Algebra

Algebra — Study Notes

Maharashtra Police Bharti (Constable)

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Overview

Algebra forms the backbone of quantitative reasoning in the Maharashtra Police Bharti exam. This topic primarily tests your ability to solve linear equations and apply algebraic identities for quick calculations. Mastering algebra is essential because these concepts appear directly in standalone questions and indirectly in problems on profit-loss, time-work, and mensuration.

For this exam, focus on two core areas: (1) solving linear equations in one or two variables, and (2) memorising standard algebraic identities that enable rapid simplification. The questions are typically straightforward but time-sensitive, so speed through pattern recognition is critical. Candidates who can spot identity applications instantly gain a significant edge.

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Key Concepts

• Variable: A symbol (usually x, y, z) representing an unknown quantity that you need to find.
• Linear Equation: An equation where the highest power of the variable is 1. Example: 3x + 5 = 17.
• Coefficient: The number multiplied with a variable. In 7x, the coefficient is 7.
• Constant: A fixed value without any variable. In 3x + 5 = 17, both 5 and 17 are constants.
• Solution of an Equation: The value of the variable that makes the equation true. If x = 4 satisfies 3x + 5 = 17, then 4 is the solution.
• Simultaneous Equations: Two or more equations solved together to find values of multiple unknowns.
• Algebraic Identity: An equation that holds true for all values of the variables involved, used to expand or factorise expressions quickly.
• Like Terms: Terms having the same variable with the same power. Only like terms can be added or subtracted.

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Formulas / Key Facts

Linear Equations

Single Variable:

• Standard form: ax + b = c
• Solution: x = (c − b) / a

Two Variables (Simultaneous Equations):

• a₁x + b₁y = c₁
• a₂x + b₂y = c₂

Elimination Method: Make coefficients of one variable equal, then add or subtract equations.

Substitution Method: Express one variable in terms of the other, substitute into the second equation.

Cross-Multiplication Formula:

• x = (b₁c₂ − b₂c₁) / (a₁b₂ − a₂b₁)
• y = (c₁a₂ − c₂a₁) / (a₁b₂ − a₂b₁)

Standard Algebraic Identities

Quick Calculation Tricks

• (a + b)² + (a − b)² = 2(a² + b²)
• (a + b)² − (a − b)² = 4ab
• If x + 1/x = k, then x² + 1/x² = k² − 2
• If x − 1/x = k, then x² + 1/x² = k² + 2

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Worked Examples

Example 1: Simple Linear Equation

Problem: Solve 5x − 3 = 2x + 12

Solution:

• Step 1: Bring variable terms to one side: 5x − 2x = 12 + 3
• Step 2: Simplify: 3x = 15
• Step 3: Divide: x = 15/3 = 5

Answer: x = 5

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Example 2: Simultaneous Equations

Problem: Solve: 2x + 3y = 13 and 5x − 2y = 4

Solution using Elimination:

• Step 1: Multiply first equation by 2: 4x + 6y = 26
• Step 2: Multiply second equation by 3: 15x − 6y = 12
• Step 3: Add both equations: 4x + 15x + 6y − 6y = 26 + 12
• Step 4: 19x = 38, so x = 2
• Step 5: Substitute x = 2 in first equation: 2(2) + 3y = 13
• Step 6: 4 + 3y = 13, so 3y = 9, y = 3

Answer: x = 2, y = 3

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Example 3: Applying Algebraic Identity

Problem: Find the value of 105² − 95²

Solution:

• Use identity: a² − b² = (a + b)(a − b)
• Here a = 105, b = 95
• Step 1: a + b = 105 + 95 = 200
• Step 2: a − b = 105 − 95 = 10
• Step 3: Result = 200 × 10 = 2000

Answer: 2000

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Example 4: x + 1/x Type Problem

Problem: If x + 1/x = 5, find x² + 1/x²

Solution:

• Use formula: x² + 1/x² = (x + 1/x)² − 2
• Step 1: (x + 1/x)² = 5² = 25
• Step 2: x² + 1/x² = 25 − 2 = 23

Answer: 23

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Common Mistakes

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Quick Reference

• Linear equation ax + b = c → x = (c − b)/a
• (a + b)² = a² + 2ab + b² — never forget the middle term
• (a² − b²) = (a + b)(a − b) — use for quick subtraction of squares
• If x + 1/x = k, then x² + 1/x² = k² − 2
• Elimination method: equalise one variable's coefficient, then add/subtract
• Sign change rule: term crossing "=" changes sign

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