Equation Solver Linear • Quadratic • Simultaneous — Step-by-Step Working Free
Solve any algebraic equation with full step-by-step working shown free — no paywall, no account. Linear equations: enter any form like 2x + 5 = 11 or 3x - 7 = 2x + 4 — the solver rearranges and solves with verification. Quadratic equations: enter coefficients or the full equation like x² - 5x + 6 = 0 — solved by the quadratic formula with discriminant analysis, completing the square, and factor form where applicable. Simultaneous equations: 2×2 systems solved by Cramer’s Rule and elimination method; 3×3 systems solved by Gaussian elimination with back-substitution. All answers verified by substituting back into the original equation. All calculations run in your browser — nothing sent to any server.
Solve Any Equation with Step-by-Step Working
Choose the equation type, enter your values, and get the full solution with every step explained.
2x + 5 = 11, 3x - 7 = 2x + 4, -4x = 12
Equation types quick reference
Rate this tool
How to Solve a Linear Equation Step by Step
A linear equation is any equation where the highest power of the variable is 1. The standard form is ax + b = c, where a, b, and c are numbers and x is the unknown. Linear equations always have exactly one solution (unless they are contradictions with no solution or identities with infinite solutions).
How to solve ax + b = c
Step 2: Divide both sides by a: x = (c - b) / a
Example: 2x + 5 = 11
Step 1: 2x = 11 - 5 = 6
Step 2: x = 6 / 2 = 3
How to solve linear equations with x on both sides
When x appears on both sides (e.g. 3x + 7 = 2x - 4), collect all x terms on one side by subtracting the smaller x term from both sides, then solve normally.
3x - 2x = -4 - 7
x = -11
Verify: 3(-11) + 7 = -26, 2(-11) - 4 = -26 ✓
Special cases: no solution and infinite solutions
| Case | Example | After rearranging | Meaning |
|---|---|---|---|
| One solution | 2x + 3 = 7 | x = 2 | Lines intersect at one point |
| No solution (contradiction) | 2x + 3 = 2x + 7 | 0 = 4 (impossible) | Parallel lines — never intersect |
| Infinite solutions (identity) | 2x + 4 = 2(x + 2) | 0 = 0 (always true) | Same line — overlap completely |
Quadratic Equation Solver — Formula, Discriminant & Completing the Square
A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0. It always has exactly two roots (counting multiplicity), though they may be real or complex. The graph of y = ax² + bx + c is a parabola. The x-intercepts are the roots of the equation.
The quadratic formula — how to use it
x = (-b ± √(b² - 4ac)) / (2a)
x₁ = (-b + √(b² - 4ac)) / (2a)
x₂ = (-b - √(b² - 4ac)) / (2a)
The discriminant Δ = b² - 4ac — what it tells you
Calculate the discriminant before applying the full formula — it tells you immediately what type of roots the equation has.
| Discriminant (Δ) | Number of real roots | Nature | Graph |
|---|---|---|---|
| Δ > 0 | Two distinct real roots | x₁ ≠ x₂, both real | Parabola crosses x-axis twice |
| Δ = 0 | One repeated root | x = -b/2a (double root) | Parabola touches x-axis (tangent) |
| Δ < 0 | No real roots | Two complex conjugate roots | Parabola does not cross x-axis |
Quadratic formula step by step — worked example
a = 1, b = -5, c = 6
Step 1: Δ = b² - 4ac = (-5)² - 4(1)(6) = 25 - 24 = 1
Step 2: Δ > 0, so two real roots
Step 3: x = (5 ± √1) / 2 = (5 ± 1) / 2
Step 4: x₁ = (5 + 1) / 2 = 3, x₂ = (5 - 1) / 2 = 2
Verify: (3)² - 5(3) + 6 = 9 - 15 + 6 = 0 ✓
Factor form: (x - 2)(x - 3) = 0
Completing the square — method and when to use it
Completing the square converts ax² + bx + c = 0 into the form (x + h)² = k, which makes the vertex of the parabola immediately visible at (-h, -k). It is especially useful when the quadratic formula gives messy numbers.
