CBSE Class 6 Mathematics Algebra Chapter Notes

.11.1 Beginning of Algebra

It is said that algebra as a branch of Mathematics began about 1550 BC, i.e. more than 3500 years ago, when people in Egypt started using symbols to denote unknown numbers. Around 300 BC, use of letters to denote unknowns and forming expressions from them was quite common in India. Many great Indian mathematicians, Aryabhatt (born 476AD), Brahmagupta (born 598AD), Mahavira (who lived around 850AD) and Bhaskara II (born 1114AD) and others, contributed a lot to the study of algebra. They gave names such as Beeja, Varna etc. to unknowns and used first letters of colour names [e.g., ka from kala (black), nee from neela (blue)] to denote them. The Indian name for algebra, Beejaganit, dates back to these ancient Indian mathematicians. The word ‘algebra’ is derived from the title of the book, ‘Aljebar w’al almugabalah’, written about 825AD by an Arab mathematician, Mohammed Ibn Al Khowarizmi of Baghdad.

11.2 Algebra

Algebra is about finding the unknown or it is about putting real life problems into equations and then solving them. Unfortunately many textbooks go straight to the rules, procedures and formulas, forgetting that these are real life problems being solved. A branch of mathematics that substitutes letters for numbers. An algebraic equation represents a scale, what is done on one side of the scale with a number is also done to the other side of the scale. The numbers are the constants. Algebra can include real numbers, complex numbers,
matrices, vectors etc.

11.3 Why Do I Need Algebra?

Only you can answer this question. I've always said math is an opportunity gateway and you can't get to higher Maths without taking algebra. Algebra develops your thinking, specifically logic, patterns, problem solving, deductive and inductive reasoning. The more math you have, the greater the opportunity for jobs in engineering, actuary, Physics, programming etc. Higher math is often an important requirement for entrance to college or universities. Ultimately, you need to do your own homework to determine if your goals mean sticking it out in math, however, you can never go wrong if you do

11.4 Algebraic Expression
An algebraic Expressions is an expression formed from any combination of numbers and variables by using the operations of addition, subtraction, multiplication, division, exponentiation (raising to powers), or extraction of roots.

algebraic expression in certain variables, we mean an expression that contains only those variables, and by a constant, we mean an algebraic expression that contains no variables at all. If numbers are substituted for the variables in an algebraic expression, the resulting number is called the value of the expression for these values of the variables.

11.5 Algebraic Terms

The basic unit of an algebraic expression is a term. In general, a term is either a number or a product of a number and one or more variables. Below is the term –3ax.

11.6 How to Form Algebraic Expressions?
Example 1: Sarita has some marbles. Ameena has 10 more. Appu says that he has 3 more marbles than the number of marbles Sarita and Ameena together have. How do you get the number of marbles that Appu has?

Solution: Since it is not given how many marbles Sarita has, we shall take it to be x. Ameena then has 10 more, i.e., x + 10. Appu says that he has 3 more marbles than what Sarita and Ameena have together. So we take the sum of the numbers of Sarita’s marbles and Ameena’s marbles, and to this sum add 3, that is, we take the sum of x, x + 10 and 3.

Example 2: Ramu’s father’s present age is 3 times Ramu’s age. Ramu’s grandfather’s age is 13 years more than the sum of Ramu’s age and Ramu’s father’s age. How do you find Ramu’s grandfather’s age?

Solution: Since Ramu’s age is not given, let us take it to be y years. Then his father’s age is 3y years. To find Ramu’s grandfather’s age we have to take the sum of Ramu’s age (y) and his father’s age (3y) and to the sum add 13, that is, we have to take the sum of y, 3y and 13.

Example 3: In a garden, roses and marigolds are planted in square plots. The length of the square plot in which marigolds are planted is 3 meters greater than the length of the square plot in which roses are planted. How much bigger in area is the marigold plot than the rose plot?

Solution: Let us take l meters to be length of the side of the rose plot. The length of the side of the marigold plot will be (l + 3) meters. Their respective areas will be l2 and (l+ 3) 2. The difference between (l+ 3) 2 and l2 will decide how much bigger in area the marigold plot is. In all the three situations, we had to carry out addition or subtraction of algebraic expressions. There are a number of real life problems in which we need to use expressions and do arithmetic operations on them. In this section, we shall see how algebraic expressions are added and subtracted.

11.7 Addition and Subtraction of Like Terms
The simplest expressions are monomials. They consist of only one term. To begin with we shall learn how to add or subtract like terms.

Example 4: Add 3x and 4x.

