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Algebraic Equations: Definition, Types, Formulas & Solved Examples

TL;DR

An algebraic equation is a mathematical statement where two expressions are connected by an equal sign, containing variables, constants, and coefficients - like 3x+5=173x + 5 = 173x+5=17. This guide covers every type (linear through polynomial), the formulas you actually need, step-by-step solving methods, and the specific mistakes students make at each stage.

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Last updated on April 2, 2026 21 min read

What Is an Algebraic Equation?

An algebraic equation is a statement that two algebraic expressions are equal. That's it. Strip away the jargon and you get: something on the left side of an "=" sign has the same value as something on the right side.

P=QP = QP=Q

where PPP and QQQ are expressions built from variables, constants, and arithmetic operations.

Take 2x+3=112x + 3 = 112x+3=11. The expression 2x+32x + 32x+3 equals 111111. Your job is to find the value of xxx that keeps both sides balanced.

Here's the part most textbooks skip: an equation is fundamentally a question. When you write 2x+3=112x + 3 = 112x+3=11, you're really asking, "What number, when doubled and increased by 3, gives 11?" Thinking of equations as questions — not as things to mechanically "solve" — changes how students engage with algebra entirely.

Algebraic Equation vs Algebraic Expression - The Confusion That Costs Marks

About 4 out of 10 students we work with at Bhanzu mix up expressions and equations in their first algebra session. The difference is simple but critical:

An expression is like an incomplete sentence. An equation is the full statement. You can simplify 3x+73x + 73x+7, but you can't "solve" it — there's nothing to solve for until you set it equal to something.

Anatomy of an Algebraic Equation

Let's break down 5x2−3x+7=05x^2 - 3x + 7 = 05x2−3x+7=0 piece by piece:

• Variables: xxx — the unknowns you're solving for.

• Coefficients: 555 and −3-3−3 — the numbers multiplied by the variables.

• Constant: 777 — the standalone number with no variable attached.

• Degree: 222 — the highest exponent on any variable. This tells you the type of equation and how many solutions to expect.

• Operator: The === sign. Everything before it is the Left-Hand Side (LHS), everything after is the Right-Hand Side (RHS).

Quick mental check: If someone writes 5x+9>125x + 9 > 125x+9>12, is that an equation? No. The ">" makes it an inequality. Equations exclusively use the === sign.

Types of Algebraic Equations

The type of an algebraic equation depends on its degree — the highest power of the variable. The degree tells you two things: what the equation is called, and the maximum number of solutions it can have.

1. Linear Algebraic Equations (Degree 1)

A linear equation has the variable raised to the power of 1. No squares, no cubes — just the variable sitting there plainly.

General form:

ax+b=0(a≠0)ax + b = 0 \quad (a \neq 0)ax+b=0(a≠0)

Examples:

• 3x+5=203x + 5 = 203x+5=20

• 7y−14=07y - 14 = 07y−14=0

• 2x+3y=122x + 3y = 122x+3y=12 (linear equation in two variables)

Linear equations are the first real "algebra" students encounter, typically in Class 7 (NCERT Chapter 4) or Grade 7 (CCSS 7.EE.1). And they're where the most fundamental solving habit — doing the same thing to both sides — either gets built or doesn't.

What makes linear equations "linear"? If you plot every solution of a linear equation on a graph, you get a straight line. That's literally where the name comes from. The equation y=2x+1y = 2x + 1y=2x+1 draws a perfect line through the coordinate plane — no curves, no bends.

One variable vs two variables:

2. Quadratic Algebraic Equations (Degree 2)

The variable gets squared. That single change — going from xxx to x2x^2x2 — transforms everything about how the equation behaves.

General form:

ax2+bx+c=0(a≠0)ax^2 + bx + c = 0 \quad (a \neq 0)ax2+bx+c=0(a≠0)

Why a≠0a \neq 0a≠0? Because if a=0a = 0a=0, the x2x^2x2 term vanishes and you're back to a linear equation. The aaa is what makes it quadratic.

