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Algebra - Introduction to Algebra and Variables

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A variable is a quantity that does not have a fixed value and is represented by letters like x,y,z,l,m,nx, y, z, l, m, n. Unlike constants (fixed numbers like 55 or 100100), variables allow us to write general rules. Visually, think of a variable as an empty box or a placeholder where any number can be placed.

Algebraic rules can be derived from patterns, such as matchstick shapes. For example, if one letter 'L' requires 22 matchsticks, then nn letters of 'L' require 2×n=2n2 \times n = 2n sticks. Visually, as you add more shapes to a row, the total number of sticks increases by a fixed amount for every new unit added.

Variables are used to generalize geometry formulas. For a square with side length ss, the perimeter is the sum of its four equal sides. Visually, a square consists of four equal line segments, so we write the rule as P=4sP = 4s. For a rectangle with length ll and breadth bb, the perimeter is 2(l+b)2(l + b).

In arithmetic, variables help express properties of numbers concisely. The Commutative Property of addition states a+b=b+aa + b = b + a. Visually, if you combine a blue strip of length aa with a red strip of length bb, the total length is the same regardless of which strip is placed first.

The Distributive Property of multiplication over addition is expressed as a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c). This can be visualized by finding the area of a large rectangle by splitting it into two smaller rectangles and adding their individual areas.

Algebraic expressions are formed by combining variables and constants using operators like ++, -, ×\times, and ÷\div. For example, y+10y + 10 means 1010 is added to yy. Visually, on a number line, y+10y + 10 represents a point 1010 units to the right of yy.

An equation is a condition on a variable showing that two expressions are equal, separated by an '=' sign. It has a Left Hand Side (LHS) and a Right Hand Side (RHS). Visually, an equation is like a balanced weighing scale; the scale remains level only when the values on both sides are exactly equal.

The solution of an equation is the specific value of the variable that makes the LHS equal to the RHS. For example, in the equation x+5=12x + 5 = 12, the solution is x=7x = 7 because 7+5=127 + 5 = 12 makes the equation true.

📐Formulae

P=4sP = 4s (Perimeter of a square with side ss)

P=3sP = 3s (Perimeter of an equilateral triangle with side ss)

P=2(l+b)P = 2(l + b) (Perimeter of a rectangle with length ll and breadth bb)

d=2rd = 2r (Diameter of a circle with radius rr)

a+b=b+aa + b = b + a (Commutative property of addition)

a×b=b×aa \times b = b \times a (Commutative property of multiplication)

a×(b+c)=a×b+a×ca \times (b + c) = a \times b + a \times c (Distributive property of multiplication over addition)

💡Examples

Problem 1:

Find the rule for the number of matchsticks required to make a pattern of the letter 'T'. If nn represents the number of 'T's, how many sticks are needed for 5050 such letters?

Solution:

Step 1: Observe that one letter 'T' is made of 22 matchsticks (one horizontal and one vertical). Step 2: For nn letters, the number of sticks required is 2×n=2n2 \times n = 2n. Step 3: To find the sticks for 5050 letters, substitute n=50n = 50 into the rule. Step 4: 2×50=1002 \times 50 = 100 matchsticks.

Explanation:

This problem shows how to identify a constant rate of change and represent it using a variable nn to create a general rule.

Problem 2:

Sarita has 1010 more marbles than Ameena. If Ameena has xx marbles, write Sarita's marbles in terms of xx. Also, if Ameena has 2525 marbles, how many does Sarita have?

Solution:

Step 1: Let Ameena's marbles be xx. Step 2: Since Sarita has 1010 more, add 1010 to xx. Sarita's marbles =x+10= x + 10. Step 3: Given Ameena has x=25x = 25 marbles. Step 4: Substitute x=25x = 25 into the expression: 25+10=3525 + 10 = 35. Sarita has 3535 marbles.

Explanation:

This example demonstrates how to translate a word problem into an algebraic expression and then evaluate that expression for a given value.