Equation Solver — Solve for x (Nonlinear, Free)
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An equation solver finds the value of x that makes both sides equal. For 5x − 4 = 2x + 11, subtract 2x from both sides to get 3x − 4 = 11, add 4 to get 3x = 15, and divide by 3: x = 5. Check: 5(5) − 4 = 21 and 2(5) + 11 = 21.
Show steps (3)
- Subtract 2x from both sides to get the x terms on the left: 3x − 4 = 11.
- Add 4 to both sides: 3x = 15.
- Divide both sides by 3: x = 5.
Have a problem that is not an equation, like a derivative, a trig value or a fraction sum? Our free Math Solver handles algebra, trigonometry and calculus with every step shown.
How to use the equation solver
- Type the equation in the box, with one = sign. Use ^ for powers (x^2), / for fractions, |x − 3| or abs(x − 3) for absolute value, and sqrt(x + 1) or √ for square roots.
- The solver detects the type for you: linear, quadratic, polynomial, rational, absolute value, radical, exponential or logarithmic.
- Read the answer, the check in the original equation and the graph of both sides, then open Show steps for the full working.
- For equations solved numerically (such as cos x = x), set the search range; only solutions inside it are listed. Use the System tab for two equations with two unknowns.
Which equations the solver handles
| Type | Example | Method | Answer |
|---|---|---|---|
| Linear | 4(x − 2) = 2x + 6 | Simplify, collect x terms, divide | x = 7 |
| Quadratic | x² = x + 6 | Set to 0, factor or use the formula | x = −2 or 3 |
| Polynomial | x³ − 6x² + 11x − 6 = 0 | Rational Root Theorem, synthetic division | x = 1, 2 or 3 |
| Rational | 1/x + 1/(x + 2) = 3/4 | Exclude zeros of denominators, clear fractions | x = −4/3 or 2 |
| Absolute value | |2x − 3| = 7 | Split into cases | x = −2 or 5 |
| Radical | √(2x + 3) = x | Isolate the root, square, check | x = 3 |
| Exponential | 2ˣ = 10 | Logarithms | x = log₂ 10 ≈ 3.3219 |
| Other | cos x = x | Numerical (sign change and bisection) | x ≈ 0.739085 |
Solving linear equations, including variables on both sides
A linear equation has x only to the first power. Undo operations in reverse order and do the same thing to both sides. For 3x + 4 = 19, subtract 4 (3x = 15) and divide by 3: x = 5.
When x appears on both sides, first move the x terms to one side. For 4(x − 2) = 2x + 6, distribute to get 4x − 8 = 2x + 6, subtract 2x (2x − 8 = 6), add 8 (2x = 14) and divide: x = 7. If the x terms cancel and a true statement is left, every number is a solution; if a false statement is left, there is no solution.
Quadratic and polynomial equations
Move everything to one side so the equation equals 0, then factor. x² − 5x + 6 = 0 factors as (x − 2)(x − 3) = 0, so x = 2 or x = 3. The equation x² = x + 6 becomes x² − x − 6 = 0, or (x + 2)(x − 3) = 0, so x = −2 or x = 3.
When a quadratic does not factor, use the quadratic formula x = (−b ± √(b² − 4ac))/(2a). For 2x² + 3x − 4 = 0 the discriminant is 9 + 32 = 41, so x = (−3 ± √41)/4, about 0.85 and −2.35. For cubic and higher equations the solver tests rational candidates, divides out each root it finds, and finds any remaining real roots numerically.
Rational equations and excluded values
A rational equation has x in a denominator. Before solving, list the excluded values: any x that makes a denominator 0 can never be a solution.
For 1/x + 1/(x + 2) = 3/4, the excluded values are 0 and −2. Multiplying both sides by 4x(x + 2) gives 4(x + 2) + 4x = 3x(x + 2), which simplifies to 3x² − 2x − 8 = 0, or (3x + 4)(x − 2) = 0. Both x = −4/3 and x = 2 are allowed, so both are solutions.
Clearing fractions can produce an extraneous solution. For x/(x − 3) = 3/(x − 3) + 2, multiplying by x − 3 gives x = 3 + 2(x − 3), so x = 3. But 3 makes the denominator 0, so it is rejected and the equation has no solution.
