# Class 7 Maths Chapter 12 Exercise 12.1 Pdf Notes NCERT Solutions

Class 7 Maths Chapter 12 Algebraic Expression Exercise 12.1 pdf notes:-

- Exercise 12.2
- Exercise 12.3
- Exercise 12.4

**Exercise 12.1** Class 7 maths Chapter 12 Pdf Notes:-

**To see video Solution Of This Exercise Click Here**

## Ncert Solution for Class 7 Maths Chapter 12 Algebraic Expression Exercise 12.1 Tips:

**INTRODUCTION**

We have already come across simple algebraic expressions like x + 3, y – 5, 4x + 5,

10y – 5 and so on. In Class VI, we have seen how these expressions are useful in formulating

puzzles and problems. We have also seen examples of several expressions in the chapter on

simple equations.

Expressions are a central concept in algebra. This Chapter is devoted to algebraic

expressions. When you have studied this Chapter, you will know how algebraic

expressions are formed, how they can be combined, how we can find their values and

how they can be used.

**HOW ARE EXPRESSIONS FORMED?**

We now know very well what a variable is. We use letters x, y, l, m, … etc. to denote

variables. A variable can take various values. Its value is not fixed. On the other hand, a

constant has a fixed value. Examples of constants are: 4, 100, –17, etc.

We combine variables and constants to make algebraic expressions. For this, we use the

operations of addition, subtraction, multiplication and division. We have already come across

expressions like 4x + 5, 10y – 20. The expression 4x + 5 is obtained from the variable x, first

by multiplying x by the constant 4 and then adding the constant 5 to the product. Similarly,

10y – 20 is obtained by first multiplying y by 10 and then subtracting 20 from the product.

The above expressions were obtained by combining variables with constants. We can

also obtain expressions by combining variables with themselves or with other variables.

Look at how the following expressions are obtained:

x2

, 2y2

, 3×2

– 5, xy, 4xy + 7

(i) The expression x2

is obtained by multiplying the variable x by itself;

x × x = x 2

Just as 4 × 4 is written as 42

, we write x × x = x2

. It is commonly read as x squared.

(Later, when you study the chapter ‘Exponents and Powers’ you will realise that x2

may also be read as x raised to the power 2).

In the same manner, we can write x × x × x = x3

Commonly, x3

is read as ‘x cubed’. Later, you will realise that x3

may also be read

as x raised to the power 3.

x, x2

, x3

, … are all algebraic expressions obtained from x.

(ii) The expression 2y2

is obtained from y: 2y 2

= 2 × y × y

Here by multiplying y with y we obtain y2

and then we multiply y2

by the constant 2.

(iii) In (3×2

– 5) we first obtain x2

, and multiply it by 3 to get 3×2

.

From 3×2

, we subtract 5 to finally arrive at 3×2

– 5.

(iv) In xy, we multiply the variable x with another variable y. Thus,

x × y = xy.

(v) In 4xy + 7, we first obtain xy, multiply it by 4 to get 4xy and add

7 to 4xy to get the expression.

**TERMS OF AN EXPRESSION**

We shall now put in a systematic form what we have learnt above about how expressions

are formed. For this purpose, we need to understand what terms of an expression and

their factors are.

Consider the expression (4x + 5). In forming this expression, we first formed 4x

separately as a product of 4 and x and then added 5 to it. Similarly consider the expression

(3×2

- 7y). Here we first formed 3×2

separately as a product of 3, x and x. We then

formed 7y separately as a product of 7 and y. Having formed 3×2

and 7y separately, we

added them to get the expression.

You will find that the expressions we deal with can always be seen this way. They

have parts which are formed separately and then added. Such parts of an expression

which are formed separately first and then added are known as terms. Look at the

expression (4×2

– 3xy). We say that it has two terms, 4×2

and –3xy. The term 4×2

is a

product of 4, x and x, and the term (–3xy) is a product of (–3), x and y.

Terms are added to form expressions. Just as the terms 4x and 5 are added to

form the expression (4x + 5), the terms 4×2

and (–3xy) are added to give the expression

(4×2

– 3xy). This is because 4×2 - (–3xy) = 4×2

– 3xy.