Problem 20 - Algebra A Combined Function

Question

Solve each formula for the specified variable. \(D=\frac{1}{4} f k\) for \(k\)

Step-by-Step Solution

Answer

The solution for \(k\) is \(k = \frac{4D}{f}\).

Step 1: Understand the Given Formula

The problem provides us with the formula \(D=\frac{1}{4}fk\), where \(D\) is expressed as a function of \(f\) and \(k\). Our task is to solve this formula for the variable \(k\).

Step 2: Isolate the Fraction Coefficient

First, we need to eliminate the fraction from the right side. This can be done by multiplying both sides of the equation by 4, which is the denominator in \(\frac{1}{4}\). Our equation becomes: \[4D = fk\]

Step 3: Solve for k

Now that the fraction is gone, our next step is to solve for \(k\). This is done by dividing both sides of the equation by \(f\). Thus, the equation becomes: \[k = \frac{4D}{f}\]

Step 4: Conclusion

We successfully solved the given formula for the variable \(k\). The final expression shows \(k\) in terms of \(D\) and \(f\).

Key Concepts

Formula Solving

Solving a formula involves finding an expression for a specific variable in terms of the other variables present in the equation. In our given problem, we need to determine the value of the variable \(k\) in terms of \(D\) and \(f\). This process is a core skill in algebra, as it allows us to rearrange equations and understand relationships between variables.

• Identify the equation you need to work with.
• Focus on the variable you need to isolate.
• Apply algebraic operations systematically to express that variable explicitly.

When you solve formulas, remember to apply each operation to both sides of the equation. This maintains the equality and helps in manipulating the equation accurately and effectively. Formula solving is the first step, setting the groundwork for the other processes such as variable isolation and fraction elimination.

Variable Isolation

Variable isolation is crucial when solving equations for a particular variable. In the context of our problem, isolating \(k\) means manipulating the equation so that \(k\) stands alone on one side. This often involves reversing operations around the variable.In our equation:\[D = \frac{1}{4}fk\]We first aim to remove any coefficients or terms attached to \(k\) besides a possible solitary constant or numerical coefficient. Typically, you may need to:

• Subtract or add terms from both sides that do not involve the variable you want to isolate.
• Use multiplication or division to remove any coefficients directly affecting the variable.

In our step-by-step solution, we multiply both sides by 4 to simplify the fraction. This process eliminates the fractional coefficient from \(k\), leading us to a cleaner equation \(4D = fk\). Successfully isolating the variable allows us to focus on making the variable explicit, which feeds directly into the next steps.

Fraction Elimination

Fractions in equations can make solving them a bit more complex. Eliminating fractions simplifies the process, allowing for easier manipulation of the equation. This involves clearing any fractions by finding a common factor or multiplying through by the denominator.In our equation:\[D = \frac{1}{4}fk\]To eliminate the fraction, we multiply both sides by 4:\[4D = fk\]By multiplying through by the denominator, we remove the fraction part, simplifying our computation. This step ensures that we have a straightforward algebraic expression to work with next, facilitating further operations like variable isolation.

Substitution

Substitution involves inserting known values into an equation to solve for another variable. While substitution is not a primary step in our current exercise, it becomes valuable when we are dealing with equations that include known quantities or solving systems of equations with multiple variables.Once we have isolated \(k\) from our equation:\[k = \frac{4D}{f}\]You can substitute any known values of \(D\) and \(f\) to directly calculate the value of \(k\). Substitution provides tangible results from abstract equations, bridging the gap between theoretical algebra and practical application. It's a powerful tool when working through multiple complex equations, helping to simplify and solve based on specific scenarios.