Examples:

• x2−5x+6=0x^2 - 5x + 6 = 0x2−5x+6=0

• 2x2+3x−2=02x^2 + 3x - 2 = 02x2+3x−2=0

• x2=49x^2 = 49x2=49 (yes, this counts — rewrite it as x2−49=0x^2 - 49 = 0x2−49=0)

Quadratic equations can have zero, one, or two real solutions. The discriminant Δ=b2−4ac\Delta = b^2 - 4acΔ=b2−4ac tells you which scenario you're in before you even solve:

The mistake students make most often: Writing x2=49x^2 = 49x2=49 and concluding x=7x = 7x=7. They forget the negative root. The correct answer is x=7x = 7x=7 or x=−7x = -7x=−7. In our experience, roughly half of students lose marks on this in their first quadratic test — not because they don't know the concept, but because they stop thinking too early.

3. Cubic Algebraic Equations (Degree 3)

General form:

ax3+bx2+cx+d=0(a≠0)ax^3 + bx^2 + cx + d = 0 \quad (a \neq 0)ax3+bx2+cx+d=0(a≠0)

Cubic equations can have up to three real roots. Unlike quadratics, a cubic always has at least one real root — the graph of a cubic function must cross the x-axis at least once.

Examples:

• x3−6x2+11x−6=0x^3 - 6x^2 + 11x - 6 = 0x3−6x2+11x−6=0

• 2x3+3x2−1=02x^3 + 3x^2 - 1 = 02x3+3x2−1=0

Students typically encounter cubics in Class 11 (NCERT) or in competition math. The factoring approach is usually the first method taught: check whether simple values like x=1,−1,2,−2x = 1, -1, 2, -2x=1,−1,2,−2 satisfy the equation (using the Factor Theorem), then reduce the cubic to a quadratic.

4. Polynomial Algebraic Equations (Higher Degree)

Any algebraic equation of degree 4 or higher falls into this category.

anxn+an−1xn−1+…+a1x+a0=0a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0an​xn+an−1​xn−1+…+a1​x+a0​=0

Quartic example (degree 4): x4−5x2+4=0x^4 - 5x^2 + 4 = 0x4−5x2+4=0

This particular quartic is solvable by substitution — let u=x2u = x^2u=x2, and it becomes u2−5u+4=0u^2 - 5u + 4 = 0u2−5u+4=0, a quadratic. Not all higher-degree equations are this friendly, though. In fact, there's a proven mathematical result (the Abel–Ruffini theorem) showing that general equations of degree 5 or higher have no formula using basic operations. That's not a limitation of our knowledge — it's a hard mathematical fact.

5. Rational Algebraic Equations

These have variables in the denominator.

P(x)Q(x)=0\frac{P(x)}{Q(x)} = 0Q(x)P(x)​=0

Example:

x2−4x−1=0\frac{x^2 - 4}{x - 1} = 0x−1x2−4​=0

The key rule: the denominator Q(x)Q(x)Q(x) can never equal zero. So here, x≠1x \neq 1x=1. Solve the numerator x2−4=0x^2 - 4 = 0x2−4=0 to get x=2x = 2x=2 or x=−2x = -2x=−2, then verify neither makes the denominator zero. Both are valid here.

The trap with rational equations: Students solve the numerator, get valid-looking answers, and forget to check the denominator. On an exam, this is a silent mark-killer.

Algebraic Equations Formulas

These are the identities and formulas you'll use repeatedly when working with algebraic equations. Memorising them saves time; understanding them prevents errors.

