We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable Class 9

We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable Class 9

Question 1.

We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x - x², is a function of x. Can you interpret this as an input- output process? What value does the expression take when x = 6 cm? Class 9

Ans:

i. Here,

x is the input = length of the rectangle, Value of 10x - x² is the output = area of the rectangle.

For every value of x we feed in, we get one corresponding value for area as output.

So yes, expression 10x - x² can be interpreted as an input-output process.

ii. Substituting x = 6 in 10x - x², we get

10(6) - (6)² = 60 - 36

= 240

So, when the input is x = 6 cm, the value of the expression is 24 cm².

Question 2.

Find the value of the linear polynomial 5x - 3 if: Class 9

i. x = 0

ii. x = -1

iii. x = 2

Solution:

i. Substituting x = 0 in 5x - 3, we get 5(0) - 3 = 0 - 3 = -3

ii. Substituting x = -1 in 5x - 3, we get 5(-1) -3 = -5 - 3 = -8

iii. Substituting x = 2 in 5x - 3, we get 5(2) - 3 = 10 - 3 = 7