Algebra Shortcuts: Equations, Identities and Simplification for Competitive Exams

Algebra is one of those topics that either earns you quick marks or quietly drains your time. In SSC CGL, algebra contributes 4–6 questions. In CAT, it accounts for 5–6 questions. In IBPS PO Mains, algebraic simplification appears regularly in the quantitative section. Across all these exams, the questions look different on the surface but they almost always test the same small set of identities, equation types, and simplification patterns — just in varying combinations.

The problem most candidates face is not that they do not know algebra. They learned it in school. The problem is that they approach it the school way — expanding, rearranging, solving step by step — which is accurate but far too slow for exam conditions. A question that takes 3 minutes the school way can be solved in 40 seconds once you recognize the underlying pattern.

This guide is built around pattern recognition. Every concept here comes with the trigger — the thing you look for in a question — and the shortcut that follows from it. Work through each section in order, because later shortcuts build on earlier ones.

Part 1: Algebraic Identities — The Core of Every Exam

Identities are the backbone of algebra speed. Once you can recognize which identity a question is testing within the first few seconds of reading it, you never need to expand or simplify manually.

The 8 Identities Every Exam Tests

Memorize these exactly — not just the formula, but the pattern that triggers each one.

Identity 1: (a + b)² = a² + 2ab + b²

Identity 2: (a − b)² = a² − 2ab + b²

Identity 3: (a + b)² − (a − b)² = 4ab
(Trigger: you see both (a+b)² and (a−b)² in the same expression)

Identity 4: (a + b)(a − b) = a² − b²
(Trigger: difference of two squares — most commonly tested identity in SSC)

Identity 5: (a + b)³ = a³ + 3a²b + 3ab² + b³

Identity 6: (a − b)³ = a³ − 3a²b + 3ab² − b³

Identity 7: a³ + b³ = (a + b)(a² − ab + b²)

Identity 8: a³ − b³ = (a − b)(a² + ab + b²)

The Sum-Product Shortcut System

This is the single most powerful algebra shortcut for competitive exams. If you know the sum (a + b) and the product (ab) of two numbers, you can find almost anything without solving for individual values.

Worked Example 1: If x + y = 7 and xy = 10, find x² + y²
= (x + y)² − 2xy = 49 − 20 = 29

Worked Example 2: If x + y = 6 and xy = 8, find x³ + y³
= (x + y)³ − 3xy(x + y) = 216 − 3(8)(6) = 216 − 144 = 72

Worked Example 3: If a − b = 4 and ab = 12, find a² + b²
= (a − b)² + 2ab = 16 + 24 = 40

Reciprocal Identity Shortcut

CAT and SSC both frequently test questions of the form: if x + 1/x = k, find x² + 1/x².

The chain:

If x + 1/x = k:

• x² + 1/x² = k² − 2
• x³ + 1/x³ = k³ − 3k
• x⁴ + 1/x⁴ = (x² + 1/x²)² − 2

Worked Example: If x + 1/x = 5, find x³ + 1/x³
= 5³ − 3(5) = 125 − 15 = 110

Worked Example: If x + 1/x = 3, find x⁴ + 1/x⁴
Step 1: x² + 1/x² = 3² − 2 = 7
Step 2: x⁴ + 1/x⁴ = 7² − 2 = 47

If x − 1/x = k:

• x² + 1/x² = k² + 2
• x³ − 1/x³ = k³ + 3k

Part 2: Linear Equations — Speed Solving Techniques

Single Variable — Direct Isolation

The rule: Isolate the variable in the fewest possible steps. Avoid expanding unless absolutely necessary.

Worked Example: Solve 3(2x − 5) = 27

• Do NOT expand first
• Divide both sides by 3: 2x − 5 = 9
• 2x = 14 → x = 7

Common Mistake: Expanding to 6x − 15 = 27 first adds an unnecessary step.

Two Variable Systems — Elimination Speed Rules

Rule 1: If coefficients of one variable are equal or opposite — eliminate immediately without any multiplication.

Rule 2: If one equation is a simple multiple of the other — the system has infinite solutions (dependent equations). Recognize this in 5 seconds.

Rule 3: For word problems — always define variables in terms of what the question asks. If the question asks for the sum (x + y), set up equations that directly give you x + y without needing individual values.

Worked Example: 3x + 4y = 24 and 3x + 2y = 18. Find y.
Subtract equation 2 from equation 1: 2y = 6 → y = 3
(Found y without ever finding x — exactly what the question asked)

The Ratio Substitution Method

When two variables appear only as a ratio in the solution, substituting y = kx often simplifies the system dramatically.

Worked Example: If 2x + 3y = 0 and x ≠ 0, find x/y.
2x = −3y → x/y = −3/2
(No solving required — just rearrange)

Part 3: Quadratic Equations — Pattern-First Approach

Factoring — The 5-Second Check

Before using the quadratic formula, always check if the equation factors cleanly.

The check: For ax² + bx + c = 0, find two numbers that:

• Multiply to give a × c
• Add to give b

Worked Example: Solve x² + 9x + 20 = 0

• a × c = 20, b = 9
• Two numbers: 4 and 5 (4 × 5 = 20, 4 + 5 = 9)
• (x + 4)(x + 5) = 0 → x = −4 or −5

Worked Example: Solve 2x² − 7x + 3 = 0

• a × c = 6, b = −7
• Two numbers: −6 and −1 (−6 × −1 = 6, −6 + −1 = −7)
• Split: 2x² − 6x − x + 3 = 0
• 2x(x − 3) − 1(x − 3) = 0
• (2x − 1)(x − 3) = 0 → x = ½ or 3

Vieta's Formulas — Find Sum and Product Without Solving

For ax² + bx + c = 0, roots are α and β:

• α + β = −b/a (sum of roots)
• α × β = c/a (product of roots)
• α² + β² = (α + β)² − 2αβ
• α³ + β³ = (α + β)³ − 3αβ(α + β)

Worked Example: For 3x² − 7x + 2 = 0, find α² + β² without solving.

