algebraic expressions word problems class 6

Adding Algebraic Expressions Word Problems

In mathematics, just like we add many numbers as we can and find the sum, we add two or many algebraic expressions too. However, for the addition of algebraic expressions, we combine all the like terms and then add them.

Like terms are the terms that have the same power for the same variables. In like terms, one can only change the numerical coefficient. Terms that have different variables or the same variables raised to different powers are known as, unlike terms.

Word problems that include addition of algebraic expressions can be solved with ease if we know how to add them. We need to write out an algebraic expression that we can evaluate to find our answer. To write our algebraic expression, we carefully read the problem to figure out what our important numbers are and what kind of operation we are dealing with.

There are two methods to do the addition of algebraic expressions: Horizontal method of Addition of Algebraic Expressions Column method for Addition of Algebraic Expressions

Horizontal method of Addition of Algebraic Expressions 1. Write all the expressions in a horizontal line by putting them into brackets and put an addition sign in between. 2. Group all the like terms together from all the expressions and rewrite the expression so formed. 3. Add numerical coefficients of all the like terms followed by the common variable. 4. Rewrite the simplified expression, and make sure all the terms in the final answer should be unlike terms.

Column method for Addition of Algebraic Expressions 1. Write all the expressions one below the other. Make sure to like terms in one column. If there a term whose like term is not there in the second expression, then either write below it or leave that column blank. 2. Add the numerical coefficient of each column (like terms) and write below it in the same column followed by the common variable.

algebraic expressions word problems class 6

We know that the perimeter of a rhombus p = 2l + 2w

p = 2(x – 1) + 2(2x + 4) p = (2x – 2) + (4x + 8) p = 6x + 6

When x = 5, the perimeter will be:

p = 6x + 6 6(5) + 6 = 36

Linear expression in simplest form: p = 6x + 6 Perimeter = 36 cm

Practice Adding Algebraic Expressions Word Problems

Practice Problem 1

Practice Problem 2

Practice Problem 3

Adding Algebraic Expressions Word Problems Practice Problem 3

Math Vocabulary

Linear Expression – an algebraic expression in which the variable is raised to the first power, and variables are not multiplied or divided.

Term – either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or − signs.

Like terms – Terms that have the same power for the same variables. In like terms, one can only change the numerical coefficient.

Unlike terms – Terms that have different variables or the same variables raised to different powers.

Distributive Property – to multiply a sum or difference by a number, multiply each term inside the parenthesis by the number outside the parenthesis.

Constant – a term without a variable.

Variable – In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation or expression.

Coefficient – is an integer that is multiplied with the variable of a single term or the terms of a polynomial.

Pre-requisite Skills Evaluate Variable Expressions-addition and subtraction Evaluate Expressions (Multiplication and Division) Solve Algebraic Equations – Addition and Subtraction Solve Algebraic Equations – Multiplication and Division Evaluate Algebraic Expressions Write Algebraic Expressions Arithmetic Sequences Geometric Sequences The Distributive Property Simplifying Complex Algebraic Expressions

Related Skills Subtract Linear Expressions Factor Linear Expressions

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RS Aggarwal Class 6 Solutions Chapter-8 Algebraic Expressions

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Algebraic Expressions Solutions By RS Aggarwal - Free PDF Download

Vedantu provides the solutions of RS Aggarwal Class 6 Math Chapter 8. The topic of Class 6 Chapter 4 of Mathematics is Algebraic Expression. The chapter Algebraic expression in class 6 RS Aggarwal includes the basic terminology of algebra, forming expressions, and solving expressions. The way the solutions are provided by Vedantu is easily understandable. The stepwise solutions of every sum of the chapter give the student an enhanced insight into the chapter. You can download the pdf of the solutions of RS Aggarwal Chapter 8 for free from Vedantu. You can also register Online for NCERT Class 6 Science tuition on Vedantu.com to score more marks in CBSE board examination.

Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Vedantu.com is No.1 Online Tutoring Company in India Provides you Free PDF download of NCERT Math Class 6 solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter wise Questions with Solutions to help you to revise complete Syllabus and Score More marks in your examinations.

RS Aggarwal Solutions for Class 6 Math Chapter 8

We have provided step by step solutions for all exercise questions given in the pdf of Class 6 RS Aggarwal Chapter-8 Algebraic Expressions. All the Exercise questions with solutions in Chapter-8 Algebraic Expressions are given below:

Exercise (Ex 8A) 8.1

Exercise (Ex 8B) 8.2

Exercise (Ex 8C) 8.3

Exercise (Ex 8D) 8.4

At Vedantu, students can also get Class 6 Maths Revision Notes, Formula and Important Questions and also students can refer to the complete Syllabus for Class 6 Maths, Sample Paper and Previous Year Question Paper to prepare for their exams to score more marks.

To Know Algebra We Must Know The Meaning of The Following Definitions

Variables - An unknown entity is called a variable when it changes with a change in the situation.

For example, Suppose 3x+6 is an algebraic expression with variable x. Now, if x=0, the value of the algebraic expression becomes 6, and when x=1, it becomes 9.

Therefore, the value of the algebraic expression changes with a change in x.

Constant - Constant never changes its value, it remains fixed.

For example: In 3x+6, 6 is the constant.

Terms - The number of entities that are either added or subtracted is called terms.

For example, 3x+6 has 2 terms i.e 3x, and 6.

Coefficient - The number with which the variable is multiplied.

For example: In 3x+6, 3 is the coefficient of x.

Like Terms - Terms having the same variables.

For example 3y, 7y, 45y

Unlike Terms - Terms that have different variables.

For example 3xy, 6x, 7y

Addition and Subtraction In Algebra

The coefficients of all the terms are added or subtracted.

All the terms with similar variables are taken together and then, the coefficients are added or subtracted.

2xy-3x+4x+5y-6xy=?

(2xy-6xy)+(-3x+4x)+5y= -4xy+x+5y

Applications In Geometry

The perimeter of a Square- Perimeter of a square can also be calculated with the help of algebra. We know that the sum of the lengths of the sides of a polygon is the perimeter. So, if a square has a side l, then the Perimeter of a square= 4*l

The perimeter of a Rectangle– If a rectangle has l and b as length and breadth respectively, then the perimeter of the rectangle = 2l+2b= 2(l+b)

Example: Find the perimeter of a square with a length of 5 cm.

Perimeter= 4*l = 4*5= 20 cm

Number of diagonals that can be drawn from one vertex of a polygon = n-3 (n is the number of sides of the polygon)

Rules From Arithmetic

Community of Numbers

Addition: A+B=B+A

Multiplication: A*B=B*A

Distributivity of Numbers

A*(B+C)=A*B+A*C

Formation of Algebraic Expression

An algebraic expression is formed with the help of operators i.e addition, subtraction, multiplications, and divisions.

Example: Form an algebraic expression: x is first multiplied by 6 and then 7 subtracted from the product.

Solution: (6*x)-7= 6x-7

Practical Use of Expressions

Question: 2 girls go to a shop and both of them buy a pencil and an eraser. The cost of the pencil and eraser is 3and1 respectively. What is the total cost?

Answer: x= cost of a pencil

Y= cost of eraser

Total cost = 2(x + y)= 2(3 + 1) = 2*4 = 8

Hence, the total cost of a pencil and eraser for 2 girls is $8.

Preparation Tips

First, go through the chapter thoroughly and cover all the topics.

Note down all the important formulae and try to learn them.

Start solving the exercise questions without any guidance from anyone.

Try the questions at least 3 times if you are unable to solve them.

Seek help from Vedantu’s solution to check your answers and solve the ones that you were unable to solve.

FAQs on RS Aggarwal Class 6 Solutions Chapter-8 Algebraic Expressions

1. Why Should I Study Algebraic Expression from Vedantu? Can I Use the Methods Shown in the Solutions of Vedantu to Get Good Marks?

Algebra is a completely new concept that a student first learns in Class 6 but if we take a deeper look, we see that the students have unconsciously used algebra all this time.

Vedantu's study material explained to the students the basic terminologies required to form an expression based on daily applications. The study material will help them not only develop basic understanding of the topic but will also help the students in higher studies.

Yes, you can use the same methods in your exams to get good marks. The answers are prepared by a group of top educators who have gained expertise in this field over the years. You can blindly trust the solutions provided here.

The solutions are prepared to keep in mind the current CBSE marking scheme. The stepwise process of solving the sum by Vedantu helps students to clear doubts. If you practice the sums regularly from Vedantu, no one stops you from securing higher grades in classes.

2. What are the Benefits of Vedantu’s Study Material and Online Coaching?

Vedantu provides free study material for all students and given the situation of a pandemic it has been tough for the students to ask for solutions from the teachers of their school. Vedantu has been of great help to the students to clear their concepts. Vedantu has also helped to save a significant amount of money. Online coaching provided by Vedantu is very economical compared to other online or offline coaching classes.

Online coaching is provided by the best educators who have gained expertise in this field over the years. The study material is made according to the needs of the student and helps them to further understand the complex concepts in higher studies.

3. Name the topics covered in chapter 8 Algebraic Expressions?

There are 10 exercises present in chapter 8 Algebraic expressions of class 6, each exercise deals with a different topic. Students can easily prepare all these exercises and topics with a little guidance from our expert teachers at Vedantu.

All the major topics covered in this chapter are named below -

Introduction to Algebra

Matchstick patterns

The Idea of variable

Standard rules for the use of variables (rules from geometry and arithmetic)

Expressions with variables

Application of expressions

Defining equations

The solution to the equations

4. Explain some common rules for Algebraic Expressions?

In class 6, three rules of algebraic expressions are explained in detail. These three are -

Commutative property of addition

In addition to two or more variables, the order of addition does not change the resulting sum of the equation.