Solution: To add 3x and 4x, keep the variable part as it is. Just add the co-efficients of each like term. i.e., 3x + 4x = (3 × x) + (4 × x)
= (3 + 4) × x (using distributive law) = 7 × x = 7x or 3x + 4x = 7x 

Example 5: Add 8xy, 4xy and 2xy

Solution: 8xy + 4xy + 2xy = (8 + 4 + 2) × xy = 14 × xy = 14xy or 8xy + 4xy + 2xy = 14xy

Example 6 : Subtract 4n from 7n.

Solution: 7n – 4n = (7 × n) – (4 × n) = (7 – 4) × n = 3 × n = 3n or 7n – 4n = 3n

Example 7: Subtract 5ab from 11ab.

Solution: 11ab – 5ab = (11 – 5) ab = 6ab

Thus, the sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms. Similarly, the difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.

Note: Unlike terms cannot be added or subtracted the way like terms are added or subtracted. We have already seen examples of this, when 5 is added to x, we write the result as (x + 5). Observe that in (x + 5) both the terms 5 and x are retained. Similarly, if we add the unlike terms 3xy and 7, the sum is 3xy + 7. If we subtract 7 from 3xy, the result is 3xy – 7.

11.8 Adding and Subtracting General Algebraic Expressions

Let us take some examples:

Example 8: Add 3x + 11 and 7x – 5

Solution:

(3x + 11) + (7x – 5) = 3x + 11 + 7x – 5

= 3x + 7x + 11 – 5 (rearranging terms)

= (3x + 7x) + (11 – 5) (grouping like terms) = 10x + 6

Hence, 3x + 11 + 7x – 5 = 10x + 6

Example 9: Add 3x + 11 + 8z and 7x – 5.

Solution:

(3x + 11 + 8z) + (7x – 5) = 3x + 11 + 8z + 7x – 5\

= (3x + 7x) + (11 – 5) + 8z (grouping like terms) = 10 x + 6 + 8z

Therefore, the sum = 10x + 6 + 8z

Example 10: Subtract a – b from 3a – b + 4

Solution: (3a – b + 4) – (a – b) = 3a – b + 4 – a + b

= (3a – a) + (b – b) + 4 (grouping like terms )

= (3 – 1) a + (1 – 1) b + 4 (using Distributive law ) = 2a + (0) b + 4 = 2a + 4

(or) 3a – b + 4 – (a – b) = 2a + 4

Example 11: Collect like terms and simplify the expression:

12m2 – 9m + 5m – 4m2– 7m + 10

Solution: Rearranging terms, we have

12m2 – 4m2+ 5m – 9m – 7m + 10 = (12 – 4) m2 + (5 – 9 – 7) m + 10

12m 2 – 4m 2+ 5m – 9m – 7m + 10 = 8m2 + (– 4 – 7) m + 10

= 8m2 + (–11) m + 10 = 8m2 – 11m + 10

Example 12: Subtract 24ab – 10b – 18a from 30ab + 12b + 14a.

Solution:

30ab + 12b + 14a – (24ab – 10b – 18a)

= 30ab + 12b + 14a – 24ab + 10b + 18a

= 30ab – 24ab + 12b + 10b + 14a + 18a

= 6ab + 22b + 32a

Alternatively, we write the expressions one below the other with the like terms appearing exactly below like terms as:

9. Finding the Value of an Expression

We know that the value of an algebraic expression depends on the values of the variables forming the expression. There are a number of situations in which we need to find the value of an expression, such as when we wish to check whether a particular value of a variable satisfies a given equation or not. We find values of expressions, also, when we use formulas from geometry and from everyday mathematics. For example, the area of a square is l2, where l is the length of a side of the square. If l = 5 cm., the area is 52 cm2 or 25 cm2; if the side is 10 cm, the area is 102cm2or 100 cm2 and so on. We shall see more such examples in the next section.

11.10 Simple Equation

Equation: Any equation is a condition on a variable. It is satisfied only for a definite value of the variable

Note: An equation has an equal sign (=) between its two sides. The equation says that the value of the left hand side (LHS) is equal to the value of the right hand side (RHS). If the LHS is not equal to the RHS, we do not get an equation. For example , The statement 2n is greater than 10, i.e. 2n > 10 is not an equation. Similarly, the statement 2n is smaller than 10 i.e. 2n < 10 is not an equation.

Solution of an Equation: The value of the variable in an equation which satisfies the equation is called a solution to the equation. Thus, n = 5 is a solution to the equation 2 n = 10.