• α + β = 7/3, αβ = 2/3
• α² + β² = (7/3)² − 2(2/3) = 49/9 − 12/9 = 37/9

Nature of Roots — Discriminant Shortcut

D = b² − 4ac

Part 4: Simplification Techniques

BODMAS — Applied Strategically

BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction) is not just an order — it is a search pattern. Before computing, scan the expression for cancellation opportunities.

Worked Example: Simplify (64 × 36) ÷ (8 × 6)

• Before computing: recognize 64 = 8 × 8 and 36 = 6 × 6
• = (8 × 8 × 6 × 6) ÷ (8 × 6) = 8 × 6 = 48
• Never multiply 64 × 36 = 2304 first. Cancel first, then compute.

Factoring Before Computing

Any time you see a sum or difference in the numerator or denominator — factor it first.

Worked Example: Simplify (x² − 16) ÷ (x + 4)
= (x + 4)(x − 4) ÷ (x + 4) = (x − 4)

Worked Example: Simplify (a³ − b³) ÷ (a − b)
= (a − b)(a² + ab + b²) ÷ (a − b) = a² + ab + b²

The Substitution Shortcut for Complex Expressions

When an expression looks complex but has a repeated sub-expression — substitute it as a single variable.

Worked Example: Simplify (x + 1/x)² − (x − 1/x)²
Let u = x + 1/x and v = x − 1/x
= u² − v² = (u + v)(u − v)
= (x + 1/x + x − 1/x)(x + 1/x − x + 1/x)
= (2x)(2/x) = 4

Part 5: Inequalities — The Direction Rules

Linear Inequalities

Solve exactly like equations, with one critical rule:

When multiplying or dividing by a negative number — flip the inequality sign.

Worked Example: Solve −4x + 8 > 0
−4x > −8
x < 2 (sign flipped because divided by −4)

Quadratic Inequalities — The Number Line Method

For (x − a)(x − b) > 0 where a < b:

• Solution: x < a OR x > b (outside the roots)

For (x − a)(x − b) < 0 where a < b:

• Solution: a < x < b (between the roots)

Worked Example: Solve x² − 5x + 6 < 0
Factor: (x − 2)(x − 3) < 0
Since a = 2, b = 3: solution is 2 < x < 3

Worked Example: Solve x² − x − 12 > 0
Factor: (x − 4)(x + 3) > 0
Solution: x < −3 OR x > 4

Absolute Value Inequalities

|x| < k means −k < x < k

|x| > k means x < −k OR x > k

Worked Example: Solve |2x − 3| < 7
−7 < 2x − 3 < 7
−4 < 2x < 10
−2 < x < 5

Part 6: SSC CGL Specific Algebra Patterns

SSC CGL repeats the same algebra question types year after year. Recognizing these patterns lets you solve them in under 60 seconds each.

Pattern 1 — The "Find the Value" Identity Question

If x + y + z = 0, find x³ + y³ + z³

Shortcut: When a + b + c = 0, then a³ + b³ + c³ = 3abc (always)

Worked Example: If a + b + c = 0 and a = 2, b = 3, find c and a³ + b³ + c³
c = −5
a³ + b³ + c³ = 3(2)(3)(−5) = −90

Pattern 2 — Symmetric Expression Simplification

If x = 2 + √3, find x + 1/x

Method: Find 1/x by rationalizing, then add.
1/x = 1/(2+√3) × (2−√3)/(2−√3) = (2−√3)/(4−3) = 2−√3
x + 1/x = (2+√3) + (2−√3) = 4

Pattern 3 — The Cyclic Sum

If a/b + b/c + c/a = 1, find a²/b² + b²/c² + c²/a²

= (a/b + b/c + c/a)² − 2(b/a + c/b + a/c) ... use sum-product system

For exam purposes: recognize this as identity-based and apply (S² − 2P) directly.

Pattern 4 — The "If a + b = k, ab = m" Chain

This is the most frequently tested SSC pattern. The exam gives you two values and asks you to find a derived expression.

Worked Example: If p + q = 10 and pq = 21, find p³ + q³
Step 1: p³ + q³ = (p+q)³ − 3pq(p+q)
= 1000 − 3(21)(10) = 1000 − 630 = 370

Part 7: CAT-Specific Algebra Techniques

Back-Substitution for MCQ Algebra

When a CAT algebra question asks for the value of an expression and gives four answer choices — substitute the simplest possible value for x and evaluate both the expression and each answer choice.

Worked Example: Which expression equals (x² − 4)/(x − 2)?
Options: (a) x − 2, (b) x + 2, (c) x², (d) x + 4

Substitute x = 3:
Expression = (9 − 4)/(3 − 2) = 5
Check options at x = 3: (a) 1, (b) 5 ✓, (c) 9, (d) 7
Answer: (b) x + 2

Functional Equations

CAT tests functions in specific formats. The two most common:

Type 1 — Find f(f(x)):
Apply the function twice. Substitute f(x) as the input into f itself.

Worked Example: f(x) = 2x + 1. Find f(f(3))
f(3) = 7. f(7) = 15. Answer: 15

Type 2 — Find the inverse function:
Replace f(x) with y, solve for x in terms of y, then replace y with x.

Worked Example: f(x) = 3x − 4. Find f⁻¹(x)
y = 3x − 4 → x = (y + 4)/3 → f⁻¹(x) = (x + 4)/3

6-Week Algebra Mastery Plan