EVALUATE ALGEBRAIC EXPRESSIONS

Evaluate the algebraic expression for the given values.

Example 1 :

7x + 8 when x = 2

Solution :

Apply x = 2 in the function, we get

= 7(2) + 8

= 14 + 8

= 22

Example 2 :

5x − 4 when x = 6

Solution :

Apply x = 6 in the function, we get

= 5(6) - 4

= 30 - 4

= 26

Example 3 :

x² when x = 12

Solution :

Apply x = 12 in the function, we get

= 12²

= 144

Example 4 :

4^x when x = 2

Solution :

Apply x = 2 in the function, we get

= 4²

= 16

Example 5 :

x² + 3x - 7 when x = 4

Solution :

Apply x = 2 in the function, we get

= 2² + 3(2) - 7

= 4 + 6 - 7

= 10 - 7

= 3

Example 6 :

(x - y)² when x = 10 and y = 7

Solution :

Apply x = 10 and y = 7 in the function, we get

= (x - y)²

= (10 - 7)²

= 3²

= 9

Example 7 :

a² + b² when a = 3 and b = 8

Solution :

Apply a = 3 and b = 8 in the function, we get

= a² + b²

= 3² + 8²

= 9 + 64

= 73

Example 8 :

r² - s² when r = 12 and s = 5

Solution :

Apply r = 12 and s = 5 in the function, we get

= r² - s²

= 12² - 5²

= 144 - 25

= 119

Example 9 :

2L + 2W when L = 15 and W = 12

Solution :

Apply L = 15 and W = 12 in the function, we get

= 2(15) + 2(12)

= 30 + 24

= 54

Example 10 :

2x + 4y - 5 when x = 7 and y = 8

Solution :

Apply x = 7 and y = 8 in the function, we get

= 2(7) + 4(8) - 5

= 14 + 32 - 5

= 46 - 5

= 41

Example 11 :

It takes approximately 770 peanuts to product one jar of peanut butter. The total number of peanuts n is represented by the function n = 770 p, where p is the number of jars of peanut butter purchased. 

a) Determine the appropriate input values.

b) Complete the function table.

Solution :

a) n = 770 p

here p represents number of peanut jars and n number of total number of peanuts in the jar.

Number of jars cannot be decimal, so the values of p must be whole numbers.

b) 

Example 12 :

A taxi charges $2.50 plus 0.75 per mile. The function y = 0.75 x + 2.50 represents the total cost y, of the taxi ride for any number of miles. If Adriana rode 10 miles in the taxi and Alonso rode 15 miles, how much more did Alonso spend on his taxi ride ?

Solution :

y = 0.75 x + 2.50

When x = 10

y = 0.75 (10) + 2.50

= 7.5 + 2.50

= 10

Adriana spend $10

When x = 15

y = 0.75(15) + 2.5

= 11.25 + 2.5

= 13.75

Alonso spend $13.75

Difference between them = 13.75 - 10

= 3.75

So, Alonso spend $3.75 more.

Example 13 :

An elephant weighs 200 pounds at birth and gains approximately 2 pounds per day. The function w = 2d + 200 represents the weight w of an elephant on a given day d during his first year. How much more does an elephant weigh on Day 60 than it does on Day 5 ?

Solution :

w = 2d + 200

Weight of elephant on day 5 :

w = 2(5) + 200

= 10 + 200

= 210

Weight of elephant on day 60 :

w = 2(60) + 200

= 120 + 200

= 320

Difference between the weights = 320 - 210

= 110

Example 14 :

A public swimming pool holds 250,000 gallons of water and is being drained at the rate of 200 gallons per minute. The function g = -200 m + 250000 represents how many gallons remain in the pool g, after any number of minutes m, how much more water after 5 minutes than 20 minutes ?

Solution :

g = -200 m + 250000

Here m represents number of minutes.

When m = 5

g = -200(5) + 250000

= -1000 + 250000

= 249000

When m = 20

g = -200(20) + 250000

= -4000 + 250000

= 246000

= 249000 - 246000

= 3000

So, there is 3000 gallon of water remain in the pool in 5 minutes than 20 minutes.