Problem 69 - Elementary Algebra

Question

For the following problems, solve each literal equation for the designated letter. \(\frac{1}{R}=\frac{1}{E}+\frac{1}{F}\) for \(F\)

Step-by-Step Solution

Answer

Answer: \(F = \frac{EF - RE}{(E - 1)R}\)

Step 1: Find the common denominator for fractions with E and F

To find a common denominator for the fractions \(\frac{1}{E}\) and \(\frac{1}{F}\), we will use the product of \(E\) and \(F\), which is \(EF\).

Step 2: Write both terms with the same denominator

Now we will rewrite both fractions with the common denominator of \(EF\). To do this, multiply the numerator and denominator of \(\frac{1}{E}\) by \(F\) and the numerator and denominator of \(\frac{1}{F}\) by \(E\). We will get: \(\frac{1}{R} = \frac{F}{EF} + \frac{E}{EF}\)

Step 3: Combine the fractions on the right side of the equation

Now that both fractions on the right side of the equation have the same denominator, we can combine the numerators over the same denominator: \(\frac{1}{R} = \frac{F + E}{EF}\)

Step 4: Cross-multiply to clear the fractions

To get rid of the fractions, we'll cross-multiply. Multiply both sides of the equation by \(R\), and then multiply the right side by the denominator \(EF\). This gives: \(1 = R \cdot \frac{F + E}{EF}\) Now multiply both sides by \(EF\) to clear the fractions: \(EF = R(F + E)\)

Step 5: Distribute R on the right side

Distribute \(R\) to both terms inside the parenthesis on the right side of the equation: \(EF = RF + RE\)

Step 6: Isolate the F term

To isolate the \(F\) term, subtract \(RE\) from both sides of the equation: \(EF - RE = RF\)

Step 7: Solve for F

Finally, to solve for \(F\), divide both sides of the equation by \(R(E - 1)\): \(F = \frac{EF - RE}{(E - 1)R}\)

Key Concepts

Algebraic Fraction

An algebraic fraction is similar to a regular fraction, but instead of integers, its numerator or denominator contains an algebraic expression. In the given exercise, \( \frac{1}{R} \) is an algebraic fraction because it involves the variable \( R \). To work with algebraic fractions, you apply the same principles as with numerical fractions—especially when adding, subtracting, or finding equivalent fractions.

For instance, when solving for \( F \) in the equation \( \frac{1}{R} = \frac{1}{E} + \frac{1}{F} \), it's necessary to manage the algebraic fractions on the right side to combine them. This process may involve finding a common denominator, rewriting fractions, and eventually clearing the fractions to solve for the desired variable.

Common Denominator

A common denominator is essential when dealing with multiple fractions, as it allows them to be combined or compared. When two or more algebraic fractions have different denominators, you often need to express them with a common denominator before you can proceed.

In the context of the exercise, to add the fractions \( \frac{1}{E} \) and \( \frac{1}{F} \) together, you first identify a denominator that both \( E \) and \( F \) will divide into evenly—which is \( EF \), their product. This shared common denominator becomes the foundation for combining the fractions and simplifying the subsequent steps in solving for \( F \).

Cross Multiplication

Cross multiplication is a handy technique used in solving equations involving two fractions. This method allows you to eliminate the fractions and make the equation easier to solve. It involves multiplying the numerator of each fraction by the denominator of the other fraction.

In the solution provided, cross multiplication is utilized in Step 4, where \( 1 \) is multiplied by the denominator \( EF \) and the \( R \) on the left side is multiplied by the numerator \( F + E \) of the other side to clear the fractions. Cross multiplication simplifies the equation \( \frac{1}{R} = \frac{F + E}{EF} \) to \( EF = R(F + E) \) and sets the stage for further manipulation to isolate \( F \).

Literal Equation Manipulation

Literal equations contain two or more variables and require manipulation to solve for a specific variable. This manipulation includes the application of various algebraic operations such as addition, subtraction, multiplication, division, and factoring.

Using the steps provided to solve for \( F \) in the given equation, you see the manipulation process in action. Starting with finding common denominators, combining like terms, and cross multiplication, the process continues with distributing \( R \) across terms and subsequent subtraction and division to isolate \( F \). Literal equation manipulation can be intricate, but understanding each step and its purpose helps to demystify the process for anyone working to solve such equations.