Setting up a Linear Equation to Solve a Real-World Application Cont.
When faced with a real-world problem, it’s often helpful to define the unknown quantities using variables. By analyzing the relationships between these unknowns and any given information, we can construct a mathematical model in the form of a linear equation. In most cases, this involves identifying fixed and variable components, which correspond to the slope and [latex]y[/latex]-intercept in a linear equation.
In the previous example about renting a car, the weekly cost to rent a compact car was modeled by identifying both fixed and variable costs. This same approach can be applied in many other real-world situations, such as calculating monthly expenses, managing inventory costs, or predicting future earnings.
To build a linear equation, start by identifying how the total cost or outcome changes with respect to the variable, which often involves determining a rate of change (the slope). Once this is established, you can add any fixed costs (the y-intercept) to complete the model.
Slope-intercept form:
[latex]y=mx+b[/latex], where:
[latex]x \text{ and } y[/latex] represent the coordinates of any point on the line
[latex]m[/latex] represents the slope of the line
[latex]b[/latex] represents the initial value, or the y-intercept
You can solve for [latex]b[/latex] by substituting a known point for [latex]x \text{ and } y[/latex] and the slope of the line for [latex]m[/latex]. The point-slope form, [latex]y-{y}_{1}=m\left(x-{x}_{1}\right)[/latex] simplifies to slope-intercept form when solved for [latex]y[/latex]. Substitute the coordinates of a point for [latex]{y}_1 \text{ and } {x}_1[/latex] and the slope for [latex]m[/latex] then simplify.
Find a linear equation to model this real-world application: It costs ABC electronics company [latex]$2.50[/latex] per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of [latex]$350[/latex] for utilities and [latex]$3,300[/latex] for salaries.
Step 1: Identify Costs:
Variable Costs: [latex]$2.50[/latex] per unit. This cost increases linearly with each additional unit produced.
Fixed Costs: Monthly utilities are [latex]$350[/latex] and monthly salaries are [latex]$3,300[/latex]. Therefore, the total fixed monthly costs is [latex]$350 + $3,300 = $3,650[/latex].
Step 2: Develop the Linear Equation:
Let [latex]x[/latex] represents the number of units produced each month.
[latex]$2.50[/latex] is the slope of the equation, representing the additional cost per unit produced.
[latex]$3,650[/latex] is the y-intercept, representing the fixed costs that are constant regardless of production volume.
The total monthly expense ([latex]C[/latex]) for ABC Electronics can be modeled as:
[latex]C = $2.50x+$3,650[/latex]
The equation helps the company forecast financial outcomes based on production levels, assisting in budgeting, pricing strategies, and profit analysis.
There are two cell phone companies that offer different packages. Company A charges a monthly service fee of [latex]$34[/latex] plus [latex]$.05[/latex]/min talk-time. Company B charges a monthly service fee of [latex]$40[/latex] plus [latex]$.04[/latex]/min talk-time.
Write a linear equation that models the packages offered by both companies.
If the average number of minutes used each month is [latex]1,160[/latex], which company offers the better plan?
If the average number of minutes used each month is [latex]420[/latex], which company offers the better plan?
How many minutes of talk-time would yield equal monthly statements from both companies?
The model for Company A can be written as [latex]A=0.05x+34[/latex]. This includes the variable cost of [latex]0.05x[/latex] plus the monthly service charge of [latex]$34[/latex].
Company B’s package charges a higher monthly fee of [latex]$40[/latex], but a lower variable cost of [latex]0.04x[/latex]. Company B’s model can be written as [latex]B=0.04x+40[/latex].
If the average number of minutes used each month is [latex]1,160[/latex], we have the following:
So, Company B offers the lower monthly cost of [latex]$86.40[/latex] as compared with the [latex]$92[/latex] monthly cost offered by Company A when the average number of minutes used each month is [latex]1,160[/latex].
If the average number of minutes used each month is [latex]420[/latex], we have the following:
If the average number of minutes used each month is [latex]420[/latex], then Company A offers a lower monthly cost of [latex]$55[/latex] compared to Company B’s monthly cost of [latex]$56.80[/latex].
To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of [latex]\left(x,y\right)[/latex] coordinates: At what point are both the [latex]x[/latex]–value and the [latex]y[/latex]–value equal? We can find this point by setting the equations equal to each other and solving for [latex]x[/latex].
Therefore, a monthly average of [latex]600[/latex] talk-time minutes renders the plans equal.
We can also use a linear equation in one variable to solve a problem with two unknowns by writing an expression for one unknown in terms of the other.A bag is filled with green and blue marbles. There are [latex]77[/latex] marbles in the bag. If there are [latex]17[/latex] more green marbles than blue marbles, find the number of green marbles and the number of blue marbles in the bag. How many marbles of each color are in the bag?
Let [latex]G[/latex] represent the number of green marbles in the bag and [latex]B[/latex] represent the number of blue marbles. We have that [latex]G + B = 77[/latex]. But we also have that there are [latex]17[/latex] more green marbles than blue. That is, the number of green marbles is the same as the number of blue marbles plus [latex]17[/latex]. We can translate that as [latex]G = B+17[/latex]. Since we’ve found a way to express the variable [latex]G[/latex] in terms of [latex]B[/latex], we can write one equation in one variable:
[latex]B+17+B=77[/latex],
which we can solve for [latex]B[/latex].
[latex]2B=60[/latex]
So [latex]B=30[/latex]. There are [latex]30[/latex] blue marbles in the bag and [latex]30+17 = 47[/latex] is the number of green marbles.