E.g. - a + b + c = c + b + a

Commutative property of multiplication.

Similar to addition, In the multiplication of 2 or more, The order of multiplication does not have any effect.

E.g. - a b = b a

Division of multiplication over the addition of numbers.

For three variables a, b and c. this rule could be written as -

a (b + c) = a b + a c

5. What is Algebra and why is it called “Algebra”?

Algebra is an important branch of mathematics, in which letters and numbers are used to represent various numbers. It also deals with the rules of manipulating those symbols to get the required solutions. Algebra makes it easier to state a mathematical relation.

The word “Algebra” is derived from a book named “Kitab Al-jabr”. The author of this book was a 9th-century mathematician and astronomer, named “Muhammad ibn Musa al Khwarizmi” also known as the father of Algebra. Later, He was known as the Grandfather of computer science.

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RD Sharma Solutions for Class 8 Maths Chapter 6 - Algebraic Expressions and Identities

As students get familiar with numbers, their problem-solving and analytical skills improve considerably. We know that algebra plays an essential role in Maths. The algebraic expressions used in algebra consist of variables and basic operations such as addition, subtraction, multiplication and division. These operations are performed using certain laws and basic formulas which have to be remembered. Here, in RD Sharma Solutions for Class 8 Maths Chapter 6 – Algebraic Expressions and Identities, such problems are solved. Our expert team have solved the questions in a step-by-step format, which helps the students to understand the concepts better. Moreover, practising RD Sharma Class 8 Solutions will help students secure excellent scores in the annual exam.

Chapter 6 – Algebraic Expressions and Identities contains seven exercises, and the RD Sharma Solutions available on this page provide solutions for the questions in each exercise. Now, let us have a look at the concepts discussed in this chapter.

RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities

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Access Answers to RD Sharma Solutions for Class 8 Maths Chapter 6 – Algebraic Expressions and Identities

Exercise 6.1 page no: 6.2.

1. Identify the terms and their coefficients for each of the following expressions:

(i) 7x 2 yz – 5xy

(ii) x 2 + x + 1

(iii) 3x 2 y 2 – 5x 2 y 2 z 2 + z 2

(iv) 9 – ab + bc – ca

(v) a/2 + b/2 – ab

(vi) 0.2x – 0.3xy + 0.5y

Solution: (i) 7x 2 yz – 5xy

The given equation has two terms that are:

7x 2 yz and – 5xy

The coefficient of 7x 2 yz is 7

The coefficient of – 5xy is – 5

The given equation has three terms that are:

The coefficient of x 2 is 1

The coefficient of x is 1

The coefficient of 1 is 1

3x 2 y, -5x 2 y 2 z 2 and z 2

The coefficient of 3x 2 y is 3

The coefficient of -5x 2 y 2 z 2 is -5

The coefficient of z 2 is 1

The given equation has four terms that are:

9, -ab, bc, -ca

The coefficient of 9 is 9

The coefficient of -ab is -1

The coefficient of bc is 1

The coefficient of -ca is -1

The coefficient of a/2 is 1/2

The coefficient of b/2 is 1/2

0.2x, -0.3xy, 0.5y

The coefficient of 0.2x is 0.2

The coefficient of -0.3xy is -0.3

The coefficient of 0.5y is 0.5

2. Classify the following polynomials as monomials, binomials and trinomials. Which polynomials do not fit into any category?

(i) x+y (ii) 1000 (iii) x+x 2 +x 3 +x 4 (iv) 7+a+5b (v) 2b-3b 2 (vi) 2y-3y 2 +4y 3 (vii) 5x-4y+3x (viii) 4a-15a 2 (ix) xy+yz+zt+tx (x) pqr (xi) p 2 q+pq 2 (xii) 2p+2q

The given expression contains two terms, x and y

∴ It is Binomial

The given expression contains one term, 1000

∴ It is Monomial

(iii) x+x 2 +x 3 +x 4

The given expression contains four terms

∴ It belongs to none of the categories

The given expression contains three terms

∴ It is Trinomial

The given expression contains two terms

(vi) 2y-3y 2 +4y 3

(viii) 4a-15a 2

(ix) xy + yz + zt + tx

The given expression contains one term

(xi) p 2 q+pq 2

EXERCISE 6.2 PAGE NO: 6.5

1. Add the following algebraic expressions:

(i) 3a 2 b, -4a 2 b, 9a 2 b

(ii) 2/3a, 3/5a, -6/5a

(iii) 4xy 2 – 7x 2 y, 12x 2 y -6xy 2 , -3x 2 y + 5xy 2

(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c

(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy

(vi) 7/2x 3 – 1/2x 2 + 5/3, 3/2x 3 + 7/4x 2 – x + 1/3, 3/2x 2 -5/2x -2

Let us add the given expression

3a 2 b + (-4a 2 b) + 9a 2 b

3a 2 b – 4a 2 b + 9a 2 b

2/3a + 3/5a + (-6/5a)

2/3a + 3/5a – 6/5a

Let us take LCM for 3 and 5, which is 15

(2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)a

10/15a + 9/15a – 18/15a

(10a+9a-18a)/15

4xy 2 – 7x 2 y + 12x 2 y – 6xy 2 – 3x 2 y + 5xy 2

Upon rearranging

12x 2 y – 3x 2 y – 7x 2 y – 6xy 2 + 5xy 2 + 4xy 2

3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c

3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c

By taking LCM for (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20)

(9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20

23a/6 – 9b/4 + 53c/20

(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy

11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy

11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y

By taking LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10)

(11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10

51xy/14 – 19x/35 – 31y/10

(vi) 7/2x 3 – 1/2x 2 + 5/3, 3/2x 3 + 7/4x 2 – x + 1/3, 3/2x 2 -5/2x – 2

7/2x 3 – 1/2x 2 + 5/3 + 3/2x 3 + 7/4x 2 – x + 1/3 + 3/2x 2 -5/2x – 2

7/2x 3 + 3/2x 3 – 1/2x 2 + 7/4x 2 + 3/2x 2 – x – 5/2x + 5/3 + 1/3 – 2

10/2x 3 + 11/4x 2 – 7/2x + 0/6

5x 3 + 11/4x 2 -7/2x

2. Subtract: (i) -5xy from 12xy (ii) 2a 2 from -7a 2 (iii) 2a-b from 3a-5b (iv) 2x 3 – 4x 2 + 3x + 5 from 4x 3 + x 2 + x + 6 (v) 2/3y 3 – 2/7y 2 – 5 from 1/3y 3 + 5/7y 2 + y – 2 (vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z (vii) x 2 y – 4/5xy 2 + 4/3xy from 2/3x 2 y + 3/2xy 2 – 1/3xy (viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac

(i) -5xy from 12xy

Let us subtract the given expression

12xy – (- 5xy)

(ii) 2a 2 from -7a 2

(-7a 2 ) – 2a 2

(iii) 2a-b from 3a-5b

(3a – 5b) – (2a – b)

3a – 5b – 2a + b

(iv) 2x 3 – 4x 2 + 3x + 5 from 4x 3 + x 2 + x + 6

(4x 3 + x 2 + x + 6) – (2x 3 – 4x 2 + 3x + 5)

4x 3 + x 2 + x + 6 – 2x 3 + 4x 2 – 3x – 5

2x 3 + 5x 2 – 2x + 1

(v) 2/3y 3 – 2/7y 2 – 5 from 1/3y 3 + 5/7y 2 + y – 2

1/3y 3 + 5/7y 2 + y – 2 – 2/3y 3 + 2/7y 2 + 5

1/3y 3 – 2/3y 3 + 5/7y 2 + 2/7y 2 + y – 2 + 5

By grouping similar expressions, we get,

-1/3y 3 + 7/7y 2 + y + 3

-1/3y 3 + y 2 + y + 3

(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z

2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)

2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z

By grouping similar expressions we get,

LCM for (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)

(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6

-5x/6 + 11y/4 + 13z/6

(vii) x 2 y – 4/5xy 2 + 4/3xy from 2/3x 2 y + 3/2xy 2 – 1/3xy

2/3x 2 y + 3/2xy 2 – 1/3xy – (x 2 y – 4/5xy 2 + 4/3xy)

2/3x 2 y – x 2 y + 3/2xy 2 + 4/5xy 2 – 1/3xy – 4/3xy

LCM for (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)

-1/3x 2 y + 23/10xy 2 – 5/3xy

(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac

3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)

3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7

LCM for (5 and 3 is 15), (5 and 5 is 5)

(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7

184bc/15 + -10ac/5 – ab/7

– ab/7 + 184bc/15 – 2ac

(i) 6/5x 2 – 4/5x 3 + 5/6 + 3/2x from x 3 /3 – 5/2x 2 + 3/5x + 1/4

(ii) 5a 2 /2 + 3a 3 /2 + a/3 – 6/5 from 1/3a 3 – 3/4a 2 – 5/2

(iii) 7/4x 3 + 3/5x 2 + 1/2x + 9/2 from 7/2 – x/3 – x 2 /5 (iv) y 3 /3 + 7/3y 2 + 1/2y + 1/2 from 1/3 – 5/3y 2

(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc

1/3x 3 – 5/2x 2 + 3/5x + 1/4 – (6/5x 2 – 4/5x 3 + 5/6 + 3/2x)

1/3x 3 + 4/5x 3 – 5/2x 2 – 6/5x 2 + 3/5x – 3/2x + 1/4 – 5/6

LCM for (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), and (4 and 6 is 24)

17/15x 3 – 37/10x 2 – 9/10x – 14/24

17/15x 3 – 37/10x 2 – 9/10x – 7/12

1/3a 3 – 3/4a 2 – 5/2 – (5/2a 2 + 3/2a 3 + a/3 – 6/5)