11.11 Introduction

In this section we shall look at some simple equations and the methods used to find their solution. There are four basic rules:

Rule 1: An equal quantity may be added to both sides of an equation.

Rule 2: An equal quantity may be subtracted form both sides of an equation.

Rule 3: An equal quantity may multiply both sides of an equation.

Rule 4: An equal non-zero quantity may divide both sides of an equation.

11.12 Further Equations
We are now ready to move on to slightly more sophisticated examples.
Example 19: Find the solution to the equation
5 (x – 3) – 7(6 – x) = 24 – 3(8 – x) – 3
Solution: Removing the brackets from both sides first and then simplifying:
5 (x – 3) – 7(6 – x) = 24 – 3(8 – x) – 3
5x – 15 – 42 + 7x = 24 – 24 + 3x – 3
5x + 7x – 15 – 42 = 3x – 3
12x – 57 = 3x – 3
Adding 57 to both sides:
12x = 3x – 3 + 57 = 3x + 54
Substracting 3x from both sides:
12x – 3x – 3 + 57 giving x = 6

Example 20: Find the solution to each to the following equations.
a) 2x + 3 = 16 – (2x – 3)
b) 8(x – 1) + 17(x – 3) = 4(4x – 9) + 4
c) 15 (x – 1) + 4(x + 3) = 2(7 + x)
Quiz: Which of the following is the solution to the equation
5x – (4x – 7) (3x – 5) = 6 – 3(4x – 9) (x – 1)?
(a) –2 (b) –1 (c) 2 (d) 4
When fractions occur we can sometimes transform the equation to one that does not involve fractions.

Example 21: Find the solution to the equation
(4x/5) – (7/4) = (x/5) + (x/4)
Solution: The least common multiple of the denominators in the equation is 4 × 5 = 20 and we proceed as follows: 

Algebra

What is an Exponent?

Multiplying a Monomial by a Monomial

Multiplying Two Monomials

We begin with 4 × x = x + x + x + x = 4x

Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x

Now, observe the following products.

i) x × 3y = x × 3 × y = 3 × x × y = 3xy

ii) 5x × 3y = 5 × x × 3 × y = 5 × 3 × x × y = 15xy

iii) 5x × (–3y) = 5 × x × (–3) × y

= 5 × (–3) × x × y = –15xy

i.e., coefficient of product = coefficient of first monomial × coefficient of second monomial; i.e., algebraic factor of product = algebraic factor of first monomial× algebraic factor of second monomial. Some more useful examples follow.

iv) 5x × 4x² = (5 × 4) × (x × x²) = 20 × x³ = 20x³

v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz) = –20 × (x × x × yz) = –20x²yz

Multiplying three or More Monomials

i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz

ii) 4xy × 5x²y²× 6x³y³ = (4xy × 5x²y²) × 6x³y³ = 20x³y³ × 6x³y³ = 120x³y³ × x³y³

= 120 (x³ × x³) × (y³ × y³) = 120x⁶ × y⁶ = 120x⁶y⁶

^{Example 1:      Complete the table for area of a rectangle with given length and breadth}

Hence, for (i) volume = (2ax) × (3by) × (5cz)

= 2 × 3 × 5 × (ax) × (by) × (cz)

= 30abcxyz for (ii) volume = m²n × n²p × p²m

= (m² × m) × (n × n²) × (p × p²) = m³n³p³

for (iii) volume = 2q × 4q² × 8q³

= 2 × 4 × 8 × q × q² × q³ = 64q⁶

Multiplying a Monomial by a Polynomial

Multiplying a Monomial by a Binomial

Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ?

Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law,

3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x

We commonly use distributive law in our calculations.

For example:

7 × 106 = 7 × (100 + 6)

= 7 × 100 + 7 × 6 (Here, we used distributive law)

 = 700 + 42 = 742

7 × 38 = 7 × (40 – 2)

 = 7 × 40 – 7 × 2 (Here, we used distributive law)

= 280 – 14 = 266

 Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x and 5xy × (y² + 3) = (5xy × y²) + (5xy × 3) = 5xy³ + 15xy

Multiplying a Monomial b y a Trinomial

3p × (4p² + 5p + 7) = (3p × 4p²) + (3p × 5p) + (3p × 7)

 = 12p³ + 15p² + 21p

Multiply each term of the trinomial by the monomial and add products.

Observe, by using the distributive law, we are able to carry out the multiplication term by term.

