Problem 25
Question
The variables x and y vary inversely. Use the given values to write an equation that relates x and y. $$x=\frac{1}{2}, y=8$$
Step-by-Step Solution
Verified Answer
The equation that relates x and y is: y = 4 / x
1Step 1: Identify the relation
Given that x and y vary inversely, this means that the product of x and y is always constant. We can express this relationship as y = k/x where k is the constant of variation.
2Step 2: Substitute given values
We are given that x = 1/2 and y = 8. Substituting these values into the equation gives 8 = k / (1/2).
3Step 3: Solve for k
To solve for k, we multiply both sides of the equation by 1/2. Doing so, we find that k = 4.
4Step 4: Write the equation
Using the value of k, we can write the equation that relates x and y as y = 4 / x.
Key Concepts
Constant of VariationSubstitutionSolving for Constants
Constant of Variation
The constant of variation in an inverse variation relationship is similar to the glue that holds the relationship together—it's the consistent figure that does not change as the two variables do. In essence, when two quantities vary inversely, it means there's a direct negative correlation between them; as one quantity increases, the other decreases, and vice versa.
For inverse variation, represented by the equation \( y = \frac{k}{x} \), the constant of variation is denoted by \( k \). It's critical to pinpoint this constant because it allows us to create a formula that defines the relationship between the variables for all values. With knowledge of the constant, one can predict one variable when given the other, a valuable tool for solving real-world problems where such relationships occur, like gravitational force and distance, electrical resistance and current, or even in financial models.
For inverse variation, represented by the equation \( y = \frac{k}{x} \), the constant of variation is denoted by \( k \). It's critical to pinpoint this constant because it allows us to create a formula that defines the relationship between the variables for all values. With knowledge of the constant, one can predict one variable when given the other, a valuable tool for solving real-world problems where such relationships occur, like gravitational force and distance, electrical resistance and current, or even in financial models.
Substitution
Substitution is a method frequently used in algebra to isolate a specific variable or to insert known values into equations to solve for unknown quantities. It's a process of exchanging variables for their corresponding values. For example, if we know that \( x = \frac{1}{2} \) and \( y = 8 \) in an equation that represents an inverse variation like \( y = \frac{k}{x} \), we apply substitution by plugging those values into the equation in place of the variables.
When performed carefully, substitution can lead you directly to the values of constants or other variables. It acts as a bridge between the known and the unknown, simplifying equations to a point where they become solvable.
When performed carefully, substitution can lead you directly to the values of constants or other variables. It acts as a bridge between the known and the unknown, simplifying equations to a point where they become solvable.
Solving for Constants
When working with equations that include constants, such as the constant of variation in inverse relationships, we often need to solve for these constants to fully utilize the equation for further calculations. Solving for constants typically involves isolating the constant on one side of the equation. This can include basic algebraic manipulations: adding, subtracting, multiplying, dividing both sides of an equation, or even performing more complex operations depending on the equation's form.
In the context of the inverse relationship where \( y = \frac{k}{x} \), and you're given specific values for \( x \) and \( y \), you can solve for \( k \) by rearranging the equation to take the form \( k = xy \). This process turns an abstract concept into a tangible figure, unlocking the door to a deeper understanding of the mathematical relationship at hand, and paving the way for applying this relationship in various scenarios.
In the context of the inverse relationship where \( y = \frac{k}{x} \), and you're given specific values for \( x \) and \( y \), you can solve for \( k \) by rearranging the equation to take the form \( k = xy \). This process turns an abstract concept into a tangible figure, unlocking the door to a deeper understanding of the mathematical relationship at hand, and paving the way for applying this relationship in various scenarios.
Other exercises in this chapter
Problem 25
Solve the percent problem. How much is \(8.2 \%\) of 800 tons?
View solution Problem 25
Simplify the expression. $$\frac{x^{2}-36}{-5 x^{2}} \div(x-6)$$
View solution Problem 26
Simplify the expression. $$\frac{3 x+10}{7 x-4}-\frac{x}{4 x+3}$$
View solution Problem 26
Solve the equation. $$\frac{1}{4}+\frac{4}{x}=\frac{1}{x}$$
View solution