Problem Solving
Linear equations can be applied to many real world applications. They are useful to solve practical problems we face everyday. First, we need to identify if a scenario is linear by checking if the change in variables is constant.
Here are a couple examples:
- A car drives \(60 \; [\text{km/h}]\). Every hour, the distance of the car changes by \(60 \; [\text{km}]\).
- The cost of a soccer ball is \($29.99\). The total cost changes by \($29.99\) for every ball you purchase.
- A tub is filled with water at a rate of \(5 \; [\text{L/s}]\). The volume of the tub increases by \(5 \; [\text{L}]\) every second.
If you are given a table of values, you can always check that the first difference is constant meaning the relationship is linear.
After you confirm the scenario is linear, we will need to write a linear equation in the form:
\(y = mx + b\)
- \(y\) is the dependent variable
- \(x\) is the independent variable
- \(m\) is the slope or rate of change
- \(b\) is the \(y\)-intercept or starting point (when \(x = 0 \))
It is a good idea to use variables that match the scenario (for example, \(t\) for time). It is also good practice to write out what each variable represents and also the units. For example, \('t\) represents the time in hours.
The cost of a rental car is \($22\) plus \($12\) per day.
- Write an equation to model the relationship
- Calculate the cost of a one week car rental
i. First, since the cost per day is constant (\($12/[\text{day}]\)), we can determine this is a linear equation.
Next, we can determine that the cost of the car rental represents the dependent variable while the number of rental days represents the independent variable.
Then, we need to write an equation in slope-intercept form, or:
\(y = mx + b\)
First, we can determine the slope, which represents the rate of change. Since the cost increases by \($12\) each day, we can determine that the slope is \(m=12\).
Next, we can determine the y-intercept, which represents the starting price (or a rental of \(0\) days). The rental car is going to cost \($22\) even if you don't drive it for a full day. Therefore, we can determine that the \(y\)-intercept is \(b = 22\).
Finally, we can write the equation as such:
\(C = 12d + 22\)
where \(C\) is the cost of the rental in dollars and \(d\) is the number of days.
ii. We need to calculate the cost which means we need to substitute a value for \(d\) and solve for \(C\). Although the rental lasts for one week, \(d\) represents the number of days!
Let's set \(d=7\) and solve for the total cost:
\(C = 12d + 22\)
\(C = 12(7) + 22\)
\(C = 84 + 22\)
\(C = $106\)
Therefore, we can determine that a one-week car rental will cost \(\boldsymbol{$106}\).
Throughout the week starting Monday, there were \(12\) sodas in Mei's fridge. Mei would drink them everyday and by Tuesday there were only \(10\) sodas left and only \(8\) on Wednesday. This pattern countinued until Mei drank all the remaining of sodas.
Create an equation and a graph to show the relationship between the the amount of sodas left each day.
Mei seems to be drinking two sodas each day (yikes!). Since it is the same amount each day, her soda consumption represents a linear equation.
Next, we can determine that the number of sodas represents the dependent variable while the number of days represents the independent variable.
Then, we need to write an equation in slope-intercept form, or:
\(y = mx + b\)
First, we can determine the slope, which represents the rate of change. Since the number of soda decreases by \(2\) each day, we can determine that the slope is (\(m=-2\)).
Next, we can determine the y-intercept, which represents the starting point. We can say Monday is the starting points which represents \(d=0\). Since there were \(12\) sodas in the fridge on Monday, we can determine that the \(y\)-intercept is (\(b = 12\)).
Finally, we can write the equation as such:
\(S = -2d + 12\)
where \(S\) is the number of soda and \(d\) is the number of days with Monday representing \(d= 0\).
In order to graph this relation, we can first outline a table of values. Remember, \(d=0\) is Monday:
| d Value | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| S Value | 12 | 10 | 8 | 6 | 4 | 2 | 0 |
Finally, we can sketch a graph as such:
