ALGEBRA

1 Introduction to Identities

In earlier classes, we have learnt to construct algebraic expressions using exponential notations. For example, consider the algebraic expression (a+b)2 = a2+2ab+b2 . Let us try to find the values of the expression for the given values of a and b.

When a = 3 and b = 5,

L.H.S =

R.H.S

Thus, for a =3 and b =5 , L.H.S = R.H.S

Similarly, when a = 4 and b = 7,

L.H.S =

R.H.S

Also, for a = 4 and b =7 , L.H.S = R.H.S

Thus, we shall find that for any value of ‘a’ and ‘b’ L.H.S = R.H.S. Such an equality, which is true for every value of the variable in it is called an identity. Thus, we observe that the equation (a+b)2 = a2+2ab+b2 is an identity.

In general, algebraic equalities which hold true for all the values of the variables are called Identities.

2 Geometrical Approach to Multiplication of Monomials

We have already learnt that the operation of multiplication can be modelled in different ways. The one that we use in this unit is the representation of multiplication as a product table that is similar to area.

For example, the product 53  can be represented as shown in Fig. 3.1, which has five rows and three columns and it comprises 15 small squares. From Fig. 3.2, it is also clear that the product of 53  is the same as the product of 35  [Since, multiplication is commutative].

                   

Here, multiplication is represented using grid model. The same can be represented using area model.

The area model helps us to understand multiplication by decomposing the area of large rectangle into areas of smaller rectangles. Also the same example which is discussed above may be decomposed as

This decomposition model is very useful when we are finding the product of large numbers.

3 Geometrical proof of Identities

By using this concept of multiplication of monomials, let us try to prove the identities geometrically, which are very much useful in solving algebraic problems.

3.1 Identity-1:

Consider four regions. One region is square shaped with dimension 33  (Grey). Also, the other three regions are rectangle in shape with dimensions 43  (yellow), 32  (red) and 42  (Blue). Arrange these four regions to form a rectangular shape as shown in the Fig. 3.8.

By observing Fig. 3.8, we can note that,

Area of the bigger rectangle = Area of a square (Grey ) + Area of Three rectangles

Where, LHS is 

RHS is 

Therefore, LHS = RHS.

In the similar way as explained above, let us check for another set of four regions as shown in the Fig. 3.9.

By observing Fig. 3.9, we can note that,

Area of the bigger rectangle = Area of a square (Grey ) + Area of Three rectangles

Therefore, LHS = RHS.

Thus, equation (1) and (2) is true for given set of any three values.

By generalising those three values as ‘x’, ‘a’ and ‘b’

we get,

 That is,  

Hence, is an identity.

Now, let us prove this identity, geometrically.

Let one side of a rectangle be (x +a) and the other side be (x + b) units.

Then, the total area of the rectangle  

From the Fig. 3.10, we can see that the

Area of the rectangle ABCD = area of the square AFIE + area of the rectangle FBGI + area of the       

rectangle EIHD + area of the rectangle IGCH

From (3) and (4) we get,

Hence, is an identity.

 

Example 1 Simplify the following using the identity

           

Solution

Let us represent the expression geometrically, as shown in the Fig. 3.11.

In the rectangle with length (x +5) and breadth (x +3 ), we get,

Area of bigger rectangle = Area of a square + Area of three rectangles

Therefore,

Let us represent the expression geometrically, as shown in Fig. 3.12.

In the rectangle with length   (y +6) and breadth (y +8) units, we get,

Area of bigger rectangle = Area of square + Area of three rectangles

Therefore, 

            We know the identity  

            Taking, x = 40, a=3 and b= –4, we get

           

3.4 Identity-4:

In the given figure, AB = AD = a.

So, area of square ABCD =a2.

Also, SB = DP = b.

Area of the rectangle SBCT = ab.

Similarly, area of the rectangle DPRC = ab.

Also, area of the square TQRC = b2.

Area of the rectangle DPQT = ab – b2.

Now, AS = PQ = (a − b) and AP = SQ = (a + b).

Hence, area of the rectangle APQS       =         area of square ABCD – area of rectangle STCB

         (the shaded rectangle)                                          + area of rectangle DPQT

                                                            = a2ab + (ab - b2)

                                                            = a2ab + ab - b2

                                                            = a2 – b2

Hence, (a + b) (a – b) = a2 – b2

Example 2 Simplify by using the identity (a + b) (a – b) = a2 – b2

                        (i) (3x + 4)(3x − 4)           (ii) 53 × 47.

Solution

(i)             

 

Substitute, a = 3x and b = 4 in the identity (a + b) (a – b) = a2 – b2, we get,

(ii)          53 × 47 = (50 + 3) × (50 − 3).