x² - 4x = 5
x² - 4x + 4 = 5 + 4 (add (b/2)² = 4 to both sides)
(x - 2)² = 9
x - 2 = ±3
x = 5 or x = -1
Vertex: (2, -9)
Factoring quadratics — when it works
If a = 1 and the discriminant is a perfect square, the quadratic factors neatly. Find two numbers that multiply to c and add to b. For x² + bx + c: look for r and s where r + s = b and r × s = c, giving (x + r)(x + s) = 0.
| Equation | a, b, c | Δ | Solution |
|---|---|---|---|
| x² - 5x + 6 = 0 | 1, -5, 6 | 1 | x = 2 or x = 3 |
| x² + 3x - 10 = 0 | 1, 3, -10 | 49 | x = 2 or x = -5 |
| x² - 4x + 4 = 0 | 1, -4, 4 | 0 | x = 2 (repeated) |
| x² + x + 1 = 0 | 1, 1, 1 | -3 | No real roots |
| 2x² - 3x - 2 = 0 | 2, -3, -2 | 25 | x = 2 or x = -1/2 |
| 3x² + 5x - 2 = 0 | 3, 5, -2 | 49 | x = 1/3 or x = -2 |
Simultaneous Equations Solver — Substitution, Elimination & Cramer’s Rule
Simultaneous equations are a set of equations with multiple unknowns that must all be satisfied at the same time. Two equations in two unknowns represent two straight lines in a plane. The solution is the intersection point of those lines.
Method 1: Substitution — how to solve simultaneous equations
Step 1: From eq.2: x = 1 + y
Step 2: Substitute into eq.1: 2(1 + y) + 3y = 12
Step 3: 2 + 2y + 3y = 12 ⟹ 5y = 10 ⟹ y = 2
Step 4: x = 1 + 2 = 3
Solution: x = 3, y = 2
Verify eq.1: 2(3) + 3(2) = 12 ✓ Verify eq.2: 3 - 2 = 1 ✓
Method 2: Elimination — step by step
Step 1: Multiply eq.1 by 2: 6x + 4y = 32
Step 2: Subtract eq.2: (6x + 4y) - (x + 4y) = 32 - 14
Step 3: 5x = 18 ⟹ x = 18/5 = 3.6
Step 4: Back-substitute: 3.6 + 4y = 14 ⟹ y = 10.4/4 = 2.6
Solution: x = 3.6, y = 2.6
Method 3: Cramer’s Rule — using determinants
D = a₁b₂ - a₂b₁ (main determinant)
Dₓ = c₁b₂ - c₂b₁ (x determinant)
Dᵧ = a₁c₂ - a₂c₁ (y determinant)
x = Dₓ/D, y = Dᵧ/D (if D ≠ 0)
Special cases in simultaneous equations
| Case | Determinant D | Dx, Dy | Geometry | Result |
|---|---|---|---|---|
| Unique solution | D ≠ 0 | Any | Lines intersect at one point | x = Dx/D, y = Dy/D |
| No solution | D = 0 | Dx or Dy ≠ 0 | Parallel lines | Inconsistent system |
| Infinite solutions | D = 0 | Dx = Dy = 0 | Coincident lines | Dependent system |
How to solve 3 simultaneous equations (3 unknowns)
For three equations in three unknowns, use Gaussian elimination. Set up the augmented coefficient matrix, then use row operations to reduce it to upper triangular form (zeros below the main diagonal), then apply back-substitution. The 3×3 solver above shows every row operation.
Solution: x = 1, y = 2, z = 3
Verify eq.1: 1 + 2 + 3 = 6 ✓
Verify eq.2: 2 + 2 - 3 = 1 ✘... (use Load example above to see full steps)
LazyTools vs Wolfram Alpha vs Other Equation Solvers
| Feature | LazyTools | Wolfram Alpha | Symbolab | Mathway |
|---|---|---|---|---|
| Step-by-step working | ✅ Always free | ❌ Requires Pro ($7/mo) | ❌ Premium required | ❌ Paid subscription |
| Linear equation solver | ✅ Any form | ✅ Yes | ✅ Yes | ✅ Yes |
| Quadratic formula + discriminant | ✅ Full analysis | ✅ Yes | ✅ Yes | ✅ Yes |
| Simultaneous 2x2 | ✅ Cramer + Elimination | ✅ Yes | ✅ Yes | ✅ Yes |
| Simultaneous 3x3 | ✅ Gaussian elimination | ✅ Yes | ✅ Yes | ✅ Yes |
| Solution verified by substitution | ✅ Every answer | ❌ No | ❌ No | ❌ No |
| Ad-free clean interface | ✅ Yes | ❌ Heavy ads | ❌ Heavy ads | ❌ Paywalled |
| No account required | ✅ Yes | ✅ Yes | ⚠ Required for steps | ❌ Account required |
Equation Solver FAQ
For ax² + bx + c = 0: x = (-b ± √(b² - 4ac)) / (2a). The discriminant Δ = b² - 4ac determines the roots: Δ > 0 = two real roots, Δ = 0 = one repeated root, Δ < 0 = no real roots.
