Problem 6 - Beginning and Intermediate Algebra

Question

Solve. $$\frac{1}{t}+\frac{1}{2 t}+\frac{1}{3 t}=5$$

Step-by-Step Solution

Answer

The solution to the equation \(\frac{1}{t}+\frac{1}{2t}+\frac{1}{3t}=5\) is \(t = \frac{11}{30}\).

Step 1: Find the common denominator

Here, we need to find a common denominator so that we can add the three fractions. The common denominator for \(t, 2t\), and \(3t\) is the least common multiple (LCM) of the three denominators, which is \(6t\).

Step 2: Combine the fractions

Now that we have a common denominator, we will express each fraction in terms of the common denominator, \(6t\), and add them together. \(\frac{1}{t}\) can be written as \(\frac{6}{6t}\) \(\frac{1}{2t}\) can be written as \(\frac{3}{6t}\) \(\frac{1}{3t}\) can be written as \(\frac{2}{6t}\) Now, we can add the fractions: \(\frac{6}{6t} + \frac{3}{6t} + \frac{2}{6t} = \frac{6+3+2}{6t} = \frac{11}{6t}\) Now, our equation is: \(\frac{11}{6t} = 5\)

Step 3: Solve for t

To find the value of t that satisfies the equation, we can cross-multiply: \(11 = 30t\) Now, divide by 30 to find the value of t: \(t = \frac{11}{30}\) The solution to the equation is \(t = \frac{11}{30}\).

Key Concepts

Least Common Multiple (LCM)

When working with algebraic fractions, like in our exercise, finding the Least Common Multiple (LCM) is a key step to combining fractions effectively. The LCM is the smallest number that is a multiple of each of the denominators involved. For denominators that involve variables, like \(t, 2t,\) and \(3t\), we consider both the numerical and variable parts.
The LCM of \(t, 2t,\) and \(3t\) must include the highest power of each prime factor and all variables involved. Here, \(t\) is a common factor in each term. Thus, to find the LCM:

• Look at the coefficients 1, 2, and 3. The LCM of these numbers is 6.
• Each term includes \(t\). The LCM must include \(t\) as well.

Combining these considerations, the LCM of \(t, 2t,\) and \(3t\) turns out to be \(6t\). This allows us to rewrite each fraction with \(6t\) as the denominator, facilitating the addition of fractions.

Solving Linear Equations

Solving linear equations is the process of finding the value of the variable that makes the equation true. In our exercise, after combining the fractions, we are left with the linear equation \(\frac{11}{6t} = 5\).
The goal is to isolate the variable \(t\). Here’s how you do it:

• First, recognize that the equation is in a form that allows us to cross-multiply to eliminate the fractions.
• Cross-multiplying involves multiplying each side of the equation by the other's denominator to clear the fraction.
• Once the fractions are cleared, you can solve the resulting equation using algebraic manipulation, keeping \(t\) by itself on one side.

In this specific problem, the cross-multiplication leads to \(11 = 30t\). Dividing both sides by 30 gives \(t = \frac{11}{30}\). This is the value of \(t\) that balances the equation.

Cross Multiplication

Cross multiplication is a handy technique often used to eliminate fractions from an equation quickly. It's especially useful when ratios or proportions are present.
In our exercise, after determining the equivalent fractions, you use cross multiplication to solve \(\frac{11}{6t} = 5\). Here's how it works:

• Cross multiplication involves multiplying the numerator of each side by the denominator of the other side.
• This turns the fraction equation \(\frac{11}{6t} = 5\) into an equivalent equation without fractions: \(11 = 30t\).

With the fractions eliminated, the equation becomes a simple linear equation that can be solved with basic algebra, as shown earlier. Cross multiplication is particularly useful in problems involving two fractions set equal to one another, helping to streamline what could otherwise be a more complex procedure.