Example 3:Simplify the expressions and evaluate them as directed:

x (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for y = –2

 Solution: i)    x (x – 3) + 2 = x² – 3x + 2

 For x = 1, x² – 3x + 2 = (1) ² – 3 (1) + 2

= 1 – 3 + 2

 = 3 – 3 = 0

ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y² – 21y – 3y + 12 – 63

= 6y² – 24y – 51

For y    = –2, 6y² – 24y – 51 = 6 (–2) ² – 24(–2) – 51

= 6 × 4 + 24 × 2 – 51

= 24 + 48 – 51 = 72 – 51 = 21

Example 4:      Add

5m (3 – m) and 6m² – 13m (ii) 4y (3y² + 5y – 7) and 2 (y³ – 4y² + 5)

Solution:  i)    First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5m²

Now adding the second expression to it,15m – 5m² + 6m² – 13m = m² + 2m

ii) The first expression

 = 4y (3y² + 5y – 7) = (4y × 3y²) + (4y × 5y) + (4y × (–7))  

= 12y³+ 20y² – 28y

The second expression = 2 (y³ – 4² + 5) = 2y³ + 2 × (– 4y²) + 2 × 5

= 2y³ – 8y² + 10

Adding the two expressions,

12y³ + 20y² – 28y+ 2y³ – 8y² + 10 = 14y³ + 12y² – 28y + 10

Example 5: Subtract 3pq (p – q) from 2pq (p + q).

3. Multiplying a Polynomial by a Polynomial

Multiplying a Binomial by a Binomial

Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this step-bystep, as we did in earlier cases, following the distributive law of multiplication,

4. What is an Identity?

Consider the equality (a + 1) (a +2) = a² + 3a + 2

We shall evaluate both sides of this equality for some value of a, say a = 10.

For a = 10,

LHS = (a + 1) (a + 2) = (10 + 1) (10 + 2) = 11 × 12 = 132

RHS = a² + 3a + 2 = 10² + 3 × 10 + 2 = 100 + 30 + 2 = 132

Thus, the values of the two sides of the equality are equal for a = 10.

Let us now take a = –5

LHS = (a + 1) (a + 2) = (–5 + 1) (–5 + 2) = (– 4) × (–3) = 12

RHS = a² + 3a + 2 = (–5)² + 3 (–5) + 2= 25 – 15 + 2 = 10 + 2 = 12

Thus, for a = –5, also LHS = RHS.

We shall find that for any value of a, LHS = RHS

Such equality, true for every value of the variable in it, is called an identity.

Thus, (a + 1) (a + 2) = a² + 3a + 2 is an identity.

An equation is true for only certain values of the variable in it. It is not true for all values of the variable.

For example, consider the equation a² + 3a + 2 = 132

It is true for a = 10, as seen above, but it is not true for a = –5 or for a = 0 etc.

6. Factorisation

Factorisation of an algebraic expression means rewriting the given expressions in terms of product of its factors. These factors may be numbers, algebraic variables or algebraic expressions.

Expressions like 3xy, x y, 2, 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. On the other hand consider expressions like 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6.

It is not obvious what their factors are. We need to develop systematic methods to factorise these expressions, i.e., to find their factors.

Methods to do Factorisation

1. Method of common factors

2. Factorisation by regrouping

3. Factorisation using identities

Example 14: Factorise 2x + 4.
Solution: 2x = 2 × x
4 = 2 × 2

Hence 2x + 4 = (2 × x) + (2 × 2) [2 is common factor to both the terms]
2x + 4 = 2 × (x + 2) [ by distributive law]
= 2 (x + 2)

Example 15: Factorise 5xy + 10x.
Solution: The irreducible factor forms of 5xy and 10x are respectively,
5xy = 5 × x × y
10x = 2 × 5 × x [5 and x are common factors to both the terms]
5xy + 10x = (5 × x × y) + (5 × x × 2)
= (5x × y) + (5x × 2)
Therefore, 5xy + 10x = 5 x (y + 2) [by distributive law]

Example 16: Factorise 12a²b + 15ab²
Solution: We have 12a2b = 2 × 2 × 3 × a × a × b
15ab² = 3 × 5 × a × b × b [3, a and b are common factors to both the terms]
Therefore, 12a²b + 15ab² = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
= 3ab × (4a + 5b)
= 3ab (4a + 5b) (required factor form)

Example 17: Factorise 10x² – 18x³ + 14x⁴
Solution: 10x² = 2 × 5 × x × x
18x³ = 2 × 3 × 3 × x × x × x
14x⁴ = 2 × 7 × x × x × x × x [2, x and x are common factors to all the three terms]
Therefore, 10x² – 18x³ + 14x⁴ = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x)
+ (2 × x × x × 7 × x × x)
= 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)]
= 2x²× (5 – 9x + 7x²) = 2x²(7x²– 9x+5)

