Problem 88 - Elementary Algebra

Question

For the following problems, solve the rational equations. Solve \(\frac{1}{R}=\frac{1}{E}+\frac{1}{F}\) for \(E\).

Step-by-Step Solution

Answer

Question: Solve for E in the given rational equation: \(\frac{1}{R} = \frac{1}{E}+\frac{1}{F}\). Answer: \(E = \frac{EF}{R+F}\)

Step 1: Determine the common denominator

To determine the common denominator, we need to find the least common multiple (LCM) of R, E, and F. The LCM of R, E, and F is REF.

Step 2: Rewrite the equation with a common denominator

Multiply both sides of the equation by the LCM, REF, to rewrite the equation with a common denominator:\[(REF)\cdot\frac{1}{R}=(REF)\cdot\left(\frac{1}{E}+\frac{1}{F}\right).\]

Step 3: Simplify the equation

Simplify the equation by canceling out the common factors in the numerator and the denominator:\[EF = RF + ER.\]

Step 4: Solve for E

To solve for E, first factor out the E on the right side of the equation:\[EF = E(R + F).\]Now, divide both sides of the equation by (R+F) to isolate E:\[E = \frac{EF}{R+F}.\] The solution for the given rational equation is \(E = \frac{EF}{R+F}\).

Key Concepts

Understanding the Least Common Multiple

When solving rational equations, one crucial step often involves finding the least common multiple (LCM) of the denominators involved. The LCM is the smallest number that is a multiple of two or more numbers. This helps combine fractions by giving them a shared base.

For instance, in the equation \( \frac{1}{R} = \frac{1}{E} + \frac{1}{F} \), the denominators are \( R, E, \) and \( F \). To clear the fractions, we find that the LCM of these values is \( REF \). By using this LCM, each term of the equation can be adjusted to have the same denominator, simplifying the process of solving further by allowing other operations.

The Role of a Common Denominator in Equations

In rational equations, a common denominator allows us to add, subtract, or compare fractions more easily since each fraction has the same base. This process is a bridge to manipulating the equation further.

• Once the LCM is determined, it is multiplied throughout the equation to achieve a common denominator for each term. This ensures each term in the equation can "speak the same language," making the equation easier to handle.
• By multiplying the entire equation by the LCM \( REF \), the original equation \( \frac{1}{R} = \frac{1}{E} + \frac{1}{F} \) becomes \( (REF)\cdot\frac{1}{R} = (REF)\cdot\left(\frac{1}{E} + \frac{1}{F}\right) \). This transforms it into a form where each part of the equation can simply be dealt with without the clutter of denominators.

Simplifying Equations through Elimination

Simplifying an equation is pivotal as it equates to peeling away layers of complexity to reveal a simple arithmetic problem.

By canceling out the denominators after achieving a common denominator, what remains is an equation involving only the numerators. In this problem, the step moves from the LCM adjusted equation to \( EF = RF + ER \) by canceling out terms.
From here, we identify similar terms and factor common elements. For example, the factor \( E \) is found in both terms on the right side, letting us rewrtie the equation as \( EF = E(R + F) \).
The final step involves isolating \( E \), which is achieved by dividing both sides by \( R+F \). You end up with the straightforward expression \( E = \frac{EF}{R+F} \), which conveys the solved expression for \( E \) in its simplest form.