Essential Algebraic Identities

(a+b)2=a2+2ab+b2 (a + b)^2 = a^2 + 2ab + b^2 (a+b)2=a2+2ab+b2

(a−b)2=a2−2ab+b2 (a - b)^2 = a^2 - 2ab + b^2 (a−b)2=a2−2ab+b2

a2−b2=(a+b)(a−b) a^2 - b^2 = (a + b)(a - b) a2−b2=(a+b)(a−b)

(a+b)3=a3+3a2b+3ab2+b3 (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 (a+b)3=a3+3a2b+3ab2+b3

(a−b)3=a3−3a2b+3ab2−b3 (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 (a−b)3=a3−3a2b+3ab2−b3

a3+b3=(a+b)(a2−ab+b2) a^3 + b^3 = (a + b)(a^2 - ab + b^2) a3+b3=(a+b)(a2−ab+b2)

a3−b3=(a−b)(a2+ab+b2) a^3 - b^3 = (a - b)(a^2 + ab + b^2) a3−b3=(a−b)(a2+ab+b2)

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca (a+b+c)2=a2+b2+c2+2ab+2bc+2ca

The Quadratic Formula

For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0:

x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x=2a−b±b2−4ac​

This formula works every single time — even when factoring doesn't. It's the universal fallback for degree-2 equations.

Where the ±\pm± does the heavy lifting: The ±\pm± symbol means you calculate twice — once with +++, once with −-−. That's why quadratics can have two roots.

Discriminant

Δ=b2−4ac \Delta = b^2 - 4ac Δ=b2−4ac

Before plugging into the full quadratic formula, compute the discriminant. It tells you whether to expect two roots, one root, or no real roots (see the table in the quadratic section above).

Sum and Product of Roots (Vieta's Formulas for Quadratics)

For ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 with roots α\alphaα and β\betaβ:

α+β=−ba \alpha + \beta = -\frac{b}{a} α+β=−ab​

α⋅β=ca \alpha \cdot \beta = \frac{c}{a} α⋅β=ac​

These are incredibly useful for checking your answers quickly. Solved a quadratic and got roots 222 and 333? Sum should be 555, product should be 666. If the equation is x2−5x+6=0x^2 - 5x + 6 = 0x2−5x+6=0, that checks out: −(−5)/1=5−(−5)/1=5−(−5)/1=5 and 6/1=66/1=66/1=6.

For Cubic Equations ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0ax3+bx2+cx+d=0 with Roots α,β,γ\alpha, \beta, \gammaα,β,γ:

α+β+γ=−ba \alpha + \beta + \gamma = -\frac{b}{a} α+β+γ=−ab​

αβ+βγ+γα=ca \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} αβ+βγ+γα=ac​

α⋅β⋅γ=−da \alpha \cdot \beta \cdot \gamma = -\frac{d}{a} α⋅β⋅γ=−ad​

How to Solve Algebraic Equations - Step by Step

Different equation types need different methods. But one golden rule applies to every single one: whatever you do to one side, you must do to the other.

Solving Linear Equations

Method: Isolate the variable using inverse operations.

Example 1: Solve 3x+7=223x + 7 = 223x+7=22

3x+7=223x + 7 = 223x+7=22

Subtract 7 from both sides:

3x=22−73x = 22 - 73x=22−7

3x=153x = 153x=15

Divide both sides by 3:

x=153=5x = \frac{15}{3} = 5x=315​=5

Verification: 3(5)+7=15+7=223(5) + 7 = 15 + 7 = 223(5)+7=15+7=22 ✓

Example 2: Solve 2x−53=7\frac{2x - 5}{3} = 732x−5​=7

Multiply both sides by 3:

2x−5=212x - 5 = 212x−5=21

Add 5 to both sides:

2x=262x = 262x=26

x=13x = 13x=13

Verification: 2(13)−53=26−53=213=7\frac{2(13) - 5}{3} = \frac{26 - 5}{3} = \frac{21}{3} = 732(13)−5​=326−5​=321​=7 ✓

Example 3 (Variables on both sides): Solve 5x−9=2x+125x - 9 = 2x + 125x−9=2x+12

Move all xxx terms to the left:

5x−2x=12+95x - 2x = 12 + 95x−2x=12+9

3x=213x = 213x=21

x=7x = 7x=7

Common student error here: When moving −9-9−9 from LHS to RHS, students write 12−912 - 912−9 instead of 12+912 + 912+9. The sign flips when a term crosses the equals sign. This single mistake accounts for more lost marks in Class 7–8 algebra than any other.

