We are asked how we can use a linear function to model and analyze a real-life situation.

Theory

Before writing the function to model and analyze the situation, we first need to define the variables. Then, we will write the equation that connects them. Finally, we use the model to find the desired values. Let's see a real-life situation.

Real Life Situation

Paula is spending her holidays in Portugal, and wants to rent a bike for a few days. She saw an advertisement at a local shop. Calculate the price of renting a bike for one week. How many days can she rent a bike if she plans on spending $$ 58 ?$

Rent a bike!

$ 5

per day plus a one-time payment of

$ 8

Let's use a linear function to model the above situation. Let $x$ be the number of days Paula rents the bike, and $y$ the cost of renting the bike for $x$ days.

Verbal Phrase	Algebraic Phrase
$The total cost$ $equals$ $the price per day$ $times$ $the number of days$ , $plus$ $the one - time payment$	$y$ $=$ $5$ $\times$ $x$ $+$ $8$

We have found the linear function to model the situation.

y = 5 \times x + 8 \Leftrightarrow y = 5 x + 8

To find the cost of renting a bike for

7

days, we substitute

7

for

x

in the above formula, and solve for

y .

y = 5 x + 8

Substitute

$x = 7$

y = 5 (7) + 8

Multiply

y = 35 + 8

AddTerms

Add terms

y = 43

The cost of renting a bike for a week is

$ 43 .

To find the number of days that a bike can be rented for

$ 58,

we substitute

58

for

y,

and solve for the

x -

variable.

y = 5 x + 8

Substitute

$y = 58$

58 = 5 x + 8

▼

Solve for

x

SubEqn

$LHS - 8 = RHS - 8$

50 = 5 x

DivEqn

$LHS / 5 = RHS / 5$

10 = x

RearrangeEqn

Rearrange equation

x = 10

With

$ 58,

Paula can rent a bike for

10

days.

Exercise