For 2x + 5 = 11: Step 1: Subtract 5 from both sides: 2x = 6. Step 2: Divide by 2: x = 3. Verify: 2(3) + 5 = 11 ✓. General rule: isolate x by applying the same operation to both sides.
Three methods: (1) Substitution — solve one equation for one variable, substitute into the other. (2) Elimination — multiply equations to match coefficients, add or subtract to eliminate one variable. (3) Cramer’s Rule — compute determinants. Use the solver above for step-by-step working.
The discriminant Δ = b² - 4ac. If Δ > 0: two distinct real roots. If Δ = 0: one repeated root. If Δ < 0: no real roots (two complex roots). It tells you the nature of solutions before you solve the equation.
Method 1 (Factoring): (x - 2)(x - 3) = 0, so x = 2 or x = 3. Method 2 (Quadratic formula): a=1, b=-5, c=6. Δ = 25 - 24 = 1. x = (5 ± 1)/2. x₁ = 3, x₂ = 2.
2x + 5 = 11. Subtract 5: 2x = 6. Divide by 2: x = 3. Verify: 2(3) + 5 = 11 ✓. Click “Linear” tab above and enter “2x + 5 = 11” for the full step-by-step.
For x² - 4x - 5 = 0: Move constant: x² - 4x = 5. Add (b/2)² = 4: x² - 4x + 4 = 9. Write as square: (x-2)² = 9. Square root: x-2 = ±3. Solutions: x = 5 or x = -1.
For a₁x + b₁y = c₁ and a₂x + b₂y = c₂: D = a₁b₂ - a₂b₁. Dx = c₁b₂ - c₂b₁. Dy = a₁c₂ - a₂c₁. x = Dx/D, y = Dy/D. If D = 0: no unique solution.
Enter any linear (2x + 5 = 11), quadratic (x²-5x+6=0), or simultaneous equation (2x+3y=12, x-y=1) in the tool above. Full step-by-step working is shown free — no account, no subscription, no paywall.
Use Gaussian elimination: write the augmented matrix, use row operations to get upper triangular form, then back-substitute. The 3×3 solver above handles this automatically, showing every row operation and substitution step.
The roots are the x-values where ax²+bx+c = 0. Found by x = (-b±√Δ)/2a. Sum of roots = -b/a. Product of roots = c/a. Nature depends on Δ: two roots (Δ>0), one (Δ=0), none real (Δ<0).
Example: 3x + 7 = 2x - 4. Subtract 2x: x + 7 = -4. Subtract 7: x = -11. Verify: 3(-11)+7=-26 and 2(-11)-4=-26 ✓. The linear solver above handles any arrangement automatically.
Linear: highest power of x is 1 (e.g. 2x+5=11). Straight line graph. Always one solution. Quadratic: highest power is 2 (e.g. x²-5x+6=0). Parabola graph. Can have 0, 1, or 2 real solutions. Needs the quadratic formula or factoring.
Substitute the solution back into the original equation. For x²-5x+6=0 with x=2: (2)²-5(2)+6 = 4-10+6 = 0 ✓. For simultaneous equations, check BOTH equations. This verification is shown automatically in the solver above.
Write coefficients as an augmented matrix. Use row operations to create zeros below each pivot (upper triangular form). Then back-substitute: solve z from row 3, substitute into row 2 for y, then substitute both into row 1 for x.
For ax²+bx+c=0: Step 1: find a, b, c. Step 2: Δ = b² - 4ac. Step 3: if Δ<0 no real roots. Step 4: x = (-b±√Δ)/(2a). Step 5: x₁ = (-b+√Δ)/(2a), x₂ = (-b-√Δ)/(2a). Example: x²+3x-10=0. Δ=49. x=(- 3±7)/2. x=2 or x=-5.