Algebra is the branch of mathematics that uses symbols (usually letters like x, y, a, b) to represent numbers and quantities in formulas and equations. It allows us to generalize arithmetic operations and solve problems where some values are unknown or variable.
Core Concepts of Algebra:β
Variables : Symbols that stand in for unknown values (e.g., π₯, π¦)
Constants : Fixed values (e.g., 2, -5, 0.75)
Expressions : Combinations of variables, constants, and operations (e.g., 2π₯ + 3)
Equations : Statements that two expressions are equal (e.g., π₯ + 5 = 12)
Operations : Addition, subtraction, multiplication, division, powers, etc.
Example in Daily Life Problem:β
A father is 30 years older than his son. Their combined age is 60. Find their age.
Let π₯ be the son's age.
Then father's age = π₯ + 30 π₯ + 30 x + 30
Equation: π₯ + ( π₯ + 30 ) = 60 π₯ + ( π₯ + 30 ) = 60 x + ( x + 30 ) = 60
Solve: 2 π₯ + 30 = 60 β π₯ = 15 2π₯ + 30 = 60 β π₯ = 15 2 x + 30 = 60 β x = 15
Factorizationβ
Definition:β
Factorization is the process of expressing an algebraic expression as a product of its factors.
Techniques of Factorizationβ
1. Common Factor Method β
Factor out the greatest common factor (GCF) from all terms.
Example: 6 x 2 + 9 x = 3 x ( 2 x + 3 ) 6x^2 + 9x = 3x(2x + 3) 6 x 2 + 9 x = 3 x ( 2 x + 3 )
2. Grouping Method β
Group terms in pairs and factor each group.
Example: a x + a y + b x + b y = ( a + b ) x + ( a + b ) y = ( a + b ) ( x + y ) ax + ay + bx + by = (a + b)x + (a + b)y = (a + b)(x + y) a x + a y + b x + b y = ( a + b ) x + ( a + b ) y = ( a + b ) ( x + y )
3. Using Identities β
Apply algebraic identities to factor expressions.
Standard Identities:
( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
( a β b ) 2 = a 2 β 2 a b + b 2 (a - b)^2 = a^2 - 2ab + b^2 ( a β b ) 2 = a 2 β 2 ab + b 2
a 2 β b 2 = ( a + b ) ( a β b ) a^2 - b^2 = (a + b)(a - b) a 2 β b 2 = ( a + b ) ( a β b )
( x + a ) ( x + b ) = x 2 + ( a + b ) x + a b (x + a)(x + b) = x^2 + (a + b)x + ab ( x + a ) ( x + b ) = x 2 + ( a + b ) x + ab
Example: x 2 β 9 = ( x + 3 ) ( x β 3 ) x^2 - 9 = (x + 3)(x - 3) x 2 β 9 = ( x + 3 ) ( x β 3 )
4. Middle Term Splitting (Quadratic Trinomials)β
Split the middle term to factor quadratics.
Example: x 2 + 5 x + 6 x^2 + 5x + 6 x 2 + 5 x + 6
= x 2 + 2 x + 3 x + 6 = x^2 + 2x + 3x + 6 = x 2 + 2 x + 3 x + 6
= x ( x + 2 ) + 3 ( x + 2 ) = x(x + 2) + 3(x + 2) = x ( x + 2 ) + 3 ( x + 2 )
= ( x + 2 ) ( x + 3 ) = (x + 2)(x + 3) = ( x + 2 ) ( x + 3 )
Special Casesβ
Perfect Square Trinomials:
x 2 + 6 x + 9 = ( x + 3 ) 2 x^2 + 6x + 9 = (x + 3)^2 x 2 + 6 x + 9 = ( x + 3 ) 2
x 2 β 10 x + 25 = ( x β 5 ) 2 x^2 - 10x + 25 = (x - 5)^2 x 2 β 10 x + 25 = ( x β 5 ) 2
