Question

With only the cold water valve open, it takes 8 minutes to fill the tub of a washing machine. With both the hot and cold water valves open, it takes 5 minutes. The time it takes for the tub to fill with only the hot water valve open can be modeled by the equation \(\frac{1}{8}+\frac{1}{t}=\frac{1}{5}\) where \(t\) is the time (in minutes) for the tub to fill. How long does it take for the tub of the washing machine to fill with only the hot water valve open?

Short answer

The tub of the washing machine will take \(\frac{40}{3}\), or approximately 13 and 1/3 minutes, to fill with only the hot water valve open.

Step-by-step solution

Rewrite the Equation

First, the given equation \(\frac{1}{8}+\frac{1}{t}=\frac{1}{5}\), can be rewritten to isolate \(t\). Start by subtracting \(\frac{1}{8}\) from both sides to get \(\frac{1}{t}=\frac{1}{5}-\frac{1}{8}\)

Simplify the Right Side

Next, simplify the right side of the equation by finding a common denominator for \(\frac{1}{5}\) and \(\frac{1}{8}\). The common denominator of 5 and 8 is 40. After converting each fraction to have denominator of 40, this results into \(\frac{1}{t}= \frac{8}{40}-\frac{5}{40}\)

Further Simplify the Right Side

Simplify the right side of the equation further by performing the subtraction operation which gives \(\frac{1}{t}=\frac{3}{40}\).

Solve for t

Finally, to solve for \(t\), invert both sides of the equation. Doing so would result into \(t=\frac{40}{3}\)

Key concepts

Rational Expressions

Rational expressions are fractions in which the numerator and the denominator are both polynomials. In our exercise, the expressions \(\frac{1}{8}\) and \(\frac{1}{t}\) are examples of rational expressions, while \(\frac{8}{40}\) and \(\frac{5}{40}\) serve as their common-denominator counterparts. Understanding how to work with such expressions is crucial in algebra, especially when solving equations that denote rates or proportions, like the time it takes to fill a tub.

When solving equations involving rational expressions, we often need to add or subtract them, as was done in the exercise. To perform these operations correctly, a common denominator is necessary to combine these terms. For students, grasping this concept means recognizing that without a common base, these fractions cannot be directly combined — a fundamental rule akin to the idea that you cannot add apples and oranges because they are different units.

Common Denominators

Finding common denominators is a critical step for solving rational equations that involve adding or subtracting fractions. A common denominator refers to a shared multiple of the original denominators that allows the fractions to be combined. In our exercise, the fractions \(\frac{1}{5}\) and \(\frac{1}{8}\) needed to be combined, which required finding a common denominator.

The Least Common Denominator (LCD) is the least common multiple of the original denominators — in this case, the smallest number that both 5 and 8 can divide into without a remainder, which is 40. By converting each fraction to equivalent fractions with the common denominator of 40, we can easily combine them. This step is vital for students to understand because without setting a common ground for the distinct fractions, the addition or subtraction cannot move forward.

Inverse Operations

Inverse operations are mathematical actions that reverse each other, such as addition and subtraction or multiplication and division. They're used to isolate the variable we're solving for in an equation. In the textbook exercise, after combining the rational expressions, we ended up with \(\frac{1}{t} = \frac{3}{40}\), and from this point, we needed to use inverse operations to solve for \(t\).

To isolate \(t\), we applied the multiplicative inverse (also known as a reciprocal). Since \(\frac{1}{t}\) is the same as \(t^{-1}\), we found that the inverse of this is \(t\), and thus \(t = \frac{40}{3}\) was concluded. For students, recognizing the use of inverse operations simplifies the process of solving equations; it's the method of 'undoing' what has been done to the variable to eventually get the variable by itself.