ICSE Class 6 Maths Chapter 13 Fundamental Operations

Chapter 13: Fundamental Operations

[Related to Algebraic Expressions]

Basic Concept

In Mathematics, the operations addition (+), subtraction (-), multiplication (×) and division (÷) are the four fundamental operations.

Students are familiar with these operations as they have already studied about these operations in Arithmetic in their lower classes.

Addition of Like Terms

The addition of like terms is a single term (like to the given terms) whose coefficient is equal to the sum of the coefficients of the given (like) terms.

Thus:

(i) Addition of 3x and 8x = 3x + 8x = (3 + 8)x = 11x

(ii) Addition of 8x²y and -5x²y = 8x²y + (-5x²y) = 8x²y - 5x²y = (8 - 3)x²y = 5x²y

(iii) 2xy + 3xy + 5xy = (2 + 3 + 5) = 10xy

(iv) 7y² - 4y² + 3y² = (7 - 4 + 3)y² = (10 - 4)y² = 6y²

For addition, the terms are taken with their given signs, e.g.

(i) addition of 7xy and -3xy = 7xy - 3xy = (7 - 3)xy = 4xy

(ii) addition of -7xy and 3xy = -7xy + 3xy = (-7 + 3)xy = -4xy

and (iii) -7x - 3x = (-7 - 3)x = -10y

In the same way:

(i) addition of -3xy², -5xy² and -xy² = (-3xy²) + (-5xy²) + (-xy²) = -3xy² - 5xy² - xy² = (-3 - 5 - 1)xy² = -9xy²

(ii) addition of 7ab, -2ab, -5ab, 6ab and -ab = 7ab - 2ab - 5ab + 6ab - ab = (7 - 2 - 5 + 6 - 1)ab = (13 - 8)ab = 5ab

Addition of Unlike Terms

As shown above, the sum of two or more like terms is a single like term, but two unlike terms cannot be added together to get a single term.

For example: the unlike terms 2ab and 4bc cannot be added together to form a single term. All that can be done is to connect them by the sign of addition and leave the result in the form 2ab + 4bc.

In the same way,

(i) addition of 5x² and 8xy = 5x² + 8xy

(ii) addition of 2y³, -5xy and 3x³ = 2y³ - 5xy + 3x³ and so on

Teacher's Note

Like terms in algebra work similarly to sorting coins by denomination - you can only combine pennies with pennies and dimes with dimes, not mix them together into a single value.

Subtraction of Like Terms

For subtraction of like terms, the rules are the same as those for subtraction of integers.

For example:

Since - 4 + 2 = -2, therefore -4x + 2x = -2x

Since 3 - 7 = -4, therefore 3x - 7x = -4x

Example 1

Subtract: (i) 4x from -8x, (ii) -3x from -7x

Solution

In each subtraction, change the sign of the term to be subtracted.

(i) -8x - (4x) = -8x - 4x = -12x

(ii) -7x - (-3x) = -7x + 3x = -4x

The result of subtraction of two like terms is also a like term.

Subtraction of Unlike Terms

Just as it is with addition of unlike terms, we cannot get a single term by the subtraction of unlike terms. For example, 2ab and 4bc are two unlike terms; the subtraction of 2ab from 4bc is 4bc - 2ab, which cannot be simplified further to get a single term.

Similarly, the subtraction of 4bc from 2ab is 2ab - 4bc, which cannot be simplified further to get a single term.

Example 2

Evaluate: (i) 3x - 4x + 7x, (ii) 6ab + 3ab - 4ab, (iii) 5ax + ax - 8ax, (iv) 8a + 3a - 5a - 2a

Solution

1. Add the positive terms together and separately add the negative terms together as well.

2. Find the result of the two terms obtained.

(i) 3x - 4x + 7x = 10x - 4x = 6x

(ii) 6ab + 3ab - 4ab = 9ab - 4ab = 5ab

(iii) 5ax + ax - 8ax = 6ax - 8ax = -2ax

(iv) 8a + 3a - 5a - 2a = 11a - 7a = 4a

Example 3

Evaluate:

(i) 3x + 1\(\frac{2}{5}\)x, (ii) 5a - 2\(\frac{1}{2}\)a + 1\(\frac{1}{2}\)a, (iii) 4\(\frac{5}{6}\)xy - 2\(\frac{1}{3}\)xy - 1\(\frac{1}{2}\)xy

Solution

(i) 3x + 1\(\frac{2}{5}\)x = \(\frac{3x}{1}\) + \(\frac{7x}{5}\) = \(\frac{15x + 7x}{5}\) = \(\frac{22x}{5}\) = 4\(\frac{2}{5}\)x