Difference of Squares
a 2 β b 2 = ( a + b ) ( a β b ) a^2 - b^2 = (a + b)(a - b) a 2 β b 2 = ( a + b ) ( a β b )
4 x 2 β 9 = ( 2 x + 3 ) ( 2 x β 3 ) 4x^2 - 9 = (2x + 3)(2x - 3) 4 x 2 β 9 = ( 2 x + 3 ) ( 2 x β 3 )
( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
( a β b ) 2 = a 2 β 2 a b + b 2 (a - b)^2 = a^2 - 2ab + b^2 ( a β b ) 2 = a 2 β 2 ab + b 2
a 2 β b 2 = ( a + b ) ( a β b ) a^2 - b^2 = (a + b)(a - b) a 2 β b 2 = ( a + b ) ( a β b )
( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
( a β b ) 3 = a 3 β 3 a 2 b + 3 a b 2 β b 3 (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 ( a β b ) 3 = a 3 β 3 a 2 b + 3 a b 2 β b 3
a 3 + b 3 = ( a + b ) β ( a 2 β a b + b 2 ) a^3 + b^3 = (a + b)\,(a^2 - ab + b^2) a 3 + b 3 = ( a + b ) ( a 2 β ab + b 2 )
a 3 β b 3 = ( a β b ) β ( a 2 + a b + b 2 ) a^3 - b^3 = (a - b)\,(a^2 + ab + b^2) a 3 β b 3 = ( a β b ) ( a 2 + ab + b 2 )
Examples:
2 3 + 3 3 = ( 2 + 3 ) ( 2 2 β 2 β
3 + 3 2 ) = 5 ( 4 β 6 + 9 ) = 35 2^3 + 3^3 = (2 + 3)(2^2 - 2\cdot3 + 3^2) = 5(4 - 6 + 9) = 35 2 3 + 3 3 = ( 2 + 3 ) ( 2 2 β 2 β
3 + 3 2 ) = 5 ( 4 β 6 + 9 ) = 35
3 3 β 2 3 = ( 3 β 2 ) ( 3 2 + 3 β
2 + 2 2 ) = 1 ( 9 + 6 + 4 ) = 19 3^3 - 2^3 = (3 - 2)(3^2 + 3\cdot2 + 2^2) = 1(9 + 6 + 4) = 19 3 3 β 2 3 = ( 3 β 2 ) ( 3 2 + 3 β
2 + 2 2 ) = 1 ( 9 + 6 + 4 ) = 19
HCF and LCM of Algebraic Expressionsβ
Definitionsβ
HCF (Highest Common Factor) : The greatest factor common to all expressions.
LCM (Lowest Common Multiple) : The smallest expression divisible by all given expressions.
Method: Factorizationβ
Example:
Find HCF and LCM of: x 2 β 4 x x^2 - 4x x 2 β 4 x x 2 β 16 x^2 - 16 x 2 β 16
Factorizing, x 2 β 4 x x^2 - 4x x 2 β 4 x
= x ( x β 4 ) = x(x - 4) = x ( x β 4 )
Factorizing, x 2 β 16 x^2 - 16 x 2 β 16
= ( x β 4 ) ( x + 4 ) = (x - 4)(x + 4) = ( x β 4 ) ( x + 4 )
HCF: ( x β 4 ) (x - 4) ( x β 4 )
LCM: x ( x β 4 ) ( x + 4 ) x(x - 4)(x + 4) x ( x β 4 ) ( x + 4 )
Example Problems β Factorizationβ
Factorizing, x 2 + 7 x + 10 x^2 + 7x + 10 x 2 + 7 x + 10
= x 2 + 2 x + 5 x + 10 = x^2 + 2x + 5x + 10 = x 2 + 2 x + 5 x + 10
= x ( x + 2 ) + 5 ( x + 2 ) = x(x + 2) + 5(x + 2) = x ( x + 2 ) + 5 ( x + 2 )
= ( x + 2 ) ( x + 5 ) = (x + 2)(x + 5) = ( x + 2 ) ( x + 5 )
Factorizing, x 2 β 6 x β 16 x^2 - 6x - 16 x 2 β 6 x β 16
= x 2 β 8 x + 2 x β 16 = x^2 - 8x + 2x - 16 = x 2 β 8 x + 2 x β 16
= x ( x β 8 ) + 2 ( x β 8 ) = x(x - 8) + 2(x - 8) = x ( x β 8 ) + 2 ( x β 8 )
= ( x β 8 ) ( x + 2 ) = (x - 8)(x + 2) = ( x β 8 ) ( x + 2 )
Factorizing by grouping, 6 x 3 + 15 x 2 + 4 x + 10 6x^3 + 15x^2 + 4x + 10 6 x 3 + 15 x 2 + 4 x + 10
Group: ( 6 x 3 + 15 x 2 ) + ( 4 x + 10 ) (6x^3 + 15x^2) + (4x + 10) ( 6 x 3 + 15 x 2 ) + ( 4 x + 10 )
= 3 x 2 ( 2 x + 5 ) + 2 ( 2 x + 5 ) = 3x^2(2x + 5) + 2(2x + 5) = 3 x 2 ( 2 x + 5 ) + 2 ( 2 x + 5 )
= ( 2 x + 5 ) ( 3 x 2 + 2 ) = (2x + 5)(3x^2 + 2) = ( 2 x + 5 ) ( 3 x 2 + 2 )
Sum of cubes, 8 x 3 + 27 8x^3 + 27 8 x 3 + 27
Let a = 2 x , β
β b = 3 a = 2x,\; b = 3 a = 2 x , b = 3 a 3 + b 3 = ( a + b ) ( a 2 β a b + b 2 ) a^3 + b^3 = (a + b)(a^2 - ab + b^2) a 3 + b 3 = ( a + b ) ( a 2 β ab + b 2 )
= ( 2 x + 3 ) ( 4 x 2 β 6 x + 9 ) = (2x + 3)(4x^2 - 6x + 9) = ( 2 x + 3 ) ( 4 x 2 β 6 x + 9 )
Perfect-square trinomial, x 2 + 12 x + 36 x^2 + 12x + 36 x 2 + 12 x + 36
Recognize ( x + 6 ) 2 (x + 6)^2 ( x + 6 ) 2
= ( x + 6 ) ( x + 6 ) = (x + 6)(x + 6) = ( x + 6 ) ( x + 6 )
HCF and LCM β pair A: x 2 β 9 x^2 - 9 x 2 β 9 x 2 + 2 x β 15 x^2 + 2x - 15 x 2 + 2 x β 15
Factor each: x 2 β 9 = ( x β 3 ) ( x + 3 ) x^2 - 9 = (x - 3)(x + 3) x 2 β 9 = ( x β 3 ) ( x + 3 )
x 2 + 2 x β 15 = ( x + 5 ) ( x β 3 ) x^2 + 2x - 15 = (x + 5)(x - 3) x 2 + 2 x β 15 = ( x + 5 ) ( x β 3 )
HCF: ( x β 3 ) (x - 3) ( x β 3 )
LCM: ( x β 3 ) ( x + 3 ) ( x + 5 ) (x - 3)(x + 3)(x + 5) ( x β 3 ) ( x + 3 ) ( x + 5 )
HCF and LCM β pair B (with numeric coefficients): 6 x 2 β 15 x 6x^2 - 15x 6 x 2 β 15 x 4 x 2 β 10 x 4x^2 - 10x 4 x 2 β 10 x
Factor each: 6 x 2 β 15 x = 3 x ( 2 x β 5 ) 6x^2 - 15x = 3x(2x - 5) 6 x 2 β 15 x = 3 x ( 2 x β 5 )
4 x 2 β 10 x = 2 x ( 2 x β 5 ) 4x^2 - 10x = 2x(2x - 5) 4 x 2 β 10 x = 2 x ( 2 x β 5 )
HCF: x ( 2 x β 5 ) x(2x - 5) x ( 2 x β 5 )
LCM: 6 x ( 2 x β 5 ) 6x(2x - 5) 6 x ( 2 x β 5 )
Practice: try similar problems by changing signs or coefficients (e.g., replace 7 with 11, or use 9 x 2 β 24 x + 16 9x^2 - 24x + 16 9 x 2 β 24 x + 16
Linear Equations
Meaning of Linear Equationβ
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. It represents a straight line when graphed.
General Form: a x + b y = c ax + by = c a x + b y = c
Methods of Solving Linear Equationsβ
Substitution Methodβ
Steps:
Solve one equation for one variable.
Substitute that expression into the other equation.
Solve for the second variable.
Back-substitute to find the first variable.
Example:
Solve:
x + y = 5 x + y = 5 x + y = 5
x β y = 1 x - y = 1 x β y = 1
Step 1: From first equation: x = 5 β y x = 5 - y x = 5 β y
Step 2: Substitute into second:( 5 β y ) β y = 1 β 5 β 2 y = 1 β y = 2 (5 - y) - y = 1 \Rightarrow 5 - 2y = 1 \Rightarrow y = 2 ( 5 β y ) β y = 1 β 5 β 2 y = 1 β y = 2
Step 3: x = 5 β 2 = 3 x = 5 - 2 = 3 x = 5 β 2 = 3
Solution: x = 3 , y = 2 x = 3, y = 2 x = 3 , y = 2
Elimination Methodβ
Steps:
Multiply equations (if needed) to align coefficients.
Add or subtract equations to eliminate one variable.
Solve for the remaining variable.
Substitute back to find the other variable.