Factorisation By Regrouping Terms

Example 17: Factorise 2xy + 2y + 3x + 3.
Solution: Look at the expression 2xy + 2y + 3x + 3.
Notice that the first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms.
Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)
Similarly, 3x + 3 = (3 × x) + (3 × 1)
= 3 × (x + 1) = 3 ( x + 1)
Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)
= (x + 1) (2y + 3)

Example 6: Factorise 6xy – 4y + 6 – 9x.
Solution: Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) (a)
–9x + 6 = –3 (3x) + 3 (2)

= – 3 (3x – 2) (b)
Step 3 Putting (a) and (b) together,
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3)
Factorisation Using Identities
Standard Identities:
1. (a + b)2 = a2 + 2ab + b2
2. (a – b) 2= a2 – 2ab + b2
3. (a + b) (a – b) = a2 – b2
4. ( x + a ) (x + b) = x2 + (a + b) x + ab
.
Example7: Factorise x2 + 8x + 16
Solution: Observe the expression; it has three terms. Therefore, it does not fit Identity 3.
Also, it’s first and third terms are perfect squares with a positive sign before the middle term. So, it is of the form a² + 2ab + b² where a = x and b = 4 such that a²+ 2ab + b² = x²+ 2 (x) (4) + 42 = x² + 8x + 16
Since a² + 2ab + b² = (a + b) 2, by comparison x² + 8x + 16 = ( x + 4) 2 (the required factorisation)

Example 8: Factorise 4y2 – 12y + 9
Solution: Observe 4y²= (2y) 2, 9 = 32 and 12y = 2 × 3 × (2y)
Therefore, 4y² – 12y + 9 = (2y) 2 – 2 × 3 × (2y) + (3) 2
= (2y – 3) 2 (required factorisation)

Example 10: Factorise a² – 2ab + b² – c²
Solution: The first three terms of the given expression form (a – b) 2. The fourth term is a square. So the expression can be reduced to a difference of two squares.
Thus, a² – 2ab + b² – c² = (a – b)²– c² (Applying Identity 2)
= [(a – b) – c) ((a – b) + c)] (Applying Identity 3)
= (a – b – c) (a – b + c) (required factorisation)

Example 11: Factorise m4 – 256
Solution: Note m⁴ = (m²) 2and 256 = (16)²
Thus, the given expression fits Identity 3.
Therefore, m⁴ – 256 = (m²)² – (16)²
= (m² –16) (m² +16) [(using Identity 3]
Now, (m² + 16) cannot be factorised further, but (m² –16) is factorisable again as per Identity 3.
m² –16 = m² – 42 = (m – 4) (m + 4)
Therefore, m² – 256 = (m – 4) (m + 4) (m² +16)

Example 12: Factorise x² + 5x + 6
Solution: If we compare the R.H.S. of Identity 4 with x² + 5x + 6, we find ab = 6, and a + b = 5. From this, we must obtain a and b. The factors then will be (x + a) and (x + b).
If ab = 6, it means that a and b are factors of 6.
Let us try a = 6, b = 1. For these values a + b = 7, and not 5,.
So this choice is not right.
Let us try a = 2, b = 3. For this a + b = 5 exactly as required.
The factorised form of this given expression is then (x +2) (x + 3).

Example 13: Find the factors of y2 –7y +12.
Solution: We note 12 = 3 × 4 and 3 + 4 = 7. Therefore,
y² – 7y+ 12 = y²  – 3y – 4y + 12
= y (y –3) – 4 (y –3) = (y –3) (y – 4)

Example 14: Obtain the factors of z2 – 4z – 12.
Solution: Here a b = –12 ; this means one of a and b is negative. Further, a + b = – 4, this
means the one with larger numerical value is negative.
We try a = – 4, b = 3; but this will not work, since a + b = –1.
Next possible values are a = – 6, b = 2, so that a + b = – 4 as required.
Hence, z² – 4z –12 = z² – 6z + 2z –12
= z(z – 6) + 2(z – 6 )
= (z – 6) (z + 2)

Example 15: Find the factors of 3m² + 9m + 6.
Solution: We notice that 3 is a common factor of all the terms.
Therefore, 3m² + 9m + 6 = 3(m² + 3m + 2)
Now, m² + 3m + 2 = m² + m + 2m + 2 (as 2 = 1 × 2)
= m(m + 1)+ 2( m + 1)
= (m + 1) (m + 2)
Therefore, 3m2 + 9m + 6 = 3(m + 1) (m + 2)