(ii) 5a - 2\(\frac{1}{2}\)a + 1\(\frac{1}{2}\)a = \(\frac{5a}{1}\) - \(\frac{5a}{2}\) + \(\frac{3a}{2}\) = \(\frac{10a - 5a + 3a}{2}\) = \(\frac{13a - 5a}{2}\) = \(\frac{8a}{2}\) = 4a

(iii) 4\(\frac{5}{6}\)xy - 2\(\frac{1}{3}\)xy - 1\(\frac{1}{2}\)xy = \(\frac{29}{6}\)xy - \(\frac{7}{3}\)xy - \(\frac{3}{2}\)xy = \(\frac{29xy - 14xy - 9xy}{6}\) = \(\frac{29xy - 23xy}{6}\) = \(\frac{6xy}{6}\) = xy

Teacher's Note

Combining fractions with variables mirrors how you would combine fractional measurements like cups and tablespoons in a recipe - you need a common denominator.

Exercise 13(A)

1. Fill in the blanks:

(i) 5 + 4 = ......................... and 5x + 4x = .............................

(ii) 12 + 18 = ......................... and 12x²y + 18x²y = .............................

(iii) 7 + 16 = ......................... and 7a + 16b = ........................... .

(iv) 1 + 3 = ......................... and x²y + 3xy² = .............................

(v) 7 - 4 = ......................... and 7ab - 4ab = .............................

(vi) 12 - 5 = ......................... and 12x - 5y = .............................

(vii) 35 - 16 = ......................... and 35ab - 16ba = .............................

(viii) 28 - 13 = ......................... and 28ax² - 13a²x = .............................

2. Fill in the blanks:

(i) The sum of -2 and -5 = ................... and the sum of -2x and -5x = ....................

(ii) The sum of 8 and -3 = .................... and the sum of 8ab and -3ab = ....................

(iii) The sum of -15 and -4 = ................ and the sum of -15x and -4y = .....................

(iv) 15 + 8 + 3 = ..................... and 15x + 8y + 3x = ......................

(v) 12 - 9 + 15 = ..................... and 12ab - 9ab + 15ba = .....................

(vi) 25 - 7 - 9 = ..................... and 25xy - 7xy - 9yx = .....................

(vii) -4 - 6 - 5 = ..................... and -4ax - 6ax - 5ay = ......................

3. Add:

(i) 8xy and 3xy, (ii) 2xyz, xyz and 6xyz, (iii) 2a, 3a and 4b

(iv) 3x and 2y, (v) 5m, 3n and 4p, (vi) 6a, 3a and 9ab

(vii) 3p, 4q and 9q, (viii) 5ab, 4ba and 6b, (ix) 50pq, 30pq and 10pr

(x) -2y, -y and -3y, (xi) -3b and -b, (xii) 5b, -4b and -10b

(xiii) -2c, -c and -5c

4. Evaluate:

(i) 6a -a - 5a - 2a, (ii) 2b - 3b - b + 4b, (iii) 3x - 2x - 4x + 7x

(iv) 5ab + 2ab - 6ab + ab, (v) 8x - 5y - 3x + 10y

5. Evaluate:

(i) -7x + 9x + 2x - 2x, (ii) 5ab - 2ab - 8ab + 6ab

(iii) -8a - 3a + 12a + 13a - 6a, (iv) 19abc - 11abc - 12abc + 14abc

6. Evaluate:

(i) 3\(\frac{1}{2}\)x + 4x - 5\(\frac{1}{2}\)x, (ii) 7b - 2\(\frac{1}{3}\)b - 1\(\frac{1}{3}\)b, (iii) 5ax + 3\(\frac{3}{5}\)ax + \(\frac{2}{5}\)ax

(iv) 9y - 7\(\frac{3}{4}\)y + 2\(\frac{1}{4}\)y, (v) 3xy - 2\(\frac{1}{2}\)xy + 1\(\frac{2}{3}\)xy

7. Subtract the first term from the second:

(i) 4ab, 6ba, (ii) 4-8b, 6-8b, (iii) 3-5abc, 10-5abc, (iv) 3\(\frac{1}{2}\)mn, 8\(\frac{1}{2}\)nm