1/3a 5 – 3/2a 3 – 3/4a 2 – 5/2a 2 – a/3 – 5/2 + 6/5

LCM for (3 and 2 is 6), (4 and 2 is 4), and (2 and 5 is 10)

(2a 3 – 9a 3 )/6 – (3a 2 + 10a 2 )/4 – a/3 + (-25+12)/10

-7/6a 3 – 13/4a 2 – a/3 – 13/10

(iii) 7/4x 3 + 3/5x 2 + 1/2x + 9/2 from 7/2 – x/3 – x 2 /5

7/2 – x/3 – 1/5x 2 – (7/4x 3 + 3/5x 2 + 1/2x + 9/2)

-7/4x 3 – 1/5x 2 – 3/5x 2 – x/3 – x/2 + 7/2 – 9/2

LCM for (3 and 2 is 6)

-7/4x 3 – 4/5x 2 – (2x-3x)/6 + (7-9)/2

-7/4x 3 – 4/5x 2 – 5/6x – 1

(iv) y 3 /3 + 7/3y 2 + 1/2y + 1/2 from 1/3 – 5/3y 2

1/3 – 5/3y 2 – (1/3y 3 + 7/3y 2 + 1/2y + 1/2)

-1/3y 3 – 5/3y 2 – 7/3y 2 – 1/2y + 1/3 – 1/2

LCM for (3 and 3 is 3), (3 and 2 is 6)

-1/3y 3 + (-5y 2 – 7y 2 )/3 – 1/2y + (2-3)/6

-1/3y 3 – 12/3y 2 – 1/2y – 1/6

3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)

3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc

LCM for (2 and 7 is 14), (4 and 3 is 12), and (6 and 3 is 6)

(21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6

31/14ab – 29/12ac – 3/2bc

4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z.

The sum of x – 3y + 2z and -4x + 9y – 11z is

(x – 3y + 2z) + (-4x + 9y – 11z)

x – 4x – 3y + 9y + 2z – 11z

-3x + 6y – 9z

Now, Let us subtract the given expression from -3x + 6y – 9z

(-3x + 6y – 9z) – (3x – 4y – 7z)

-3x – 3x + 6y + 4y – 9z + 7z

-6x + 10y – 2z

5. Subtract the sum of 3l – 4m – 7n 2 and 2l + 3m – 4n 2 from the sum of 9l + 2m – 3n 2 and -3l + m + 4n 2 ….

Sum of 3l – 4m – 7n 2 and 2l + 5m – 4n 2

3l – 4m – 7n 2 + 2l + 3m – 4n 2

3l + 2l – 4m + 3m – 7n 2 – 4n 2

5l – m – 11n 2 ……………………..equation (1)

Sum of 9l + 2m – 3n 2 and -3l + m + 4n 2

9l + 2m – 3n 2 + (-3l + m + 4n 2 )

9l – 3l + 2m + m – 3n 2 + 4n 2

6l + 3m + n 2 ……………………….equation (2)

Let us subtract equation (i) from (ii), and we get

6l + 3m + n 2 – (5l – m – 11n 2 )

6l – 5l + 3m + m + n 2 + 11n 2

6. Subtract the sum of 2x – x 2 + 5 and -4x – 3 + 7x 2 from 5.

Sum of 2x – x 2 + 5 and -4x – 3 + 7x 2 is

2x – x 2 + 5 + (-4x – 3 + 7x 2 )

2x – x 2 + 5 – 4x – 3 + 7x 2

– x 2 + 7x 2 + 2x – 4x + 5 – 3

6x 2 -2x + 2 ………….equation (i)

Let us subtract equation (i) from 5 we get,

5 – (6x 2 -2x + 2)

5 – 6x 2 + 2x – 2

3 + 2x – 6x 2

7. Simplify each of the following:

(i) x 2 – 3x + 5 – 1/2(3x 2 – 5x + 7)

(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)

(iii) 11/2x 2 y – 9/4xy 2 + 1/4xy – 1/14y 2 x + 1/15yx 2 + 1/2xy

(iv) (1/3y 2 – 4/7y + 11) – (1/7y – 3 + 2y 2 ) – (2/7y – 2/3y 2 + 2)

(v) -1/2a 2 b 2 c + 1/3ab 2 c – 1/4abc 2 – 1/5cb 2 a 2 + 1/6cb 2 a – 1/7c 2 ab + 1/8ca 2 b

x 2 – 3/2x 2 – 3x + 5/2x + 5 – 7/2

LCM for (1 and 2 is 2)

(2x 2 – 3x 2 )/2 – (6x + 5x)/2 + (10-7)/2

-1/2x 2 – 1/2x + 3/2

5 – 3x + 2y – 2x + y – 3x + 7y – 9

– 3x – 2x – 3x + 2y + y + 7y + 5 – 9

-8x + 10y – 4

11/2x 2 y + 1/15x 2 y – 9/4xy 2 – 1/14xy 2 + 1/4xy + 1/2xy

LCM for (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4)

(165x 2 y + 2x 2 y)/30 + (-126xy 2 – 4xy 2 )/56 + (xy + 2xy)/4

167/30x 2 y – 130/56xy 2 + 3/4xy

167/30x 2 y – 65/28xy 2 + 3/4xy

1/3y 2 – 2y 2 – 2/3y 2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2

LCM for (3, 1 and 3 is 3), (7, 7 and 7 is 7)

(y 2 – 6y 2 + 2y 2 )/3 – (4y – y – 2y)/7 + 12

-3/3y 2 – 7/7y + 12

-y 2 – y + 12

-1/2a 2 b 2 c – 1/5a 2 b 2 c + 1/3ab 2 c + 1/6ab 2 c – 1/4abc 2 – 1/7abc 2 + 1/8a 2 bc

LCM for (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28)

-7/10a 2 b 2 c + 1/2ab 2 c – 11/28abc 2 + 1/8a 2 bc

EXERCISE 6.3 PAGE NO: 6.13

Find each of the following products:

1. 5x 2 × 4x 3

Let us simplify the given expression

5 × x × x × 4 × x × x × x

5 × 4 × x 1+1+1+1+1

2. -3a 2 × 4b 4

– 3 × a 2 × 4 × b 4

-12 × a 2 × b 4

3. (-5xy) × (-3x 2 yz)

(-5) × (-3) × x × x 2 × y × y × z

15 × x 1+2 × y 1+1 × z

4. 1/2xy × 2/3x 2 yz 2

1/2 × 2/3 × x × x 2 × y × y × z 2

1/3 × x 1+2 × y 1+1 × z 2

5. (-7/5xy 2 z) × (13/3x 2 yz 2 )

-7/5 × 13/3 × x × x 2 × y 2 × y × z × z 2

-91/15 × x 1+2 × y 2+1 × z 1+2

-91/15x 3 y 3 z 3

6. (-24/25x 3 z) × (-15/16xz 2 y)

-24/25 × -15/16 × x 3 × x × z × z 2 × y

18/20 × x 3+1 × z 1+2 × y

7. (-1/27a 2 b 2 ) × (9/2a 3 b 2 c 2 )

-1/27 × 9/2 × a 2 × a 3 × b 2 × b 2 × c 2

-1/6 x a 2+3 × b 2+2 × c 2

-1/6a 5 b 4 c 2

8. (-7xy) × (1/4x 2 yz)

-7 × 1/4 × x × y × x 2 × y × z

-7/4 × x 1+2 × y 1+1 × z

9. (7ab) × (-5ab 2 c) × (6abc 2 )

7 × -5 × 6 × a × a × a × b × b 2 × b × c × c 2

210 × a 1+1+1 × b 1+2+1 × c 1+2

10. (-5a) × (-10a 2 ) × (-2a 3 )

(-5) × (-10) × (-2) × a × a 2 × a 3

-100 × a 1+2+3

11. (-4x 2 ) × (-6xy 2 ) × (-3yz 2 )

(-4) × (-6) – (-3) × x 2 × x × y 2 × y × z 2

– 72 × x 2+1 × y 2+1 × z 2

12. (-2/7a 4 ) × (-3/4a 2 b) × (-14/5b 2 )

-2/7 × -3/4 × -14/5 × a 4 × a 2 × b × b 2

-6/10 × a 4+2 × b 1+2

13. (7/9ab 2 ) × (15/7ac 2 b) × (-3/5a 2 c)

7/9 × 15/7 × -3/5 × a × a × a 2 × b 2 × b × c 2 × c

– a 1+1+2 × b 2+1 × c 2+1

14. (4/3u 2 vw) × (-5uvw 2 ) × (1/3v 2 wu)

4/3 × -5 × 1/3 × u 2 × u × u × v × v × v 2 × w × w 2 × w

-20/9 × u 2+1+1 × v 1+1+2 × w 1+2+1

-20/9u 4 v 4 w 4

15. (0.5x) × (1/3xy 2 z 4 ) × (24x 2 yz)

0.5 × 1/3 × 24 × x × x × y 2 × y × x 2 × z 4 × z

12/3 × x 1+1+2 × y 2+1 × z 4+1

4x 4 × y 3 × z 5

16. (4/3pq 2 ) × (-1/4p 2 r) × (16p 2 q 2 r 2 )

4/3 × 1/4 × 16 × p × p 2 × p 2 × q 2 × q 2 × r × r 2

-16/3 × p 1+2+2 × q 2+2 × r 1+2

-16/3p 5 q 4 r 3

17. (2.3xy) × (0.1x) × (0.16)

2.3 × 0.1 × 0.16 × x × x × y

0.0368 × x 1+1 × y

Express each of the following products as monomials and verify the result in each case for x=1:

18. (3x) × (4x) × (-5x)

3 × 4 × -5 × x × x × x

LHS = (3 × 1) × (4 × 1) × (-5 × 1)

= 3 × 4 × – 5

RHS = -60 (1) 3 = – 60

Therefore, LHS = RHS.