Solving Quadratic Equations

Three methods. Use the one that fits the equation best.

Method 1: Factoring

Works when the quadratic factors neatly into two binomials.

Example 4: Solve x2−7x+12=0x^2 - 7x + 12 = 0x2−7x+12=0

Find two numbers that multiply to 121212 and add to −7-7−7. Those numbers are −3-3−3 and −4-4−4.

(x−3)(x−4)=0(x - 3)(x - 4) = 0(x−3)(x−4)=0

By the zero-product property (if A×B=0A \times B = 0A×B=0, then A=0A = 0A=0 or B=0B = 0B=0):

x−3=0⇒x=3x - 3 = 0 \quad \Rightarrow \quad x = 3x−3=0⇒x=3

x−4=0⇒x=4x - 4 = 0 \quad \Rightarrow \quad x = 4x−4=0⇒x=4

Verification:

• x=3x = 3x=3: (3)2−7(3)+12=9−21+12=0(3)^2 - 7(3) + 12 = 9 - 21 + 12 = 0(3)2−7(3)+12=9−21+12=0 ✓

• x=4x = 4x=4: (4)2−7(4)+12=16−28+12=0(4)^2 - 7(4) + 12 = 16 - 28 + 12 = 0(4)2−7(4)+12=16−28+12=0 ✓

Method 2: Quadratic Formula

Works for every quadratic equation — use this when factoring isn't obvious.

Example 5: Solve 2x2+5x−3=02x^2 + 5x - 3 = 02x2+5x−3=0

Here a=2a = 2a=2, b=5b = 5b=5, c=−3c = -3c=−3.

Δ=b2−4ac=25−4(2)(−3)=25+24=49\Delta = b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49Δ=b2−4ac=25−4(2)(−3)=25+24=49

x=−5±492×2=−5±74x = \frac{-5 \pm \sqrt{49}}{2 \times 2} = \frac{-5 \pm 7}{4}x=2×2−5±49​​=4−5±7​

x=−5+74=24=12x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}x=4−5+7​=42​=21​

x=−5−74=−124=−3x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3x=4−5−7​=4−12​=−3

Solutions: x=12x = \frac{1}{2}x=21​ or x=−3x = -3x=−3

Quick check with Vieta's formulas:

• Sum of roots: 12+(−3)=−52=−ba=−52\frac{1}{2} + (-3) = -\frac{5}{2} = -\frac{b}{a} = -\frac{5}{2}21​+(−3)=−25​=−ab​=−25​ ✓

• Product of roots: 12×(−3)=−32=ca=−32\frac{1}{2} \times (-3) = -\frac{3}{2} = \frac{c}{a} = \frac{-3}{2}21​×(−3)=−23​=ac​=2−3​ ✓

Method 3: Completing the Square

Useful when you need the vertex form or when dealing with equations that don't factor cleanly.

Example 6: Solve x2+6x+2=0x^2 + 6x + 2 = 0x2+6x+2=0

Move the constant:

x2+6x=−2x^2 + 6x = -2x2+6x=−2

Take half the coefficient of xxx (which is 6÷2=36 \div 2 = 36÷2=3), square it (32=93^2 = 932=9), and add to both sides:

x2+6x+9=−2+9x^2 + 6x + 9 = -2 + 9x2+6x+9=−2+9

(x+3)2=7(x + 3)^2 = 7(x+3)2=7

x+3=±7x + 3 = \pm\sqrt{7}x+3=±7​

x=−3+7orx=−3−7x = -3 + \sqrt{7} \quad \text{or} \quad x = -3 - \sqrt{7}x=−3+7​orx=−3−7​

x≈−0.354orx≈−5.646x \approx -0.354 \quad \text{or} \quad x \approx -5.646x≈−0.354orx≈−5.646

Solving Cubic Equations

Strategy: Find one root by trial (using the Factor Theorem), then reduce to a quadratic.

