Problem 50 - Elementary and Intermediate Algebra

Question

Solve each formula for the specified variable. $$ \frac{S}{k+h}=E \text { for } k $$

Step-by-Step Solution

Answer

The solution for \( k \) is \( k = \frac{S - Eh}{E} \).

Step 1: Understand the equation

We have the formula \( \frac{S}{k+h} = E \). We need to solve this equation for \( k \). To do that, we need to isolate \( k \) on one side of the equation.

Step 2: Multiply both sides by \( k+h \)

To eliminate the fraction, multiply both sides of the equation by \( k+h \). This gives us: \[ S = E(k+h) \] We have removed the fraction, so now we can work towards isolating \( k \).

Step 3: Distribute \( E \) on the right-hand side

Distribute \( E \) among the terms inside the parentheses on the right-hand side: \[ S = Ek + Eh \] This gives us an equation without parentheses, making it easier to isolate \( k \).

Step 4: Subtract \( Eh \) from both sides

Subtract \( Eh \) from both sides to move terms not involving \( k \) to the left-hand side: \[ S - Eh = Ek \] This step isolates the \( k \) term by itself on the right-hand side.

Step 5: Divide by \( E \)

To solve for \( k \), divide both sides of the equation by \( E \): \[ k = \frac{S - Eh}{E} \] This gives us the expression for \( k \) in terms of \( S \), \( h \), and \( E \).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in mathematics where we rearrange and simplify equations to make them easier to work with. In the context of this exercise, we start with the equation \( \frac{S}{k+h} = E \). We need to isolate \( k \), meaning we must use algebraic techniques to change the format of the equation without altering its meaning.

The first step is eliminating the fraction by multiplying both sides by \( k+h \). This use of multiplication is a key algebraic manipulation technique. After multiplying, the equation becomes \( S = E(k+h) \). By doing this, we are simplifying the equation from a fraction to a linear form.

Next, you distribute \( E \) across \( (k+h) \), which leads to \( S = Ek + Eh \). Distributing is another essential algebraic tool that involves multiplying each term inside the parenthesis individually by the term outside. This step is crucial because it simplifies the equation further, making it easier to isolate \( k \) in the subsequent steps.

Isolation of Variables

Isolation of variables is the process of rearranging an equation so that the unknown variable is by itself on one side of the equation. This is exactly what we do in our exercise to solve for \( k \). After transforming the equation to \( S = Ek + Eh \), the goal is to isolate \( k \).

To begin isolating \( k \), we subtract \( Eh \) from both sides, arriving at \( S - Eh = Ek \). This subtraction is strategic because it moves terms involving the variable \( k \) to one side, while everything else goes to the opposite side. The key to isolation is ensuring that only the variable you're solving for (\( k \) in this case) remains on one side.

Finally, to completely isolate \( k \), we divide both sides by \( E \) to obtain \( k = \frac{S - Eh}{E} \). Dividing by \( E \) ensures \( k \) is alone, providing a solution in terms of the other given variables. Effective isolation helps clarify the relationship between variables in an equation.

Literal Equations

Literal equations, or formulas, have multiple variables and are used in various fields such as physics, chemistry, and mathematics. These equations express relationships between quantities. In this exercise, \( \frac{S}{k+h}=E \) is a literal equation because it contains more than one variable.

Solving literal equations involves isolating one of the variables based on given conditions. This is beneficial when needing to express one variable in terms of others for practical applications. For example, in physics, you might need to solve for time in terms of distance and speed.

By working through this exercise, we transformed \( \frac{S}{k+h}=E \) to a solution for \( k \) as \( k = \frac{S - Eh}{E} \). This illustrates how you can convert a complex equation into a more usable form, showcasing the power of algebraic manipulation and isolation techniques. Understanding how to manipulate literal equations prepares you to tackle real-world problems where multiple changing variables are involved.