19. (4x 2 ) × (-3x) × (4/5x 3 )

4 × -3 × 4/5 × x 2 × x × x 3

-48/5 × x 2+1+3

LHS = 4 × 1 2 × – 3 × 1 × 4/5 × 1 3

RHS = – 48/5 × 1 6 = – 48/5

20. (5x 4 ) × (x 2 ) 3 × (2x) 2

5 × x 4 × x 6 × 4 × x 2

5 × 4 × x 4 × x 6 × x 2

LHS = (5 × 1 4 ) × (1 2 ) 3 × (2 × 1) 2

RHS = 20 × 1 12 = 20

21. (x 2 ) 3 × (2x) × (-4x) × (5)

x 6 × 2 × x × -4 × x × 5

2 × -4 × 5 × x 6 × x × x

LHS = (1 2 ) 3 × (2 × 1) × (-4 × 1) × 5

RHS = – 40 × 1 8 = – 40

22. Write down the product of -8x 2 y 6 and -20xy verify the product for x = 2.5, y = 1

-8 × -20 × x 2 × x × y 6 × y

160 × x 2+1 × y 6+1

Now let us verify when, x = 2.5 and y = 1

160 (2.5) 3 × (1) 7

For -8x 2 y 6 and -20xy

-8 × 2.5 2 × 1 6 × -20 × 1 × 2.5

Hence, the given expression is verified.

23. Evaluate (3.2x 6 y 3 ) × (2.1x 2 y 2 ) when x = 1 and y = 0.5

3.2 × 2.1 × x 6 × x 2 × y 3 × y 2

6.72 × x 6+2 × y 3+2

Now let us substitute when, x = 1 and y = 0.5

For 6.72x 8 y 5

6.72 × 1 8 × 0.5 5

24. Find the value of (5x 6 ) × (-1.5x 2 y 3 ) × (-12xy 2 ) when x = 1, y = 0.5

5 × -1.5 × -12 × x 6 × x 2 × x × y 3 × y 2

90 × x 6+2+1 × y 3+2

90 × (1) 9 × (0.5) 5

25. Evaluate (2.3a 5 b 2 ) × (1.2a 2 b 2 ) when a = 1 and b = 0.5

2.3a 5 b 2 × 1.2a 2 b 2

2.3 × 1.2 × a 5 × a 2 × b 2 × b 2

2.76 × a 5+2 × b 2+2

Now let us substitute when, a = 1 and b = 0.5

For 2.76 a 7 b 4

2.76 (1) 7 (0.5) 4

2.76 × 1 × 0.0025

26. Evaluate (-8x 2 y 6 ) × (-20xy) for x = 2.5 and y = 1

-8 × – 20 × x 2 × x × y 6 × y

Now let us substitute when, x = 2.5 and y = 1

160 × (2.5) 3 × (1) 7

Express each of the following products as monomials and verify the result for x = 1, y = 2:

27. (-xy 3 ) × (yx 3 ) × (xy)

-x × y 3 × y × x 3 × x × y

-x 1+3+1 × y 3+1+1

Now let us substitute when, x = 1 and y = 2

28. (1/8x 2 y 4 ) × (1/4x 4 y 2 ) × (xy) × 5

1/8 × 1/4 × 5 × x 2 × x 4 × x × y 4 × y 2 × y

5/32 × x 2+4+1 × y 4+2+1

5/32 × 1 7 × 2 7

29. (2/5a 2 b) × (-15b 2 ac) × (-1/2c 2 )

2/5 × -15 × -1/2 × a 2 × a × b × b 2 × c × c 2

3 × a 2+1 × b 1+2 × c 1+2

30. (-4/7a 2 b) × (-2/3b 2 c) × (-7/6c 2 a)

-4/7 × -2/3 × -7/6 × a 2 × a × b × b 2 × c × c 2

-4/9 × a 2+1 × b 2+1 × c 1+2

-4/9a 3 b 3 c 3

31. (4/9abc 3 ) × (-27/5a 3 b 2 ) × (-8b 3 c)

4/9 × -27/5 × -8 × a × a 3 × b × b 2 × b 3 × c 3 × c

96/5 × a 1+3 × b 1+2+3 × c 3+1

96/5a 4 b 6 c 4

Evaluate each of the following when x = 2, and y = -1.

32. (2xy) × (x 2 y/4) × (x 2 ) × (y 2 )

2 × 1/4 × x × x 2 × x 2 × y × y 2 × y

1/2x 1+2+2 y 1+2+1

Now let us substitute when, x = 2 and y = -1

1/2 × (2) 5 × (-1) 4

33. (3/5x 2 y) × (-15/4xy 2 ) × (7/9x 2 y 2 )

3/5 × -15/4 × 7/9 × x 2 × x × x 2 × y × y 2 × y 2

-7/4 × x 2+1+2 × y 1+2+2

For -7/4x 5 y 5

-7/4 × (2) 5 (-1) 5

-7/4 × 32 × -1

EXERCISE 6.4 PAGE NO: 6.21

Find the following products:

1. 2a 3 (3a + 5b)

(2a 3 × 3a) + (2a 3 × 5b)

6a 3+1 + 10a 3 b

2. -11a (3a + 2b)

(-11a × 3a) + (-11a × 2b)

-33a 2 – 22ab

3. -5a (7a – 2b)

-5a (7a – 2b)

(-5a × 7a) – (-5a × 2b)

4. -11y 2 (3y + 7)

-11y 2 (3y + 7)

(-11y 2 × 3y) + (-11y 2 × 7)

-33y 3 – 77y 2

5. 6x/5(x 3 + y 3 )

6/5x (x 3 + y 3 )

(6/5x × x 3 ) + (6/5x × y 3 )

6/5x 4 + 6/5xy 3

6. xy (x 3 – y 3 )

xy (x 3 – y 3 )

(xy × x 3 ) – (xy × y 3 )

7. 0.1y (0.1x 5 + 0.1y)

0.1y (0.1x 5 + 0.1y)

(0.1y × 0.1x 5 ) + (0.1y × 0.1y)

0.01x 5 y + 0.01y 2

8. (-7/4ab 2 c – 6/25a 2 c 2 ) (-50a 2 b 2 c 2 )

(-7/4ab 2 c – 6/25a 2 c 2 ) (-50a 2 b 2 c 2 )

(-7/4ab 2 c × -50a 2 b 2 c 2 ) – (6/25a 2 c 2 × -50a 2 b 2 × c 2 )

350/4a 3 b 4 c 3 + 12a 4 b 2 c 4

175/2a 3 b 4 c 3 + 12a 4 b 2 c 4

9. -8/27xyz (3/2xyz 2 – 9/4xy 2 z 3 )

-8/27xyz (3/2xyz 2 – 9/4xy 2 z 3 )

(-8/27xyz × 3/2xyz 2 ) – (-8/27xyz × 9/4xy 2 z 3 )

-4/9x 2 y 2 z 3 + 2/3x 2 y 3 z 4

10. -4/27xyz (9/2x 2 yz – 3/4xyz 2 )

-4/27xyz (9/2x 2 yz – 3/4xyz 2 )

(-4/27xyz × 9/2x 2 yz) – (-4/27xyz × 3/4xyz 2 )

-2/3x 3 y 2 z 2 + 1/9x 2 y 2 z 3

11. 1.5x (10x 2 y – 100xy 2 )

1.5x (10x 2 y – 100xy 2 )

(1.5x × 10x 2 y) – (1.5x × 100xy 2 )

15x 3 y – 150x 2 y 2

12. 4.1xy (1.1x – y)

4.1xy (1.1x – y)

(4.1xy × 1.1x) – (4.1xy × y)

4.51x 2 y – 4.1xy 2

13. 250.5xy (xz + y/10)

250.5xy (xz + y/10)

(250.5xy × xz) + (250.5xy × y/10)

250.5x 2 yz + 25.05xy 2

14. 7/5x 2 y (3/5xy 2 + 2/5x)

7/5x 2 y (3/5xy 2 + 2/5x)

(7/5x 2 y × 3/5xy 2 ) + (7/5x 2 y × 2/5x)

21/25x 3 y 3 + 14/25x 3 y

15. 4/3a (a 2 + b 2 – 3c 2 )

4/3a (a 2 + b 2 – 3c 2 )

(4/3a × a 2 ) + (4/3a × b 2 ) – (4/3a × 3c 2 )

4/3a 3 + 4/3ab 2 – 4ac 2

16. Find the product 24x 2 (1-2x) and evaluate its value for x = 3

24x 2 (1 – 2x)

(24x 2 × 1) – (24x 2 × 2x)

24x 2 – 48x 3

Now let us evaluate the expression when x = 3

24 × (3) 2 – 48 × (3) 3

24 × (9) – 48 × (27)

17. Find the product -3y (xy+y 2 ) and evaluate its value for x = 4 and y = 5

(-3y × xy) + (-3y × y 2 )

-3xy 2 – 3y 3

Now let us evaluate the expression when x = 4 and y = 5

-3 × (4) × (5) 2 – 3 × (5) 3

18. Multiply -3/2x 2 y 3 by (2x-y) and verify the answer for x = 1 and y = 2

-3/2x 2 y 3 by (2x-y)

(-3/2x 2 y 3 × 2x) – (-3/2x 2 y 3 × y)

-3x 3 y 3 + 3/2x 2 y 4

Now let us evaluate the expression when x = 1 and y = 2

-3 × (1) 4 × (2) 3 + 3/2 × (1) 2 × (2) 4

– 3 × (8) + 3 (8)

19. Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.005: (i) 15y 2 (2 – 3x) (ii) -3x (y 2 + z 2 ) (iii) z 2 (x – y) (iv) xz (x 2 + y 2 )

