Problem 15
Question
The variables \(x\) and \(y\) vary directly. Use the given values to write an equation that relates \(x\) and \(y .\) $$ x=8, y=24 $$
Step-by-Step Solution
Verified Answer
The equation that relates \(x\) and \(y\) is \(y=3x\).
1Step 1: Understanding the direct variation formula
For a direct variation, we use the formula \(y = kx\), where \(k\) is the constant of variation.
2Step 2: Substitute known values into the formula
Insert the given values of \(x = 8\) and \(y =24 \) into the formula so it becomes \(24=k \cdot 8\).
3Step 3: Solve for the constant
To find the value of the constant \(k\), we need to divide 24 by 8. This gives us \(k = 3\).
4Step 4: Write the direct variation formula with the found constant
Now that we have found \(k\), we can insert it into the original equation for a direct variation to get the formula that relates \(x\) and \(y\), which is \(y=3x\).
Key Concepts
Constant of VariationWriting EquationsSolving for a Variable
Constant of Variation
In the world of mathematics, especially when dealing with direct variations, understanding the constant of variation is crucial. A direct variation between two variables implies that they change in a consistent relationship which can be defined by a simple formula.
For direct variation, we express the relationship between two variables, say, \(x\) and \(y\), using the formula:
The constant \(k\) remains the same as \(x\) and \(y\) change, defining a direct relationship between these variables. For example, if you know that when \(x = 8\), \(y = 24\), this tells us that 24 is three times 8, which means our constant of variation, \(k\), is 3.
Recognizing the constant of variation is essential in defining and predicting relationships accurately in mathematical problems.
For direct variation, we express the relationship between two variables, say, \(x\) and \(y\), using the formula:
- \(y = kx\)
The constant \(k\) remains the same as \(x\) and \(y\) change, defining a direct relationship between these variables. For example, if you know that when \(x = 8\), \(y = 24\), this tells us that 24 is three times 8, which means our constant of variation, \(k\), is 3.
Recognizing the constant of variation is essential in defining and predicting relationships accurately in mathematical problems.
Writing Equations
Turning real-world relationships into mathematical expressions can seem daunting, but it's all about understanding the relationships and applying the right formulas. When given that two variables vary directly, this implies a proportional relationship.
To express this direct variation, here's what you do:
\(24 = k \cdot 8\)
Now you have a mathematical representation of how \(y\) changes with \(x\). This step is vital because it transforms a real-world situation into a general equation that describes the relationship. Once the equation is written with the correct constant, it can be used to predict the value of one variable based on the other, showcasing the utility of writing equations.
To express this direct variation, here's what you do:
- Identify the relationship: Are the variables directly related?
- Use the direct variation formula: \(y = kx\)
- Substitute known values into the formula to identify the constant.
\(24 = k \cdot 8\)
Now you have a mathematical representation of how \(y\) changes with \(x\). This step is vital because it transforms a real-world situation into a general equation that describes the relationship. Once the equation is written with the correct constant, it can be used to predict the value of one variable based on the other, showcasing the utility of writing equations.
Solving for a Variable
Solving for a variable is a foundational skill in algebra. Once you have your equation, understanding how to isolate and solve for a variable is key. This often involves basic arithmetic operations to manipulate the equation and find the unknown variable.
Let's revisit our direct variation formula \(24 = k \cdot 8\). The goal is to find \(k\). Start by isolating \(k\) on one side of the equation. To do this, divide both sides of the equation by 8:
\[ k = \frac{24}{8} = 3\]
Now, you've solved for \(k\). What you've done is performed operations that keep the relationship balanced to solve for the constant of variation. This constant allows the variable \(x\) and \(y\) to form the equation \(y = 3x\), which you can now use for future calculations.
Solving for a variable not only helps in finding constants but also enables you to rework the equation to solve for either \(x\) or \(y\) when new values are presented. Mastering this skill is crucial in tackling more complex mathematical relationships.
Let's revisit our direct variation formula \(24 = k \cdot 8\). The goal is to find \(k\). Start by isolating \(k\) on one side of the equation. To do this, divide both sides of the equation by 8:
\[ k = \frac{24}{8} = 3\]
Now, you've solved for \(k\). What you've done is performed operations that keep the relationship balanced to solve for the constant of variation. This constant allows the variable \(x\) and \(y\) to form the equation \(y = 3x\), which you can now use for future calculations.
Solving for a variable not only helps in finding constants but also enables you to rework the equation to solve for either \(x\) or \(y\) when new values are presented. Mastering this skill is crucial in tackling more complex mathematical relationships.
Other exercises in this chapter
Problem 14
Find the quotient. $$ \text { Divide }\left(x^{2}+5 x+6\right) \text { by }(x+3) $$
View solution Problem 15
ADDING RATIONAL EXPRESSIONS. Simplify the expression. $$ \frac{4}{x+1}+\frac{2 x-2}{x+1} $$
View solution Problem 15
Solve the equation by cross multiplying. Check your solutions. \(\frac{7}{x+1}=\frac{5}{x-3}\)
View solution Problem 15
Find the least common denominator of the pair of rational expressions. $$ \frac{x+1}{15 x}, \frac{25}{18 x^{3}} $$
View solution