Problem 11 - Intermediate Algebra

Question

Solve equation. \(\frac{1}{a}=\frac{1}{3}-\frac{2}{3 a}\)

Step-by-Step Solution

Answer

The solution is \(a = 5\).

Step 1: Find a Common Denominator

The equation is given as \( \frac{1}{a} = \frac{1}{3} - \frac{2}{3a} \). To simplify, find a common denominator for the fractions on the right side. The common denominator for 3 and \(3a\) is \(3a\). Rewrite each fraction with the denominator \(3a\):\[\frac{1}{3} = \frac{a}{3a}, \quad \frac{2}{3a} = \frac{2}{3a}\] This gives:\[\frac{1}{a} = \frac{a}{3a} - \frac{2}{3a}\]

Step 2: Simplify the Right Side

Combine the fractions on the right side of the equation:\[\frac{1}{a} = \frac{a - 2}{3a}\]

Step 3: Eliminate the Fractions

To eliminate the fractions, cross-multiply:\[1 \cdot 3a = a(a - 2)\]This simplifies to:\[3a = a^2 - 2a\]

Step 4: Rearrange the Equation

Rearrange the equation to set it to zero:\[a^2 - 2a - 3a = 0\]Combine like terms:\[a^2 - 5a = 0\]

Step 5: Factor the Equation

Factor out \(a\) from the expression:\[a(a - 5) = 0\]

Step 6: Solve for a

Set each factor equal to zero and solve for \(a\):1. \(a = 0\)2. \(a - 5 = 0\) which gives \(a = 5\)

Step 7: Verify the Solutions

Check the solutions in the original equation. For \(a = 0\), \(\frac{1}{0}\) is undefined, so \(a = 0\) is not a valid solution. For \(a = 5\), substitute back to verify:\[\frac{1}{5} = \frac{1}{3} - \frac{2}{15} = \frac{5}{15} - \frac{2}{15} = \frac{3}{15} = \frac{1}{5}\]The equation holds true, so \(a = 5\) is a valid solution.

Key Concepts

Common Denominator

When dealing with equations that have fractions, it's helpful to rewrite the fractions with a common denominator. This makes it easier to simplify and combine the terms. For instance, the equation \( \frac{1}{a} = \frac{1}{3} - \frac{2}{3a} \) needs a common denominator on the right side.
Common denominators are like a common language for fractions—they help bring different fractions onto the same page, or in this case, onto the same denominator line.

• In this equation, the denominators are \(3\) and \(3a\).
• The smallest common denominator that both denominators can fit into without any leftovers is \(3a\).

Now, rewrite each fraction so they all share this common denominator of \(3a\):

• Rewrite \( \frac{1}{3} \) as \( \frac{a}{3a} \).
• Keep \( \frac{2}{3a} \) the same since it already has \(3a\) as its denominator.

This transformation allows you to easily subtract one fraction from the other.

Cross-Multiplication

Once fractions share a common denominator, another powerful tool to solve equations is cross-multiplication. This technique helps to get rid of fractions altogether, simplifying the equation into a more manageable form.
In the equation after finding the common denominator, we have:\[\frac{1}{a} = \frac{a - 2}{3a}\]Cross-multiplication involves multiplying diagonally across the equals sign:

• Multiply the left side's numerator by the right side's denominator: \(1 \times 3a \).
• Multiply the right side's numerator by the left side's denominator: \(a \times (a-2) \).

Setting these products equal gives the equation:\[3a = a^2 - 2a\]Cross-multiplication simplifies the equation by removing the fractions, making it easier to solve.

Factoring

After cross-multiplying, you often find yourself with a polynomial equation. Factoring is the next strategy. Factoring breaks down complex expressions into simpler ones to find solutions for the variable.
Take the equation obtained from simplifying and cross-multiplying:\[a^2 - 5a = 0\]To solve by factoring:

• Look for common factors in the equation. Here, \(a\) is common to both terms, so you can factor it out.
• This gives \(a(a - 5) = 0\), which means either \(a = 0\) or \(a - 5 = 0\).

Factoring simplifies the polynomial, breaking it into two simpler factors. This makes finding solutions straightforward, only requiring you to set each factor to zero and solve for the variable \(a\).

Verifying Solutions

The final essential step is verifying the solutions you find. Even if you've followed all the correct mathematical steps, some solutions might not work in the original equation, especially when it involves variables in the denominator.
In our equation:

• We found solutions \(a = 0\) and \(a = 5\).
• Plug these solutions back into the original equation \(\frac{1}{a} = \frac{1}{3} - \frac{2}{3a} \) to check if both sides of the equation are equal.
• If you substitute \(a = 0\), you get \(\frac{1}{0}\), which is undefined, so \(a = 0\) is not valid.
• For \(a = 5\), the equation holds as \(\frac{1}{5} = \frac{1}{5}\).

Checking your work ensures that the solutions are not just mathematically correct, but also meaningful within the context of the equation.