(i) 15y 2 (2 – 3x)

30y 2 – 45xy 2

By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000

30 × (25/100) 2 – 45 × (-1) × (25/100) 2

30 (1/16) + 45 (1/16)

(ii) -3x (y 2 + z 2 )

-3xy 2 + -3xz 2

-3× (-1) × (25/100) 2 – 3 × (-1) × (5/1000) 2

(3×25×25/100×100) + (3×5×5/1000×1000)

(iii) z 2 (x – y)

z 2 x – z 2 y

(5/1000) 2 (-1 – 25/100)

(1/40000) (-100-25/100)

(1/40000) (-125/100)

(1/40000) (-5/4)

(iv) xz (x 2 + y 2 )

(-1) 3 × (5/1000) + (-1) × (5/1000) × (25/100) 2

-1/200 – 1/16 × 1/200

-1/200 – 1/3200

By taking LCM as 3200

(i) 2x 2 (x 3 – x) – 3x (x 4 + 2x) – 2 (x 4 – 3x 2 )

(ii) x 3 y (x 2 – 2x) + 2xy (x 3 – x 4 )

(iii) 3a 2 + 2 (a+2) – 3a (2a+1)

(iv) x (x+4) + 3x (2x 2 -1) + 4x 2 + 4

(v) a (b-c) – b (c-a) – c (a-b)

(vi) a (b-c) +b (c-a) + c (a-b)

(vii) 4ab (a-b) – 6a 2 (b-b 2 ) -3b 2 (2a 2 -a) + 2ab (b-a)

(viii) x 2 (x 2 + 1) – x 3 (x + 1) – x (x 3 – x)

(ix) 2a 2 + 3a (1 – 2a 3 ) + a (a + 1)

(x) a 2 (2a – 1) + 3a + a 3 – 8

(xi) 3/2x 2 (x 2 – 1) + 1/4x 2 (x 2 + x) – 3/4x (x 3 – 1)

(xii) a 2 b (a-b 2 ) + ab 2 (4ab – 2a 2 ) – a 3 b(1-2b)

(xiii) a 2 b (a 3 – a + 1) – ab(a 4 – 2a 2 + 2a) – b(a 3 – a 2 – 1)

2x 5 – 2x 3 – 3x 5 – 6x 2 – 2x 4 + 6x 2

2x 5 – 3x 5 – 2x 3 – 2x 4 – 6x 2 + 6x 2

-x 5 – 2x 4 – 2x 3

x 5 y – 2x 4 y + 2x 4 y – 2x 5 y

-x 5 y – 2x 5 y

3a 2 + 2a + 4 – 6a 2 – 3a

3a 2 – 6a 2 + 2a – 3a + 4

-3a 2 – a + 4

x 2 + 4x + 6x 3 – 3x + 4x 2 + 4

6x 3 + 5x 2 + x + 4

ab – ac – bc + ab – ca + bc

ab – ac + bc – ab + ac – bc

4a 2 b – 4ab 2 – 6a 2 b + 6a 2 b 2 – 6a 2 b 2 + 3ab 2 + 2ab 2 – 2a 2 b

4a 2 b – 6a 2 b– 2a 2 b – 4ab 2 + 3ab 2 + 2ab 2 + 6a 2 b 2 – 6a 2 b 2

x 4 + x 2 – x 4 – x 3 – x 4 + x 2

x 4 – x 4 – x 4 – x 3 + x 2 + x 2

– x 4 – x 3 + 2x 2

2a 2 + 3a – 6a 4 + a 2 + a

-6a 4 + 3a 2 + 4a

2a 3 – a 2 + 3a + a 3 – 8

3a 3 – a 2 + 3a – 8

3/2x 4 – 3/2x 2 + 1/4x 4 + 1/4x 3 – 3/4x 4 + 3/4x

3/2x 4 + 1/4x 4 – 3/4x 4 – 3/2x 2 + 1/4x 3 + 3/4x

4/4x 4 + 1/4x 3 – 3/2x 2 + 3/4x

x 4 + 1/4x 3 – 3/2x 2 + 3/4x

a 3 b – a 2 b 3 + 4a 2 b 3 – 2a 3 b 2 – a 3 b + 2a 3 b 2

-a 2 b 3 + 4a 2 b 3

a 5 b – a 3 b + a 2 b – a 5 b + 2a 3 b – 2a 2 b – ba 3 + a 2 b + b

a 5 b – a 5 b – a 3 b + 2a 3 b – ba 3 + a 2 b – 2a 2 b + a 2 b + b

EXERCISE 6.5 PAGE NO: 6.30

1. (5x + 3) by (7x + 2)

Now, let us simplify the given expression

(5x + 3) × (7x + 2)

5x (7x + 2) + 3 (7x + 2)

35x 2 + 10x + 21x + 6

35x 2 + 31x + 6

2. (2x + 8) by (x – 3)

(2x + 8) × (x – 3)

2x (x – 3) + 8 (x – 3)

2x 2 – 6x + 8x – 24

2x 2 + 2x – 24

3. (7x + y) by (x + 5y)

(7x + y) × (x + 5y)

7x (x + 5y) + y (x + 5y)

7x 2 + 35xy + xy + 5y 2

7x 2 + 36xy + 5y 2

4. (a – 1) by (0.1a 2 + 3)

(a – 1) × (0.1a 2 + 3)

a (0.1a 2 + 3) -1 (0.1a 2 + 3)

0.1a 3 + 3a – 0.1a 2 – 3

0.1a 3 – 0.1a 2 + 3a – 3

5. (3x 2 + y 2 ) by (2x 2 + 3y 2 )

(3x 2 + y 2 ) × (2x 2 + 3 y 2 )

3x 2 × (2x 2 + 3y 2 ) + y 2 × (2x 2 + 3y 2 )

6x 4 + 9x 2 y 2 + 2x 2 y 2 + 3y 4

6x 4 + 11x 2 y 2 + 3y 4

6. (3/5x + 1/2y) by (5/6x + 4y)

(3/5x + 1/2y) × (5/6x + 4y)

3/5x × (5/6x + 4y) + 1/2y × (5/6x + 4y)

15/30x 2 + 12/5xy + 5/12xy + 4/2y 2

1/2x 2 + 169/60xy + 2y 2

7. (x 6 – y 6 ) by (x 2 + y 2 )

(x 6 – y 6 ) × (x 2 + y 2 )

x 6 × (x 2 + y 2 ) – y 6 × (x 2 + y 2 )

x 8 + x 6 y 2 – x 2 y 6 – y 8

8. (x 2 + y 2 ) by (3a + 2b)

(x 2 + y 2 ) × (3a + 2b)

x 2 × (3a + 2b) + y 2 × (3a + 2b)

3ax 2 + 3ay 2 + 2bx 2 + 2by 2

9. (- 3d – 7f) by (5d + f)

(- 3d – 7f) × (5d + f)

-3d (5d + f) – 7f (5d + f)

– 15d 2 – 3df – 35df – 7f 2

– 15d 2 – 38df – 7f 2

10. (0.8a – o.5b) by (1.5a – 3b)

(0.8a – 0.5b) × (1.5a – 3b)

0.8a (1.5a – 3b) – 0.5b (1.5a – 3b)

1.2a 2 – 2.4ab – 0.75ab + 1.5b 2

1.2a 2 – 3.15ab + 1.5b 2

11. (2x 2 y 2 – 5xy 2 ) by (x 2 – y 2 )

(2x 2 y 2 – 5xy 2 ) × (x 2 – y 2 )

2x 2 y 2 (x 2 – y 2 ) – 5xy 2 (x 2 – y 2 )

2x 4 y 2 – 5x 3 y 2 – 2x 2 y 4 + 5xy 4

12. (x/7 + x 2 /2) by (2/5 + 9x/4)

(x/7 + x 2 /2) × (2/5 + 9x/4)

x/7 (2/5 + 9x/4) + x 2 /2 (2/5 + 9x/4)

2x/35 + (9 x 2 )/28 + x 2 /5 + (9 x 3 )/8

9/8x 3 + 73/140x 2 + 2/35x

13. (-a/7 + a 2 /9) by (b/2 – b 2 /3)

(-a/7 + a 2 /9) × (b/2 – b 2 /3)

-a/7 (b/2 – b 2 /3) + a 2 /9 (b/2 – b 2 /3)

-ab/14 + ab 2 /21 + a 2 b/18 – a 2 b 2 /27

14. (3x 2 y – 5xy 2 ) by (1/5x 2 + 1/3y 2 )

(3x 2 y – 5xy 2 ) × (1/5x 2 + 1/3y 2 )

3x 2 y (1/5x 2 + 1/3y 2 ) – 5xy 2 (1/5x 2 + 1/3y 2 )

3/5x 4 y + 3/3x 2 y 3 – x 3 y 2 + 5/3xy 4

3/5x 4 y + x 2 y 3 – x 3 y 2 + 5/3xy 4

15. (2x 2 – 1) by (4x 3 + 5x 2 )

(2x 2 – 1) × (4x 3 + 5x 2 )

2x 2 (4x 3 + 5x 2 ) – 1 (4x 3 + 5x 2 )

8x 5 + 10x 4 – 4x 3 – 5x 2

16. (2xy + 3y 2 ) by (3y 2 – 2)

(2xy + 3y 2 ) × (3y 2 – 2)

2xy (3y 2 – 2) + 3y 2 (3y 2 – 2)

6xy 3 – 4xy + 9y 4 – 6y 2

Find the following products and verify the results for x = -1, y = -2:

17. (3x – 5y) (x + y)

(3x – 5y) × (x + y)

x (3x – 5y) + y (3x – 5y)

3x 2 – 5xy + 3xy – 5y 2

3x 2 – 2xy – 5y 2

Let us substitute the given values x = – 1 and y = – 2, then

(-3+10) × (-1-2)

3 (-1) 2 – 2 (-1) (-2) – 5 (-2) 2

∴ the given expression is verified.