8. Simplify:

(i) 2a²b² + 5ab² + 8a²b² - 3ab², (ii) 4a + 3b - 2a - b

(iii) 2xy + 4yz + 5xy + 3yz - 6xy, (iv) ab + 15ab - 11ab - 2ab

(v) 6a² - 3b² + 2a² + 5b² - 4a², (vi) 8abc + 2ab - 4abc + ab

(vii) 9xyz + 15xyz - 10zyx - 2zxy, (viii) 13pqr + 2p + 4q - 6pqr + 5pqr

(ix) 4ab + 0 - 2ba, (x) 6x²y - 2xy² + 5x²y - xy²

(xi) 6-4a + 5-3b - 2-4a - 2-2b, (xii) 2-5a + 4-6b + 1-2a - 3-6b

(xiii) 22m -12\(\frac{1}{2}\)n -15p + 16n, (xiv) 6p + \(\frac{2}{3}\)q - 1\(\frac{1}{2}\)p + \(\frac{1}{3}\)q + 2q

(xv) 2\(\frac{2}{3}\)xy - 3\(\frac{1}{2}\)xy + 3\(\frac{1}{3}\)xy - 2\(\frac{1}{2}\)xy

More About Addition and Subtraction

Addition of Polynomials

Example 4

Add: 4a + 2b, 3a - 3b + c and -2a + 4b + 2c.

Solution

First Method (Row Method)

Steps

1. Write all the given polynomials in a row.

2. Group the like terms.

3. Add the like terms.

The required addition = (4a + 2b) + (3a - 3b + c) + (-2a + 4b + 2c) = 4a + 2b + 3a - 3b + c - 2a + 4b + 2c = 4a + 3a - 2a + 2b - 3b + 4b + c + 2c = 5a + 3b + 3c

Second Method (Column Method)

Arrange the given polynomials so that the like terms of the polynomials are one below the other in a vertical column; then add.

4a + 2b, 3a - 3b + c, -2a + 4b + 2c

In general, the column method is preferred.

5a + 3b + 3c

Example 5

(i) Add: 3x³ - 5x² + 8x + 10, 15x³ - 6x - 23 and 9x² - 4x + 15.

(ii) Add: 3ab² - 2b² + a², 5a²b - 2ab² - 3a² and 8a² - 5b².

Solution

Using the row method:

(i) (3x³ - 5x² + 8x + 10) + (15x³ - 6x - 23) + (9x² - 4x + 15) = 3x³ - 5x² + 8x + 10 + 15x³ - 6x - 23 + 9x² - 4x + 15 = 3x³ + 15x³ - 5x² + 9x² + 8x - 6x - 4x + 10 - 23 + 15 = 18x³ + 4x² + 8x - 10x + 25 - 23 = 18x³ + 4x² - 2x + 2

(ii) (3ab² - 2b² + a²) + (5a²b - 2ab² - 3a²) + (8a² - 5b²) = 3ab² - 2b² + a² + 5a²b - 2ab² - 3a² + 8a² - 5b² = 5a²b + 3ab² - 2ab² + a² - 3a² + 8a² - 2b² - 5b² = 5a²b + ab² + 6a² - 7b²

Using the column method:

(i) 3x³ - 5x² + 8x + 10, 15x³ - 6x - 23, 9x² - 4x + 15

18x³ + 4x² - 2x + 2

(ii) 3ab² - 2b² + a², 5a²b - 2ab² - 3a², -5b² + 8a²

5a²b + ab² - 7b² + 6a²

Teacher's Note

Adding polynomials is like combining inventory from multiple stores - you group items of the same type (like terms) and count them all together.

Subtraction in Polynomials

Steps (for the row method)

1. Enclose the expression to be subtracted in brackets with a minus sign prefixed.

2. Remove the bracket by changing the sign of each term kept in the bracket.

Examples:

(i) (2x - y) - (x + 5y) = 2x - y - x - 5y

(ii) (3a + b - c) - (2a - 3b + c) = 3a + b - c - 2a + 3b - c, etc.

3. Combine the like terms and add.

Example 6

Subtract: 3a - 4b + 5c from 4a - b + 6c.

Solution

4a - b + 6c - (3a - 4b + 5c) = 4a - b + 6c - 3a + 4b - 5c = 4a - 3a - b + 4b + 6c - 5c = a + 3b + c

Row method is used

Whenever there is a negative sign before a bracket, open (remove) the bracket and, at the same time, change the sign of each term inside the bracket.

e.g. (x + y) - (x - y + z) = x + y - x + y - z = 2y - z

Alternative Method (Column Method)

Steps (for the column method)

1. Rewrite the given expressions in two lines (rows) such that the lower line is the expression to be subtracted and like terms of both the expressions are one below the other.

2. Change the sign of each term in the lower line, i.e. change the sign of each term of the expression to be subtracted.

3. Add column-wise.

Thus, for Example 6 given above, we have:

Step 1: 4a - b + 6c, 3a - 4b + 5c

Step 2: - + -

Step 3: a + 3b + c

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