Example:
Solve:
Step 1: Add both equations:
( 2 x + 3 y ) + ( 4 x β 3 y ) (2x + 3y) + (4x - 3y) ( 2 x + 3 y ) + ( 4 x β 3 y )
= 12 + 6 β 6 x 12 + 6 \Rightarrow 6x 12 + 6 β 6 x
= 18 18 18
β x = 3 \Rightarrow x = 3 β x = 3
Step 2: Substitute into first:
2 ( 3 ) + 3 y 2(3) + 3y 2 ( 3 ) + 3 y
= 12 β 6 + 3 y 12 \Rightarrow 6 + 3y 12 β 6 + 3 y
= 12 12 12
β y = 2 \Rightarrow y = 2 β y = 2
Solution: x = 3 , y = 2 x = 3, y = 2 x = 3 , y = 2
Practice Problems β Linear Equations (with step-by-step solutions)β
Solve using the substitution method.
Problem:x + y = 5 x + y = 5 x + y = 5
2 x β y = 4 2x - y = 4 2 x β y = 4
Solution (substitution):
From x + y = 5 β x = 5 β y x + y = 5 \Rightarrow x = 5 - y x + y = 5 β x = 5 β y
Substitute into 2 x β y = 4 2x - y = 4 2 x β y = 4
2 ( 5 β y ) β y = 4 β 10 β 2 y β y = 4 2(5 - y) - y = 4 \Rightarrow 10 - 2y - y = 4 2 ( 5 β y ) β y = 4 β 10 β 2 y β y = 4
Simplify: β 3 y = β 6 β y = 2 -3y = -6 \Rightarrow y = 2 β 3 y = β 6 β y = 2
Back-substitute: x = 5 β 2 = 3 x = 5 - 2 = 3 x = 5 β 2 = 3
Answer: ( x , y ) = ( 3 , 2 ) (x,y) = (3,2) ( x , y ) = ( 3 , 2 )
Solve using the elimination method.
Problem:
2 x + 3 y = 12 2x + 3y = 12 2 x + 3 y = 12
4 x β 3 y = 6 4x - 3y = 6 4 x β 3 y = 6
Solution (elimination):
Add the two equations to eliminate y y y ( 2 x + 3 y ) + ( 4 x β 3 y ) = 12 + 6 (2x + 3y) + (4x - 3y) = 12 + 6 ( 2 x + 3 y ) + ( 4 x β 3 y ) = 12 + 6
This gives 6 x = 18 β x = 3 6x = 18 \Rightarrow x = 3 6 x = 18 β x = 3
Substitute into 2 x + 3 y = 12 2x + 3y = 12 2 x + 3 y = 12 2 ( 3 ) + 3 y = 12 β 6 + 3 y = 12 2(3) + 3y = 12 \Rightarrow 6 + 3y = 12 2 ( 3 ) + 3 y = 12 β 6 + 3 y = 12
Solve: 3 y = 6 β y = 2 3y = 6 \Rightarrow y = 2 3 y = 6 β y = 2
Answer: ( x , y ) = ( 3 , 2 ) (x,y) = (3,2) ( x , y ) = ( 3 , 2 )
Solve the same system by both substitution and elimination (demonstrate both methods).
Problem:x + y = 4 x + y = 4 x + y = 4 2 x + 3 y = 11 2x + 3y = 11 2 x + 3 y = 11
Solution (method A β substitution):
From x + y = 4 β x = 4 β y x + y = 4 \Rightarrow x = 4 - y x + y = 4 β x = 4 β y
Substitute into 2 x + 3 y = 11 2x + 3y = 11 2 x + 3 y = 11
2 ( 4 β y ) + 3 y = 11 β 8 β 2 y + 3 y = 11 2(4 - y) + 3y = 11 \Rightarrow 8 - 2y + 3y = 11 2 ( 4 β y ) + 3 y = 11 β 8 β 2 y + 3 y = 11
Simplify: 8 + y = 11 β y = 3 8 + y = 11 \Rightarrow y = 3 8 + y = 11 β y = 3
Back-substitute: x = 4 β 3 = 1 x = 4 - 3 = 1 x = 4 β 3 = 1
Answer: ( x , y ) = ( 1 , 3 ) (x,y) = (1,3) ( x , y ) = ( 1 , 3 )
Solution (method B β elimination):
Multiply first equation by β 2 -2 β 2 β 2 x β 2 y = β 8 -2x - 2y = -8 β 2 x β 2 y = β 8
Add to second equation:( 2 x + 3 y ) + ( β 2 x β 2 y ) = 11 + ( β 8 ) (2x + 3y) + (-2x - 2y) = 11 + (-8) ( 2 x + 3 y ) + ( β 2 x β 2 y ) = 11 + ( β 8 )
This gives y = 3 y = 3 y = 3
Substitute back into x + y = 4 x + y = 4 x + y = 4 x = 1 x = 1 x = 1
Answer: ( x , y ) = ( 1 , 3 ) (x,y) = (1,3) ( x , y ) = ( 1 , 3 )
Indices
What Are Indices?β
Indices (or exponents) represent repeated multiplication of a number by itself.
Example:β
2 4 = 2 β
2 β
2 β
2 = 16 2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16 2 4 = 2 β
2 β
2 β
2 = 16
Why Do We Need Indices?β
To simplify large multiplications
To express powers and roots compactly
To solve exponential equations
To handle scientific notation and algebraic expressions efficiently
Laws of Indicesβ
1. Multiplication Lawβ
Rule: a m β
a n = a m + n a^m \cdot a^n = a^{m+n} a m β
a n = a m + n
Example: 3 2 β
3 4 = 3 2 + 4 = 3 6 3^2 \cdot 3^4 = 3^{2+4} = 3^6 3 2 β
3 4 = 3 2 + 4 = 3 6
2. Division Lawβ
Rule: a m a n = a m β n ( a β 0 ) \dfrac{a^m}{a^n} = a^{m-n} \quad (a \ne 0) a n a m β = a m β n ( a ξ = 0 )
Example: 5 6 5 2 = 5 6 β 2 = 5 4 \dfrac{5^6}{5^2} = 5^{6-2} = 5^4 5 2 5 6 β = 5 6 β 2 = 5 4
3. Power of a Power Lawβ
Rule: ( a m ) n = a m β
n (a^m)^n = a^{m \cdot n} ( a m ) n = a m β
n
Example: ( 2 3 ) 2 = 2 3 β
2 = 2 6 (2^3)^2 = 2^{3 \cdot 2} = 2^6 ( 2 3 ) 2 = 2 3 β
2 = 2 6
4. Law of Negative Indexβ
Rule: a β n = 1 a n ( a β 0 ) a^{-n} = \dfrac{1}{a^n} \quad (a \ne 0) a β n = a n 1 β ( a ξ = 0 )
Example: 4 β 2 = 1 4 2 = 1 16 4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16} 4 β 2 = 4 2 1 β = 16 1 β
5. Law of Zero Indexβ
Rule: a 0 = 1 ( a β 0 ) a^0 = 1 \quad (a \ne 0) a 0 = 1 ( a ξ = 0 )
Example: 7 0 = 1 7^0 = 1 7 0 = 1
6. Root Law of Indicesβ
Rule:
( a ) 1 n = a n (a)^{\dfrac{1}{n}} = \sqrt[n]{a} ( a ) n 1 β = n a β Example: ( 16 ) 1 2 = 16 = 4 (16)^{\dfrac{1}{2}} = \sqrt{16} = 4 ( 16 ) 2 1 β = 16 β = 4
Also:( a ) m n = a m n (a)^{\dfrac{m}{n}} = \sqrt[n]{a^m} ( a ) n m β = n a m β
Example: ( 8 ) 2 3 = 8 2 3 = 64 3 = 4 (8)^{\dfrac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4 ( 8 ) 3 2 β = 3 8 2 β = 3 64 β = 4
Common Problemsβ
Simplify:β
2 3 β
2 4 = 2 7 = 128 2^3 \cdot 2^4 = 2^7 = 128 2 3 β
2 4 = 2 7 = 128
9 5 9 2 = 9 3 = 729 \dfrac{9^5}{9^2} = 9^3 = 729 9 2 9 5 β = 9 3 = 729
( 5 2 ) 3 = 5 6 = 15625 (5^2)^3 = 5^6 = 15625 ( 5 2 ) 3 = 5 6 = 15625
10 β 1 = 1 10 10^{-1} = \dfrac{1}{10} 1 0 β 1 = 10 1 β
3 0 + 4 0 = 1 + 1 = 2 3^0 + 4^0 = 1 + 1 = 2 3 0 + 4 0 = 1 + 1 = 2
( 27 ) 1 3 = 27 3 = 3 (27)^{\dfrac{1}{3}} = \sqrt[3]{27} = 3 ( 27 ) 3 1 β = 3 27 β = 3
( 81 ) 3 4 = 81 3 4 = 531441 4 = 27 (81)^{\dfrac{3}{4}} = \sqrt[4]{81^3} = \sqrt[4]{531441} = 27 ( 81 ) 4 3 β = 4 8 1 3 β = 4 531441 β = 27
Real-World Applicationsβ
Scientific notation: 6.02 Γ 10 23 6.02 \times 10^{23} 6.02 Γ 1 0 23
Compound interest and population growth
Physics formulas involving powers and roots