18. (x 2 y – 1) (3 – 2x 2 y)

(x 2 y – 1) × (3 – 2x 2 y)

x 2 y (3 – 2x 2 y) – 1 (3 – 2x 2 y)

3x 2 y – 2x 4 y 2 – 3 + 2x 2 y

5x 2 y – 2x 4 y 2 – 3

(-2 – 1) × (3 + 4)

– 8 – 10 – 3

19. (1/3x – y 2 /5) (1/3x + y 2 /5)

(1/3x – y 2 /5) × (1/3x + y 2 /5)

(1/3x) 2 – (y 2 /5) 2

(1/3x – y 2 /5) (1/3x + y 2 /5)

1/9x 2 – 1/25y 4

(1/3(-1) – (-2) 2 /5) × (1/3(-1) + (-2) 2 /5)

(-17/15) × (7/15)

1/9 (-1) 2 – 1/25 (-2) 4

20. x 2 (x + 2y) (x – 3y)

x 2 (x + 2y) (x – 3y)

x 2 (x 2 – 3xy + 2xy – 3y 2 )

x 2 (x 2 – xy – 6y 2 )

x 4 – x 3 y – 6x 2 y 2

21. (x 2 – 2y 2 ) (x + 4y)x 2 y 2

(x 2 – 2y 2 ) (x + 4y)x 2 y 2

(x 3 + 4x 2 y – 2xy 2 – 8y 3 ) × x 2 y 2

x 5 y 2 + 4x 4 y 3 – 2x 3 y 4 – 8x 2 y 5

22. a 2 b 2 (a + 2b) (3a + b)

a 2 b 2 (a + 2b) (3a + b)

a 2 b 2 (3a 2 + ab + 6ab + 2b 2 )

a 2 b 2 (3a 2 + 7ab + 2b 2 )

3a 4 b 2 + 7a 3 b 3 + 2a 2 b 4

23. x 2 (x – y) y 2 (x + 2y)

x 2 (x – y) y 2 (x + 2y)

x 2 y 2 (x 2 + 2xy – xy – 2y 2 )

x 2 y 2 (x 2 + xy – 2y 2 )

x 4 y 2 + x 3 y 3 – 2x 2 y 4

24. (x 3 – 2x 2 + 5x – 7) (2x – 3)

(x 3 – 2x 2 + 5x – 7) (2x – 3)

2x 4 – 4x 3 + 10x 2 – 14x – 3x 3 + 6x 2 – 15x + 21

2x 4 – 7x 3 + 16x 2 – 29x + 21

25. (5x + 3) (x – 1) (3x – 2)

(5x + 3) (x – 1) (3x – 2)

(5x 2 – 2x – 3) (3x – 2)

15x 3 – 6x 2 – 9x – 10x 2 + 4x + 6

15x 3 – 16x 2 – 5x + 6

26. (5 – x) (6 – 5x) (2 – x)

(5 – x) (6 – 5x) (2 – x)

(x 2 – 7x + 10) (6 – 5x)

-5x 3 + 35x 2 – 50x + 6x 2 – 42x + 60

60 – 92x + 41x 2 – 5x 3

27. (2x 2 + 3x – 5) (3x 2 – 5x + 4)

(2x 2 + 3x – 5) (3x 2 – 5x + 4)

6x 4 + 9x 3 – 15x 2 – 10x 3 – 15x 2 + 25x + 8x 2 + 12x – 20

6x 4 – x 3 – 22x 2 + 37x – 20

28. (3x – 2) (2x – 3) + (5x – 3) (x + 1)

(3x – 2) (2x – 3) + (5x – 3) (x + 1)

6x 2 – 9x – 4x + 6 + 5x 2 + 5x – 3x – 3

11x 2 – 11x + 3

29. (5x – 3) (x + 2) – (2x + 5) (4x – 3)

(5x – 3) (x + 2) – (2x + 5) (4x – 3)

5x 2 + 10x – 3x – 6 – 8x 2 + 6x – 20x + 15

-3x 2 – 7x + 9

30. (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)

(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)

12x 2 + 9xy + 8xy

12x 2 + 9xy + 8xy + 6y 2 – 14x 2 + 6xy + 7xy – 3y 2

-2x 2 + 3y 2 + 30xy

31. (x 2 – 3x + 2) (5x – 2) – (3x 2 + 4x – 5) (2x – 1)

(x 2 – 3x + 2) (5x – 2) – (3x 2 + 4x – 5) (2x – 1)

5x 3 – 15x 2 + 10x – 2x 2 + 6x – 4 – (6x 3 + 8x 2 – 10x – 3x 2 – 4x + 5)

5x 3 – 6x 3 – 15x 2 – 2x 2 – 5x 2 + 16x + 14x – 4 – 5

– x 3 – 22x 2 + 30x – 9

32. (x 3 – 2x 2 + 3x – 4) (x – 1) – (2x – 3) (x 2 – x + 1)

(x 3 – 2x 2 + 3x – 4) (x – 1) – (2x – 3) (x 2 – x + 1)

x 4 – 2x 3 + 3x 2 – 4x – x 3 + 2x 2 – 3x + 4 – (2x 3 – 2x 2 + 2x – 3x 2 + 3x – 3)

x 4 – 3x 3 + 5x 2 – 7x + 4 – 2x 3 + 5x 2 – 5x + 3

x 4 – 5x 3 + 10x 2 – 12x + 7

EXERCISE 6.6 PAGE NO: 6.43

1. Write the following squares of binomials as trinomials:

(ii) (8a + 3b) 2

(iii) (2m + 1) 2

(iv) (9a + 1/6) 2

(v) (x + x 2 /2) 2

(vi) (x/4 – y/3) 2

(vii) (3x – 1/3x) 2

(viii) (x/y – y/x) 2

(ix) (3a/2 – 5b/4) 2

(x) (a 2 b – bc 2 ) 2

(xi) (2a/3b + 2b/3a) 2

(xii) (x 2 – ay) 2

Let us express the given expression in trinomial

x 2 + 2 (x) (2) + 2 2

(8a) 2 + 2 (8a) (3b) + (3b) 2

64a 2 + 48ab + 9b 2

(2m) 2 + 2 (2m) (1) + 1 2

(9a) 2 + 2 (9a) (1/6) + (1/6) 2

81a 2 + 3a + 1/36

(x) 2 + 2 (x) (x 2 /2) + (x 2 /2) 2

x 2 + x 3 + 1/4x 4

(x/4) 2 – 2 (x/4) (y/3) + (y/3) 2

1/16x 2 – xy/6 + 1/9y 2

(3x) 2 – 2 (3x) (1/3x) + (1/3x) 2

9x 2 – 2 + 1/9x 2

(x/y) 2 – 2 (x/y) (y/x) + (y/x) 2

x 2 /y 2 – 2 + y 2 /x 2

(3a/2) 2 – 2 (3a/2) (5b/4) + (5b/4) 2

9/4a 2 – 15/4ab + 25/16b 2

(a 2 b) 2 – 2 (a 2 b) (bc 2 ) + (bc 2 ) 2

a 4 b 2 – 2a 2 b 2 c 2 + b 2 c 4

(2a/3b) 2 + 2 (2a/3b) (2b/3a) + (2b/3a) 2

4a 2 /9b 2 + 8/9 + 4b 2 /9a 2

(x 2 ) 2 – 2 (x 2 ) (ay) + (ay) 2

x 4 – 2x 2 ay + a 2 y 2

2. Find the product of the following binomials:

(i) (2x + y) (2x + y)

(ii) (a + 2b) (a – 2b)

(iii) (a 2 + bc) (a 2 – bc)

(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)

(v) (2x + 3/y) (2x – 3/y)

(vi) (2a 3 + b 3 ) (2a 3 – b 3 )

(vii) (x 4 + 2/x 2 ) (x 4 – 2/x 2 )

(viii) (x 3 + 1/x 3 ) (x 3 – 1/x 3 )

Let us find the product of the given expression

2x (2x + y) + y (2x + y)

4x 2 + 2xy + 2xy + y 2

4x 2 + 4xy + y 2

a (a – 2b) + 2b (a – 2b)

a 2 – 2ab + 2ab – 4b 2

a 2 (a 2 – bc) + bc (a 2 – bc)

a 4 – a 2 bc + bca 2 – b 2 c 2

a 4 – b 2 c 2

4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)

16/25x 2 + 12/20yx – 12/20xy – 9y 2 /16

16/25x 2 – 9/16y 2

2x (2x – 3/y) + 3/y (2x – 3/y)

4x 2 – 6x/y + 6x/y – 9/y 2

2a 3 (2a 3 – b 3 ) + b 3 (2a 3 – b 3 )