Absolute value equations
|A| = c means A = c or A = −c, as long as c is not negative. For |2x − 3| = 7: 2x − 3 = 7 gives x = 5, and 2x − 3 = −7 gives x = −2. An equation such as |x + 1| = −3 has no solution, because an absolute value is a distance and is never negative.
When x is also outside the bars, check each answer against its case. For |x − 1| = 2x + 4, the case x − 1 ≥ 0 gives x = −5, which breaks that case’s condition, while the case x − 1 < 0 gives x = −1, which fits. The only solution is x = −1.
Radical equations and extraneous roots
Isolate the square root, square both sides, solve, and then check every answer in the original equation. For √(2x + 3) = x, squaring gives 2x + 3 = x², so x² − 2x − 3 = 0 and x = 3 or x = −1. Checking: √9 = 3 works, but at x = −1 the right side is −1 while √1 = 1. Squaring created x = −1, so it is extraneous. With two square roots, as in √(x + 5) − √x = 1, isolate one root, square, isolate the other and square again; the answer is x = 4.
Exponential, logarithmic and other equations
For an exponential equation, take logarithms. 2ˣ = 10 gives x = log₂ 10 = ln 10 / ln 2 ≈ 3.3219. A logarithm is undone with a power: ln x = 1 gives x = e, and log(2x) = 2 gives 2x = 100, so x = 50.
Equations such as cos x = x or eˣ = x + 2 cannot be rearranged to isolate x. The solver writes f(x) = left side − right side, scans the search range for places where f changes sign, and narrows each one down by bisection. This is the graphical method done precisely: the solutions are where the graphs of the two sides cross.
Common mistakes
- Doing an operation to only one side, or to only one term on a side.
- Dividing both sides by x, which loses the solution x = 0. Factor instead.
- Keeping a value that makes a denominator 0 in a rational equation.
- Skipping the check after squaring both sides, so extraneous roots stay in the answer.
- Splitting |x − 4| = 3 into x − 4 = 3 and −x − 4 = 3. The second case is −(x − 4) = 3: the minus sign applies to the whole expression inside the bars.
Equation Solving practice quiz: 10 questions
Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
Find two numbers that multiply to 12 and add to −7: −3 and −4. So (x − 3)(x − 4) = 0.
A denominator cannot be 0: x − 2 = 0 at x = 2 and x + 1 = 0 at x = −1.
Multiply both sides by x − 1: 4 = 2(x − 1), so 2x = 6 and x = 3. It is not the excluded value 1.
Multiplying by x − 4 gives x + 2 = 6, so x = 4. But 4 makes the denominator 0, so it is extraneous.
Squaring gives x + 6 = x², so x = 3 or x = −2. At x = −2 the right side is negative while √4 = 2, so −2 is extraneous.
Quiz complete
Frequently asked questions
What is equation solving?
Solving an equation means finding every value of the unknown that makes the two sides equal. Those values are the solutions, or roots. For 3x + 4 = 19 the only solution is x = 5.
How do you solve an equation step by step?
Simplify each side, move the x terms to one side and the numbers to the other, then divide by the coefficient of x. For harder types, follow the matching method above (factor, clear fractions, split cases or square), and always check the answer in the original equation.
How do you solve equations with variables on both sides?
Subtract one of the x terms from both sides so x appears on one side only. For 5x − 4 = 2x + 11, subtracting 2x gives 3x − 4 = 11, so 3x = 15 and x = 5. Both sides then equal 21.
How do you solve an equation graphically?
Graph y = left side and y = right side; the x-coordinates where the graphs cross are the solutions. For x² = x + 6 the parabola and the line cross at x = −2 and x = 3. The solver draws both sides for you.
Where is the equation solver on the TI-84?
Press MATH and scroll to Solver (option 0 or B, depending on the model and operating system). Enter the equation (older versions use the form 0 = left side − right side), type a guess for X, and choose SOLVE. It returns one root near your guess, so try other guesses to find the rest.
Is solving equations algebra?
Yes. Solving equations is the core skill of algebra: using inverse operations and properties of equality to find an unknown. Exponential and trigonometric equations extend the same idea in algebra 2 and precalculus.
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