Sometimes we would like to assign definite values to the variables in an expression to find the value of that expression. This situation arises in many real life problems.

**Value
of an algebraic expression**

Sometimes we would like to assign definite values
to the variables in an expression to find the value of that expression. This situation
arises in many real life problems.

For example, if the teacher of class 7 wants to
select 10 students for a competition, for which she can choose any number of boys
and girls.

If the number of boys are* x *and the number of girls are *y*,
then the required algebraic expression to find the total number of participants
is *x* + *y*.

Here, the total number of students to be selected
is 10 and if only two girls are interested to participate in the competition, then
how many boys should be selected? Obviously, if *y* = 2 then *x* + 2 = 10. The value *x* = 8 satisfies the equation. So, the number of boys to be selected is 8.

Follow the steps to obtain the value**.**

Step − 1: Study the problem. Fix the variable and
write the algebraic expression.

Step − 2: Replace each variable by the given numerical
value to obtain an arithmetical expression.

Step − 3: Simplify the arithmetical expression by
**BIDMAS** method.

Step − 4: The value so obtained is the required
value of the expression.

**Try this**

Try to find the value of the following
expressions, if *p* = 5 and
*q* = 6.

i) *p* + *q* ii) *q* − *p* iii) 2*p +* 3*q* iv) *pq* − *p* − *q* v) 5*pq* – 1

i) *p + q* = **5 + 6 = 11**

ii) *q – p* = **6 – 5 = 1**

(iii) *2p + 3q* = 2(5)+
3(6)

** 10 + 18 = 28**

iv) *pq – p – q* = 5 × 6
–5 – 6

** 30 – 11 = 19**

v) *5pq – l* = 5 × 5 × 6
– 1

** 150 – 1 = 149**

** **

__Example 3.3__

If *x* = 3, *y* = 2 find the value of (i) 4*x +* 7*y* (ii) 3*x* + 2*y* − 5 (iii) *x* – *y*

*Solution*

(i) 4*x* + 7*y* = 4 (3) + 7 (2) = 12 + 14 = 26

(ii) 3*x* + 2*y* – 5 = 3 (3) + 2 (2) – 5 = 9 + 4 – 5 = 8

(iii)* x
*–*
y *= 3 – 2 = 1

** **

__Example 3.4__

Find the value of (i) 3*m* + 2*n* (ii) 2*m* − *n* (iii) *mn* − 1, given that *m* = 2, *n* = − 1.

*Solution*

(i) 3*m* + 2*n* = 3(2) + 2( −1) = 6 −2 = 4

(ii) 2*m* − *n* = 2(2) − ( −1) = 4 + 1 = 5

(iii) *mn
*−1 = (2) ( −1) −1 = −2 −1 = –3

Tags : Algebra | Term 1 Chapter 3 | 7th Maths , 7th Maths : Term 1 Unit 3 : Algebra

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7th Maths : Term 1 Unit 3 : Algebra : Value of an algebraic expression | Algebra | Term 1 Chapter 3 | 7th Maths

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