4a 6 – 2a 3 b 3 + 2a 3 b 3 – b 6

x 4 (x 4 – 2/x 2 ) + 2/x 2 (x 4 – 2/x 2 )

x 8 – 2x 2 + 2x 2 – 4/x 4

(x 8 – 4/x 4 )

x 3 (x 3 – 1/x 3 ) + 1/x 3 (x 3 – 1/x 3 )

x 6 – 1 + 1 – 1/x 6

3. Using the formula for squaring a binomial, evaluate the following:

We can express 102 as 100 + 2

So, (102) 2 = (100 + 2) 2

Upon simplification, we get,

(100 + 2) 2 = (100) 2 + 2 (100) (2) + 2 2

= 10000 + 400 + 4

We can express 99 as 100 – 1

So, (99) 2 = (100 – 1) 2

(100 – 1) 2 = (100) 2 – 2 (100) (1) + 1 2

= 10000 – 200 + 1

We can express 1001 as 1000 + 1

So, (1001) 2 = (1000 + 1) 2

(1000 + 1) 2 = (1000) 2 + 2 (1000) (1) + 1 2

= 1000000 + 2000 + 1

We can express 999 as 1000 – 1

So, (999) 2 = (1000 – 1) 2

(1000 – 1) 2 = (1000) 2 – 2 (1000) (1) + 1 2

= 1000000 – 2000 + 1

We can express 700 as 700 + 3

So, (703) 2 = (700 + 3) 2

(700 + 3) 2 = (700) 2 + 2 (700) (3) + 3 2

= 490000 + 4200 + 9

4. Simplify the following using the formula: (a – b) (a + b) = a 2 – b 2 :

(i) (82) 2 – (18) 2

(ii) (467) 2 – (33) 2

(iii) (79) 2 – (69) 2

(iv) 197 × 203

(vii) 1.8 × 2.2

(viii) 9.8 × 10.2

Let us simplify the given expression using the formula (a – b) (a + b) = a 2 – b 2

(82) 2 – (18) 2 = (82 – 18) (82 + 18)

(467) 2 – (33) 2 = (467 – 33) (467 + 33)

(79) 2 – (69) 2 = (79 + 69) (79 – 69)

We can express 203 as 200 + 3 and 197 as 200 – 3

197 × 203 = (200 – 3) (200 + 3)

= (200) 2 – (3) 2

We can express 113 as 100 + 13 and 87 as 100 – 13

113 × 87 = (100 – 13) (100 + 13)

= (100) 2 – (13) 2

= 10000 – 169

We can express 95 as 100 – 5 and 105 as 100 + 5

95 × 105 = (100 – 5) (100 + 5)

= (100) 2 – (5) 2

We can express 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2

1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)

= (2) 2 – (0.2) 2

We can express 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2

9.8 × 10.2 = (10 – 0.2) (10 + 0.2)

= (10) 2 – (0.2) 2

5. Simplify the following using the identities:

(i) ((58) 2 – (42) 2 )/16

(ii) 178 × 178 – 22 × 22

(iii) (198 × 198 – 102 × 102)/96

(iv) 1.73 × 1.73 – 0.27 × 0.27

(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726

((58) 2 – (42) 2 )/16 = ((58-42) (58+42)/16)

= ((16) (100)/16)

178 × 178 – 22 × 22 = (178) 2 – (22) 2

= (178-22) (178+22)

(198 × 198 – 102 × 102)/96 = ((198) 2 – (102) 2 )/96

= ((198-102) (198+102))/96

= (96 × 300)/96

1.73 × 1.73 – 0.27 × 0.27 = (1.73) 2 – (0.27) 2

= (1.73-0.27) (1.73+0.27)

(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63) 2 – (1.37) 2 )/0.726

= ((8.63-1.37) (8.63+1.37))/0.726

= (7.26 × 10)/0.726

6. Find the value of x, if:

(i) 4x = (52) 2 – (48) 2

(ii) 14x = (47) 2 – (33) 2

(iii) 5x = (50) 2 – (40) 2

Let us simplify to find the value of x by using the formula (a – b) (a + b) = a 2 – b 2

4x = (52) 2 – (48) 2

4x = (52 – 48) (52 + 48)

14x = (47) 2 – (33) 2

14x = (47 – 33) (47 + 33)

5x = (50) 2 – (40) 2

5x = (50 – 40) (50 + 40)

7. If x + 1/x =20, find the value of x 2 + 1/ x 2 .

We know that x + 1/x = 20

So when squaring both sides, we get

(x + 1/x) 2 = (20) 2

x 2 + 2 × x × 1/x + (1/x) 2 = 400

x 2 + 2 + 1/x 2 = 400

x 2 + 1/x 2 = 398

8. If x – 1/x = 3, find the values of x 2 + 1/ x 2 and x 4 + 1/ x 4 .

We know that x – 1/x = 3

So, when squaring both sides, we get

(x – 1/x) 2 = (3) 2

x 2 – 2 × x × 1/x + (1/x) 2 = 9

x 2 – 2 + 1/x 2 = 9

x 2 + 1/x 2 = 9+2

x 2 + 1/x 2 = 11

Now, again when we square on both sides we get,

(x 2 + 1/x 2 ) 2 = (11) 2

x 4 + 2 × x 2 × 1/x 2 + (1/x 2 ) 2 = 121

x 4 + 2 + 1/x 4 = 121

x 4 + 1/x 4 = 121-2

x 4 + 1/x 4 = 119

∴ x 2 + 1/x 2 = 11

9. If x 2 + 1/x 2 = 18, find the values of x + 1/ x and x – 1/ x.

We know that x 2 + 1/x 2 = 18

When adding 2 on both sides, we get

x 2 + 1/x 2 + 2 = 18 + 2

x 2 + 1/x 2 + 2 × x × 1/x = 20

(x + 1/x) 2 = 20

x + 1/x = √20

When subtracting 2 from both sides, we get

x 2 + 1/x 2 – 2 × x × 1/x = 18 – 2

(x – 1/x) 2 = 16

x – 1/x = √16

10. If x + y = 4 and xy = 2, find the value of x 2 + y 2

We know that x + y = 4 and xy = 2

Upon squaring on both sides of the given expression, we get

(x + y) 2 = 4 2

x 2 + y 2 + 2xy = 16

x 2 + y 2 + 2 (2) = 16 (since xy=2)

x 2 + y 2 + 4 = 16

x 2 + y 2 = 16 – 4

11. If x – y = 7 and xy = 9, find the value of x 2 +y 2

We know that x – y = 7 and xy = 9

(x – y) 2 = 7 2

x 2 + y 2 – 2xy = 49

x 2 + y 2 – 2 (9) = 49 (since xy=9)

x 2 + y 2 – 18 = 49

x 2 + y 2 = 49 + 18

12. If 3x + 5y = 11 and xy = 2, find the value of 9x 2 + 25y 2

We know that 3x + 5y = 11 and xy = 2

(3x + 5y) 2 = 11 2

(3x) 2 + (5y) 2 + 2(3x)(5y) = 121

9x 2 + 25y 2 + 2 (15xy) = 121 (since xy=2)

9x 2 + 25y 2 + 2(15(2)) = 121

9x 2 + 25y 2 + 60 = 121

9x 2 + 25y 2 = 121-60

9x 2 + 25y 2 = 61

13. Find the values of the following expressions:

(i) 16x 2 + 24x + 9 when x = 7/4

(ii) 64x 2 + 81y 2 + 144xy when x = 11 and y = 4/3

(iii) 81x 2 + 16y 2 – 72xy when x = 2/3 and y = ¾

Let us find the values using the formula (a + b) 2 = a 2 + b 2 + 2ab

(4x) 2 + 2 (4x) (3) + 3 2

Evaluating when x = 7/4

(8x) 2 + 2 (8x) (9y) + (9y) 2 (8x + 9y)

Evaluating when x = 11 and y = 4/3

(9x) 2 + (4y) 2 – 2 (9x) (4y)

Putting x = 2/3 and y = 3/4

14. If x + 1/x = 9 find the value of x 4 + 1/ x 4 .

We know that x + 1/x = 9

(x + 1/x) 2 = (9) 2

x 2 + 2 × x × 1/x + (1/x) 2 = 81

x 2 + 2 + 1/x 2 = 81

x 2 + 1/x 2 = 81 – 2

x 2 + 1/x 2 = 79

Now, again when we square on both sides, we get,

(x 2 + 1/x 2 ) 2 = (79) 2

x 4 + 2 × x 2 × 1/x 2 + (1/x 2 ) 2 = 6241

x 4 + 2 + 1/x 4 = 6241

x 4 + 1/x 4 = 6241- 2

x 4 + 1/x 4 = 6239

∴ x 4 – 1/x 4 = 6239

15. If x + 1/x = 12 find the value of x – 1/x.

We know that x + 1/x = 12

(x + 1/x) 2 = (12) 2

x 2 + 2 × x × 1/x + (1/x) 2 = 144

x 2 + 2 + 1/x 2 = 144

x 2 + 1/x 2 = 144 – 2

x 2 + 1/x 2 = 142

x 2 + 1/x 2 – 2 × x × 1/x = 142 – 2

(x – 1/x) 2 = 140

x – 1/x = √140

16. If 2x + 3y = 14 and 2x – 3y = 2, find value of xy. [Hint: Use (2x+3y) 2 – (2x-3y) 2 = 24xy]

We know that the given equations are

2x + 3y = 14… equation (1)

2x – 3y = 2… equation (2)

Now, let us square both the equations and subtract equation (2) from equation (1), we get,

(2x + 3y) 2 – (2x – 3y) 2 = (14) 2 – (2) 2

4x 2 + 9y 2 + 12xy – 4x 2 – 9y 2 + 12xy = 196 – 4

∴ the value of xy is 8.

17. If x 2 + y 2 = 29 and xy = 2, find the value of (i) x + y (ii) x – y (iii) x 4 + y 4

We know that x 2 + y 2 = 29

x 2 + y 2 + 2xy – 2xy = 29

(x + y) 2 – 2 (2) = 29

(x + y) 2 = 29 + 4

x + y = ± √33

(x – y) 2 + 2 (2) = 29

(x – y) 2 + 4 = 29

(x – y) 2 = 25

(x – y) = ± 5

(iii) x 4 + y 4

Squaring both sides, we get

(x 2 + y 2 ) 2 = (29) 2

x 4 + y 4 + 2x 2 y 2 = 841

x 4 + y 4 + 2 (2) 2 = 841

x 4 + y 4 = 841 – 8

x 4 + y 4 = 833

18. What must be added to each of the following expressions to make it a whole square? (i) 4x 2 – 12x + 7

(ii) 4x 2 – 20x + 20

(i) 4x 2 – 12x + 7

(2x) 2 – 2 (2x) (3) + 3 2 – 3 2 + 7

(2x – 3) 2 – 9 + 7

(2x – 3) 2 – 2

∴ 2 must be added to the expression to make it a whole square.

