Problem 86 - Beginning and Intermediate Algebra

Question

Solve. \(\frac{1}{t}=\frac{1}{a}+\frac{1}{b},\) for \(t\) (A formula for work rate)

Step-by-Step Solution

Answer

In short, to solve the work rate formula \(\frac{1}{t} = \frac{1}{a} + \frac{1}{b}\) for \(t\), you can follow these steps: find a common denominator (ab), combine the fractions, and take the reciprocal of both sides. The final solution is: \(t = \frac{ab}{a+b}\).

Step 1: Find a common denominator

To find a common denominator for the fractions \(\frac{1}{a}\) and \(\frac{1}{b}\), we can multiply 'a' and 'b'. The common denominator for both fractions is 'ab'.

Step 2: Combine the fractions

Now, we will combine the right side fractions as follows: \(\frac{1}{t} = \frac{1}{a} + \frac{1}{b}\) \(\frac{1}{t} = \frac{b}{ab} + \frac{a}{ab}\) \(\frac{1}{t} = \frac{a+b}{ab}\)

Step 3: Take the reciprocal of both sides

To isolate 't', take the reciprocal of both sides of the equation. The reciprocal of a fraction is obtained by swapping the numerator and the denominator: \(\frac{1}{\frac{1}{t}} = \frac{1}{\frac{a+b}{ab}}\) \(t = \frac{ab}{a+b}\) The value for \(t\) is: \(t = \frac{ab}{a+b}\)

Key Concepts

Understanding Work Rate Problems

Work rate problems often involve understanding how multiple workers or machines can complete a task together. If one person or machine can finish a job in a certain time, you can describe their contribution to the task as a fraction of the work done per unit of time. In the given formula \( \frac{1}{t} = \frac{1}{a} + \frac{1}{b} \), \( t \) represents the total time needed to complete the work if both contribute together. It combines individual rates where \( a \) and \( b \) are times for each person to complete the work alone. This formula helps us find a shared speed without calculating time per unit manually.
Using such equations, you solve for the time it takes when resources are pooled together, making it practical for solving real-world tasks in industries like manufacturing or project management.

Solving Equations by Combining Fractions

Solving equations, especially with fractions, is a skill that involves finding a common ground to combine separate terms. When fractions share a common denominator, it's easier to add or subtract them. In our exercise, the equation \( \frac{1}{t} = \frac{1}{a} + \frac{1}{b} \) requires combining two fractions:

• First, find a common denominator (the product of \( a \) and \( b \), which is \( ab \)).
• Then rewrite the fractions: \( \frac{1}{a} \) becomes \( \frac{b}{ab} \) and \( \frac{1}{b} \) becomes \( \frac{a}{ab} \).
• Combine them into a single fraction, \( \frac{a+b}{ab} \).

This method makes complex additions look simple, showing that the abstract parts come together beautifully into an understanding that's applicable in broader scenarios.

Working with Fractions

Fractions are central to many mathematical problems. They represent parts of a whole, and their notational simplicity translates into more accessible, deeper ideas, such as division and ratios. In a work rate problem, each person’s time contribution is represented by a fraction, allowing for easy addition to find combined efforts.
Specific rules apply when adding or subtracting fractions. Key to these operations is the concept of a common denominator, the term that allows you to equate fractions in a standard form:

• This involves multiplying the denominators and adjusting numerators accordingly.
• For instance, \( \frac{1}{a} \) and \( \frac{1}{b} \) become \( \frac{b}{ab} \) and \( \frac{a}{ab} \), respectively, using \( ab \) as the common denominator.

Getting comfortable with these transforms builds a strong foundation for understanding more complex mathematics.

Mastering Reciprocals

In mathematics, the reciprocal of a number is its inverse, found by flipping its numerator and denominator. For example, the reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \). When solving our work rate equation, this concept helps to isolate \( t \).
For the equation \( \frac{1}{t} = \frac{a+b}{ab} \), making \( t \) the subject involves taking the reciprocal of both sides:

• This flips \( \frac{1}{t} \) to \( t \) and \( \frac{a+b}{ab} \) to \( \frac{ab}{a+b} \).
• Reciprocals are useful not only in solving equations but also in understanding multiplicative relationships.

Grasping this principle is essential for progressing in algebra and understanding how components interact in a fraction-driven framework.