Example 7: Solve x3−6x2+11x−6=0x^3 - 6x^2 + 11x - 6 = 0x3−6x2+11x−6=0

Step 1: Test simple integer values. Try x=1x = 1x=1:

1−6+11−6=01 - 6 + 11 - 6 = 01−6+11−6=0 ✓

So (x−1)(x - 1)(x−1) is a factor.

Step 2: Divide x3−6x2+11x−6x^3 - 6x^2 + 11x - 6x3−6x2+11x−6 by (x−1)(x - 1)(x−1) using polynomial long division or synthetic division:

x3−6x2+11x−6=(x−1)(x2−5x+6)x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)x3−6x2+11x−6=(x−1)(x2−5x+6)

Step 3: Factor the quadratic x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)x2−5x+6=(x−2)(x−3)

Solutions: x=1x = 1x=1, x=2x = 2x=2, x=3x = 3x=3

Verification with Vieta's:

• Sum: 1+2+3=6=−(−6)/11 + 2 + 3 = 6 = -(-6)/11+2+3=6=−(−6)/1 ✓

• Product: 1×2×3=6=−(−6)/11 \times 2 \times 3 = 6 = -(-6)/11×2×3=6=−(−6)/1 ✓

Solving Equations with Variables on Both Sides

Example 8: Solve 4(2x−1)=3(x+5)4(2x - 1) = 3(x + 5)4(2x−1)=3(x+5)

Expand both sides:

8x−4=3x+158x - 4 = 3x + 158x−4=3x+15

Move xxx terms left, constants right:

8x−3x=15+48x - 3x = 15 + 48x−3x=15+4

5x=195x = 195x=19

x=195=3.8x = \frac{19}{5} = 3.8x=519​=3.8

Solving Rational Equations

Example 9: Solve 3x−2+1x+1=4x2−x−2\frac{3}{x - 2} + \frac{1}{x + 1} = \frac{4}{x^2 - x - 2}x−23​+x+11​=x2−x−24​

First, notice that x2−x−2=(x−2)(x+1)x^2 - x - 2 = (x - 2)(x + 1)x2−x−2=(x−2)(x+1). So the LCD is (x−2)(x+1)(x - 2)(x + 1)(x−2)(x+1).

Restrictions: x≠2x \neq 2x=2 and x≠−1x \neq -1x=−1 (these make the denominator zero).

Multiply every term by (x−2)(x+1)(x - 2)(x + 1)(x−2)(x+1):

3(x+1)+1(x−2)=43(x + 1) + 1(x - 2) = 43(x+1)+1(x−2)=4

3x+3+x−2=43x + 3 + x - 2 = 43x+3+x−2=4

4x+1=44x + 1 = 44x+1=4

4x=34x = 34x=3

x=34x = \frac{3}{4}x=43​

Since 34≠2\frac{3}{4} \neq 243​=2 and 34≠−1\frac{3}{4} \neq -143​=−1, the solution is valid.

Common Mistakes Students Make with Algebraic Equations

These aren't hypothetical. They come from tutoring sessions — the same errors we see week after week.

Mistake 1: Sign Errors When Transposing

Wrong: 3x−8=13⟹3x=13−8=53x - 8 = 13 \implies 3x = 13 - 8 = 53x−8=13⟹3x=13−8=5

Right: 3x−8=13⟹3x=13+8=21⟹x=73x - 8 = 13 \implies 3x = 13 + 8 = 21 \implies x = 73x−8=13⟹3x=13+8=21⟹x=7

The −8-8−8 on the left becomes +8+8+8 on the right. Students who write the "move" mechanically without thinking about why the sign flips make this error constantly.

Mistake 2: Distributing to Only Part of a Bracket

Wrong: 3(x+4)=3x+43(x + 4) = 3x + 43(x+4)=3x+4

Right: 3(x+4)=3x+123(x + 4) = 3x + 123(x+4)=3x+12

The 333 multiplies everything inside the bracket. Every term. Always.