(2x) 2 – 2 (2x) (5) + 5 2 – 5 2 + 20

(2x – 5) 2 – 25 + 20

(2x – 5) 2 – 5

∴ 5 must be added to the expression to make it a whole square.

(i) (x – y) (x + y) (x 2 + y 2 ) (x 4 + y 4 )

(ii) (2x – 1) (2x + 1) (4x 2 + 1) (16x 4 + 1)

(iii) (7m – 8n) 2 + (7m + 8n) 2

(iv) (2.5p – 1.5q) 2 – (1.5p – 2.5q) 2

(v) (m 2 – n 2 m) 2 + 2m 3 n 2

By grouping the values

(x 2 – y 2 ) (x 2 + y 2 ) (x 4 + y 4 )

(x 4 – y 4 ) (x 4 – y 4 )

Let us simplify the expression by grouping

(4x 2 – 1) (4x 2 + 1) (16x 4 + 1) 1

(16x 4 – 1) (16x 4 + 1) 1

(7m) 2 + (8n) 2 – 2(7m)(8n) + (7m) 2 + (8n) 2 + 2(7m)(8n)

(7m) 2 + (8n) 2 – 112mn + (7m) 2 + (8n) 2 + 112mn

49m 2 + 64n 2 + 49m 2 + 64n 2

By grouping the similar expression, we get,

98m 2 + 64n 2 + 64n 2

(2.5p) 2 + (1.5q) 2 – 2 (2.5p) (1.5q) – (1.5p) 2 – (2.5q) 2 + 2 (1.5p) (2.5q)

6.25p 2 + 2.25q 2 – 2.25p 2 – 6.25q 2

4p 2 – 6.25q 2 + 2.25q 2

4 (p 2 – q 2 )

Upon expansion using (a + b) 2 formula

(m 2 ) 2 – 2 (m 2 ) (n 2 ) (m) + (n 2 m) 2 + 2m 3 n 2

m 4 – 2m 3 n 2 + (n 2 m) 2 + 2m 3 n 2

m 4 + n 4 m 2 – 2m 3 n 2 + 2m 3 n 2

(i) (3x + 7) 2 – 84x = (3x – 7) 2

(ii) (9a – 5b) 2 + 180ab = (9a + 5b) 2 (iii) (4m/3 – 3n/4) 2 + 2mn = 16m 2 /9 + 9n 2 /16

(iv) (4pq + 3q) 2 – (4pq – 3q) 2 = 48pq 2

(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Let us consider LHS (3x + 7) 2 – 84x

By using the formula (a + b) 2 = a 2 + b 2 + 2ab

(3x) 2 + (7) 2 + 2 (3x) (7) – 84x

(3x) 2 + (7) 2 + 42x – 84x

(3x) 2 + (7) 2 – 42x

(3x) 2 + (7) 2 – 2 (3x) (7)

(3x – 7) 2 = R.H.S

(ii) (9a – 5b) 2 + 180ab = (9a + 5b) 2

Let us consider LHS (9a – 5b) 2 + 180ab

(9a) 2 + (5b) 2 – 2 (9a) (5b) + 180ab

(9a) 2 6 (5b) 2 – 90ab + 180ab

(9a) 2 + (5b) 2 + 9ab

(9a) 2 + (5b) 2 + 2 (9a) (5b)

(9a + 5b) 2 = R.H.S

(iii) (4m/3 – 3n/4) 2 + 2mn = 16m 2 /9 + 9n 2 /16

Let us consider LHS (4m/3 – 3n/4) 2 + 2mn

(4m/3) 2 + (3n/4) 2 – 2mn + 2mn

(4m/3) 2 + (3n/4) 2

16/9m 2 + 9/16n 2 = R.H.S

Let us consider LHS (4pq + 3q) 2 – (4pq – 3q) 2

( 4pq) 2 + (3q) 2 + 2 (4pq) (3q) – (4pq) 2 – (3q) 2 + 2(4pq)(3q)

24pq 2 + 24pq 2

Let us consider LHS (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

By using the identity (a – b) (a + b) = a 2 – b 2

(a 2 – b 2 ) + (b 2 – c 2 ) + (c 2 – a 2 )

a 2 – b 2 + b 2 – c 2 + c 2 – a 2

EXERCISE 6.7 PAGE NO: 6.47

1. Find the following products:

(i) (x + 4) (x + 7)

(ii) (x – 11) (x + 4)

(iii) (x + 7) (x – 5)

(iv) (x – 3) (x – 2)

(v) (y 2 – 4) (y 2 – 3)

(vi) (x + 4/3) (x + 3/4)

(vii) (3x + 5) (3x + 11)

(viii) (2x 2 – 3) (2x 2 + 5)

(ix) (z 2 + 2) (z 2 – 3)

(x) (3x – 4y) (2x – 4y)

(xi) (3x 2 – 4xy) (3x 2 – 3xy)

(xii) (x + 1/5) (x + 5)

(xiii) (z + 3/4) (z + 4/3)

(xiv) (x 2 + 4) (x 2 + 9)

(xv) (y 2 + 12) (y 2 + 6)

(xvi) (y 2 + 5/7) (y 2 – 14/5)

(xvii) (p 2 + 16) (p 2 – 1/4)

x (x + 7) + 4 (x + 7)

x 2 + 7x + 4x + 28

x (x + 4) – 11 (x + 4)

x 2 + 4x – 11x – 44

x 2 – 7x – 44

x (x – 5) + 7 (x – 5)

x 2 – 5x + 7x – 35

x 2 + 2x – 35

x (x – 2) – 3 (x – 2)

x 2 – 2x – 3x + 6

y 2 (y 2 – 3) – 4 (y 2 – 3)

y 4 – 3y 2 – 4y 2 + 12

y 4 – 7y 2 + 12

x (x + 3/4) + 4/3 (x + 3/4)

x 2 + 3x/4 + 4x/3 + 12/12

x 2 + 3x/4 + 4x/3 + 1

x 2 + 25x/12 + 1

3x (3x + 11) + 5 (3x + 11)

9x 2 + 33x + 15x + 55

9x 2 + 48x + 55

2x 2 (2x 2 + 5) – 3 (2x 2 + 5)

4x 4 + 10x 2 – 6x 2 – 15

4x 4 + 4x 2 – 15

z 2 (z 2 – 3) + 2 (z 2 – 3)

z 4 – 3z 2 + 2z 2 – 6

z 4 – z 2 – 6

3x (2x – 4y) – 4y (2x – 4y)

6x 2 – 12xy – 8xy + 16y 2

6x 2 – 20xy + 16y 2

3x 2 (3x 2 – 3xy) – 4xy (3x 2 – 3xy)

9x 4 – 9x 3 y – 12x 3 y + 12x 2 y 2

9x 4 – 21x 3 y + 12x 2 y 2

x (x + 1/5) + 5 (x + 1/5)

x 2 + x/5 + 5x + 1

x 2 + 26/5x + 1

z (z + 4/3) + 3/4 (z + 4/3)

z 2 + 4/3z + 3/4z + 12/12

z 2 + 4/3z + 3/4z + 1

z 2 + 25/12z + 1

x 2 (x 2 + 9) + 4 (x 2 + 9)

x 4 + 9x 2 + 4x 2 + 36

x 4 + 13x 2 + 36

y 2 (y 2 + 6) + 12 (y 2 + 6)

y 4 + 6y 2 + 12y 2 + 72

y 4 + 18y 2 + 72

y 2 (y 2 – 14/5) + 5/7 (y 2 – 14/5)

y 4 – 14/5y 2 + 5/7y 2 – 2

y 4 – 73/35y 2 – 2

p 2 (p 2 – 1/4) + 16 (p 2 – 1/4)

p 4 – 1/4p 2 + 16p 2 – 4

p 4 + 63/4p 2 – 4

2. Evaluate the following:

(ii) 109 × 107

(vii) 994 × 1006

We can express 102 as 100 + 2 and 106 as 100 + 6

Now let us simplify

102 × 106 = (100 + 2) (100 + 6)

= 100 (100 + 6) + 2 (100 + 6)

= 10000 + 600 + 200 + 12

We can express 109 as 100 + 9 and 107 as 100 + 7

109 × 107 = (100 + 9) (100 + 7)

= 100 (100 + 7) + 9 (100 + 7)

= 10000 + 700 + 900 + 63

We can express 35 as 30 + 5 and 37 as 30 + 7

35 × 37 = (30 + 5) (30 + 7)

= 30 (30 + 7) + 5 (30 + 7)

= 900 + 210 + 150 + 35

We can express 53 as 50 + 3 and 55 as 50 + 5

53 × 55 = (50 + 3) (50 + 5)

= 50 (50 + 5) + 3 (50 + 5)

= 2500 + 250 + 150 + 15

We can express 103 as 100 + 3 and 96 as 100 – 4

103 × 96 = (100 + 3) (100 – 4)

= 100 (100 – 4) + 3 (100 – 4)

= 10000 – 400 + 300 – 12

= 10000 – 112

We can express 34 as 30 + 4 and 36 as 30 + 6

34 × 36 = (30 + 4) (30 + 6)

= 30 (30 + 6) + 4 (30 + 6)

= 900 + 180 + 120 + 24

We can express 994 as 1000 – 6 and 1006 as 1000 + 6

994 × 1006 = (1000 – 6) (1000 + 6)

= 1000 (1000 + 6) – 6 (1000 + 6)

= 1000000 + 6000 – 6000 – 36

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