 

Take, a = 50 and b = 3,

Substituting the values of ‘a’ and ‘b’ in the identity(a + b) (a – b) = a2 – b2, we get

4 Factorisation using identities

A factor of a number is the number that exactly divides a given number without leaving any remainder. For example, the factors of 4 are 1, 2 and 4; the factors of 12 are 1, 2, 3, 4, 6 and 12. And, the number can be expressed as product of its factors such as 12 = 1 × 12 or 2 × 6 or 3 × 4.

In the same way, algebraic expressions also have factors that divide the expression exactly. Given an algebraic expression, the factors of that algebraic expression are two or more expressions whose product is the given expression. For example,  hence the factors of the algebraic expression  are x, x and y.

Example 3 Express the following algebraic expressions as the product of its factors:

                       

Solution

 

Example  Factorise by using the identity:

                       

Solution

Example      Factorise

Solution

           

This expression is in the form of identity 

Hence,

Therefore, the factors of 

5 Inequations

An algebraic statement that shows two algebraic expressions being unequal is known as an algebraic inequation.

In general, when two expressions are compared, one might be; less than (), greater than or equal to (≥) the other.

In an inequation, the algebraic expressions are connected by one out of the four signs of inequalities, namely, >, ≥, < and ≤.

5.1 Solving linear Inequations

A simple linear equation has atmost one solution, but a linear inequation may have many solutions.

To solve an inequation, it is necessary to know the set of values that the variable symbol can be substituted with. The collection of of all such values of an inequation is known as solution of the inequation.

For example, the solution of the equation 3x − 3 = 12 is 5. (How?) Let us find the solution for the inequation 3x − 3 < 12, where x is a natural number. Note that, the solutions of this inequation are ‘natural numbers’. Now,

Add 3 on both sides, we get 3x − 3 + 3 < 12 + 3 3x < 15

Divide by 3 on both sides, we get

Hence, x takes value which is less than 5 and x is a natural number. Thus, the solution for this inequation are 1, 2, 3 and 4.

Note:

When ‘x’ is not restricted to a natural number, the solution includes all values less than 5.

Rules to solve inequation

While solving an inequation, the rules for transposition in case of inequalities are the same as for equations.

1.    Addition of the same number on both sides of the inequation does not change the value of the inequation. Example: 10 > 5 10 + 1 > 5 + 1 11 > 6.

Extending this result, when adding any number ‘x’ instead of 1, the inequality 10 + x > x + 5 remains unchanged.

2.    Subtraction of the same number from both sides of the inequation does not change the value of the inequation. Example: 10 > 5 10 1 > 5 − 1 9 > 4.

Extending this result, when subtracting any number x instead of 1, the inequatity 10− x > x − 5remains unchanged.

3.    Multiplication by the same positive number on both sides of the inequation does not change the value of the inequation. Example: 10 > 5 10 × 2 > 5 × 2 20 > 10.

Similarly, when multiplying any positive number x instead of 2, the inequatity 10 × x > x × 5 remains unchanged.

4.    Division by the same non-zero positive number on both sides of the inequation does not change the value of the inequation

 

Example  Solve: 2x + 4 < 18, where x is a natural number.

Solution

2x + 4 < 18

2x + 4 – 4 < 18 − 4 [Subtracting 4 from both sides]

2x < 14 [Divide by 2 on both sides]

x < 7

Since the solution belongs to natural numbers, that are less than 7, we take the values of the x as 1, 2, 3, 4, 5 and 6.

Therefore, the solutions are 1, 2, 3, 4, 5 and 6.

Example   Solve: 5 − 7x ≥ 33, where x is an integer.

Solution

 

x ≤−4 [since, it is divided by a negative number, the inequality is reversed]

Since, solution belongs to the set of integers, that are less than −4, we take the values of x as –4, –5, –6, ...

Therefore, the solutions are –4, –5, –6,... .

5.2 Graphical representation of Inequation

The solutions of an inequation can be represented on the number line by marking the true values of solutions with different colour on the number line.

Look into the following inequations and its graphical representation on number line. Here, we consider the solution belongs to natural numbers. That is, each and every value of the solution is a natural number.

1.    When x < 3, the solution in natural numbers are 1 and 2. Its graph on number line is shown below:

2.    When x ≥ 3, the solutions are natural numbers 3, 4, 5, ... and its graph is as shown below:

3.    To mark the values represented by the inequation 2 ≤ x ≤ 5, the solutions are set of natural numbers 2, 3, 4 and 5 and its graph is as given below:

 

Example   Represent the solutions −8 < 2x < 10 in a number line, where x is a natural number.

Solution

Since the solution belongs to the set of natural numbers, the solutions are 1, 2, 3 and 4.

It’s graph on the number line is shown below:

Example 

Solve the inequation: −2 ≤ z + 3 ≤ 5, where z is an integer. Also, represent the solution, graphically.

Solution

Since the solution belongs to integers, the solutions are –5, –4 , –3, –2, –1, 0, 1 and 2.

It’s graph on the number line is shown below: