adding to 7. Working through the factor pairs of 120, the pair 15 and −8 works.
$15 \times (- 8) = - 120, 15 + (- 8) = 7$
3
Split the middle term using that pair, then group in twos.
$6 x^{2} + 15 x - 8 x - 20 = 3 x (2 x + 5) - 4 (2 x + 5)$
4
The bracket 2x + 5 is now a common factor. Pull it out.
$(3 x - 4) (2 x + 5) = 0$
5
A product is zero only if a factor is zero, so set each bracket to zero and solve.
$x = \frac{4}{3} or x = - \frac{5}{2}$
6
Confirm with the formula. The discriminant is 49 + 480 = 529, and 529 is 23 squared, which is why the integer factoring worked at all.
$x = \frac{- 7 \pm 23}{12} = \frac{16}{12} or \frac{- 30}{12}$

Answer

x = \frac{4}{3} or x = - \frac{5}{2}

Worked example

1 + \frac{5}{x - 3} = \frac{30}{x ^{2} - 9}

1
Factor the quadratic denominator first. That exposes the shared factor and tells you which values of x are forbidden.
$x^{2} - 9 = (x - 3) (x + 3) ⟹ x \neq = 3, x \neq = - 3$
2
Multiply every term by (x − 3)(x + 3). The first term gains the whole product; the second keeps only x + 3.
$(x - 3) (x + 3) + 5 (x + 3) = 30$
3
Expand both products on the left.
$x^{2} - 9 + 5 x + 15 = 30$
4
Collect terms and move everything to one side.
$x^{2} + 5 x - 24 = 0$
5
Factor: two numbers multiplying to −24 and adding to 5 are 8 and −3.
$(x + 8) (x - 3) = 0 ⟹ x = - 8 or x = 3$
6
Now use the exclusion list. x = 3 makes the original denominators zero, so it is extraneous — an artefact of multiplying through. Only x = −8 survives, and substituting it gives 1 − 5/11 on the left and 30/55 on the right, both 6/11.
$1 - \frac{5}{11} = \frac{6}{11}, \frac{30}{55} = \frac{6}{11}$

Answer

x = - 8

Each of those ended with a check, and in the third the check is the only thing separating a right answer from a wrong one. For quadratics there is something faster than substitution. The roots of $6 x^{2} + 7 x - 20$ must sum to $- b / a = - \frac{7}{6}$ and multiply to $c / a = - \frac{10}{3}$ . Adding $\frac{4}{3}$ and $- \frac{5}{2}$ gives $- \frac{7}{6}$ ; multiplying them gives $- \frac{10}{3}$ . One addition and one multiplication confirm both roots, which is quick enough to do every time.

The habit generalises. Before writing an answer down, ask what it should satisfy that you have not yet used — the original equation, the excluded values, the sign of the discriminant, or simply whether a length came out negative.

Choosing a method before you start

Most wasted time in algebra comes from committing to a method the equation does not suit — completing the square on a trinomial that factors in one line, or squaring both sides before the radical is alone. Two questions about the shape of the equation are usually enough to settle it.

Which method fits this equation?

What is the highest power of the unknown, and where does it sit?

For quadratics it also helps to see what the algebra is describing. Drag the coefficients below: the roots are where the curve crosses the x-axis, so a negative discriminant is the same statement as a parabola floating clear of the axis.

Move a, b, and c and watch the roots

y = 1x² + 0x + -2

Where students go wrong

The negative that only reaches the first term

$- (3 x - 4)$ is $- 3 x + 4$ , not $- 3 x - 4$ . Subtracting a bracket means subtracting everything inside it. The fix is mechanical: write the subtraction as adding $- 1$ times the bracket, then distribute as usual. This single error accounts for more lost marks than any other in early algebra.

Dividing by x and losing a root

From $x^{2} = 5 x$ , dividing both sides by $x$ gives $x = 5$ and silently deletes $x = 0$ . You are only allowed to divide by something you know is non-zero. Move everything to one side and factor instead: $x (x - 5) = 0$ keeps both roots.

Cancelling terms instead of factors

In $\frac{x + 3}{3}$ the 3 on top is a term, not a factor, so nothing cancels — the expression is not $x$ . Cancelling is division applied to the whole numerator, which is why it only works when the numerator is a product. Factor first; if a common factor appears in every term, it may go.

Forgetting the inequality flip

Multiplying or dividing an inequality by a negative reverses it: $- 2 x > 6$ becomes $x < - 3$ . Adding and subtracting never flip it. If you distrust the rule, test one number from your answer set in the original inequality — $x = - 4$ gives $8 > 6$ , so the direction is right.

One line here is wrong

One of these lines is wrong. Click it.

What these four share is a rule applied outside the conditions that make it true. Cancelling is valid for factors and not terms; dividing is valid for quantities you know are non-zero; the inequality rule depends on the sign of what you divided by. When marks come back, record the condition you broke rather than the question number. A short list of the mistakes that recur in marked work is worth rereading before an exam; a list of questions you got wrong is not.

Formulas worth knowing cold

Algebra reference

Tap any formula with a derivation to see where it comes from.

Quadratics

Factoring patterns

Exponents and radicals

Practice

Four to try

Answers are checked here — nothing is sent anywhere.

1
$Solve 5 x - 3 = 2 x + 9$
2
$Factor x^{2} - 7 x + 12$
3
$Solve x^{2} - 2 x - 15 = 0$
4
$Solve - 3 x + 2 \leq 11$

Frequently asked questions

Is the algebra solver free?

Your first solution is free and needs no account — type the equation and the working appears. After that, a free account gives you three solutions a day. Gauth Plus lifts the cap for $11.99 a month with a three-day free trial.

Does it show the steps or just the answer?

Every solution is a sequence of steps with the operation named: multiply both sides by the LCD, factor the left side, discard the extraneous root. The answer sits at the end of the working rather than in place of it.

Can I photograph a question from a textbook?

Yes. A photo of a printed or handwritten equation works as well as typing it out. One clear problem per photo gives a cleaner solution than a whole exercise page.

Will it solve systems and inequalities, not just single equations?

Yes — two and three variable systems, linear and quadratic inequalities, compound inequalities, and absolute value equations. If your class requires a particular method, ask for it by name and the solution follows that route.

What if one of the steps does not make sense?

Ask a follow-up on that step. The working stays on screen while you ask, and follow-up questions do not count against your daily quota.

Can I check work I have already done?

Paste your own lines and ask where they go wrong. That is usually more useful than a fresh solution, because it points at the line that broke instead of showing a different route to the same number.

Keep going

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