Mistake 3: Cancelling Incorrectly in Fractions

Wrong: x+33=x+1\frac{x + 3}{3} = x + 13x+3​=x+1 (cancelling the 3s)

Right: x+33=x3+1\frac{x + 3}{3} = \frac{x}{3} + 13x+3​=3x​+1

You can only cancel a factor of the entire numerator with the denominator — not a term. This is a consequence of how division distributes over addition.

Mistake 4: Forgetting the Negative Root of a Square

Wrong: x2=16⟹x=4x^2 = 16 \implies x = 4x2=16⟹x=4

Right: x2=16⟹x=4x^2 = 16 \implies x = 4x2=16⟹x=4 or x=−4x = -4x=−4

A squared number is always positive, so both 42=164^2 = 1642=16 and (−4)2=16(-4)^2 = 16(−4)2=16.

Mistake 5: Dividing by Zero Without Realising It

Wrong approach to x(x−3)=0x(x - 3) = 0x(x−3)=0: Divide both sides by xxx, getting x−3=0x - 3 = 0x−3=0, so x=3x = 3x=3.

Right approach: Use the zero-product property. x=0x = 0x=0 or x−3=0x - 3 = 0x−3=0, giving x=0x = 0x=0 or x=3x = 3x=3.

Dividing by xxx silently assumes x≠0x \neq 0x=0, which eliminates one valid solution.

Mistake 6: Forgetting to Check Extraneous Solutions

When you multiply both sides of a rational equation by an expression containing the variable, you might introduce solutions that make the original denominator zero. Always substitute back and verify.

Algebraic Equations in Real Life

Algebra isn't a classroom invention. These equation types model real situations:

Linear equations: If a taxi charges ₹50 base fare plus ₹12 per kilometre, your total fare for ddd km is F=50+12dF = 50 + 12dF=50+12d. Need to stay within a ₹350 budget? Solve 50+12d=35050 + 12d = 35050+12d=350 to find you can travel 25 km.

Quadratic equations: A ball thrown upward from a 20-metre building at 15 m/s follows h=−5t2+15t+20h = -5t^2 + 15t + 20h=−5t2+15t+20, where hhh is height and ttt is time in seconds. When does it hit the ground? Solve −5t2+15t+20=0-5t^2 + 15t + 20 = 0−5t2+15t+20=0.

Cubic equations: In engineering, cubic equations model the volume of materials, the flow rate of liquids through pipes, and the deflection of beams under load.

How to Approach Algebraic Equations on Exams [A Practical Framework]

1. Identify the type. Look at the highest power. That tells you the equation type and the maximum number of solutions.

2. Count expected roots. A degree-nnn equation has at most nnn real roots. If you find fewer, that's fine. If you find more, you've made an error.

3. Pick the right method. Linear → isolate. Quadratic → try factoring first, then formula. Cubic → trial root + factor. Rational → find LCD, multiply, check for extraneous solutions.

4. Verify every solution. Substitute back into the original equation. Not the simplified version — the original. This catches sign errors, extraneous solutions, and arithmetic slips.

5. Read the question again. Did it ask for xxx, or for 2x+12x + 12x+1? Did it ask for positive values only? Students lose marks by answering a different question than the one asked.

Practice Problems

Test yourself. Solutions follow each problem — try before peeking.

Problem 1: Solve 7x−3(2x+5)=47x - 3(2x + 5) = 47x−3(2x+5)=4

Problem 2: Solve x2+5x−14=0x^2 + 5x - 14 = 0x2+5x−14=0

Problem 3: Solve 2x2−3x−9=02x^2 - 3x - 9 = 02x2−3x−9=0

Problem 4: Solve x3−7x+6=0x^3 - 7x + 6 = 0x3−7x+6=0

Problem 5: If the roots of x2−px+12=0x^2 - px + 12 = 0x2−px+12=0 are in the ratio 1:31:31:3, find ppp.