Problem 139 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ \frac{8}{b-2}+\frac{3}{2-b}=-\frac{1}{b} $$

Step-by-Step Solution

Answer

The solution is \(b = \frac{1}{3}\).

Step 1: Recognize the Common Denominator

Examine the denominators in the equation: \((b-2)\), \((2-b)\), and \(b\). Notice how \((b-2)\) and \((2-b)\) are negatives of each other. Rewrite \(\frac{3}{2-b}\) as \(-\frac{3}{b-2}\) for a common denominator.

Step 2: Rewrite the Equation

Substitute \(-\frac{3}{b-2}\) for \(\frac{3}{2-b}\). The equation becomes: \(\frac{8}{b-2} - \frac{3}{b-2} = -\frac{1}{b}\).

Step 3: Combine Fractions on the Left-Hand Side

Combine the fractions on the left-hand side since they have the same denominator: \(\frac{8}{b-2} - \frac{3}{b-2} = \frac{5}{b-2}\). This simplifies the equation to: \(\frac{5}{b-2} = -\frac{1}{b}\).

Step 4: Cross-Multiply

Cross-multiply to eliminate the fractions: \(5b = -(b-2)\).

Step 5: Solve for b

Expand and simplify the equation: \(5b = -b + 2\). Add \(b\) to both sides to get \(6b = 2\). Divide both sides by 6 to solve for \(b\): \(b = \frac{2}{6} = \frac{1}{3}\).

Step 6: Verify the Solution

Substitute \(b = \frac{1}{3}\) back into the original equation to check if it holds true. Evaluate each term: \(\frac{8}{b-2}\), \(\frac{3}{2-b}\), and \(\frac{1}{b}\) and confirm that both sides are equal. Since both sides are equal, \(b = \frac{1}{3}\) is a valid solution.

Key Concepts

Common Denominator

When solving rational equations, finding a common denominator is often the first step. This concept helps in combining fractions, making it easier to manipulate and solve the equation. Imagine you're combining fractions much like ingredients in a recipe that need a similar base to blend perfectly.

In this example, the goal is to rewrite the fractions so they share this common base. We see the expression \((b-2)\) and \((2-b)\) present in different fractions. With careful observation, note that \((2-b)\) is simply \(-(b-2)\). This means we can transform the expression \(\frac{3}{2-b}\) into \(-\frac{3}{b-2}\) which shares \((b-2)\) as a common denominator with \(\frac{8}{b-2}\).

Once denominators are the same, fractions can be summed as straightforward as basic arithmetic operations:

• \(\frac{8}{b-2} - \frac{3}{b-2} = \frac{5}{b-2}\)

You may then proceed to tackle the rest of the equation.

Cross-Multiplication

Cross-multiplication is a powerful tool used to solve equations that involve fractions, especially when two fractions are set equal to each other. By cross-multiplying, you essentially get rid of the denominators, turning a complex equation into something more manageable.

In our exercise, once the equations are simplified to \(\frac{5}{b-2} = -\frac{1}{b}\), cross-multiplication can be used to proceed. The process involves multiplying the numerator of each fraction by the denominator of the other. This gives:

• \(5 \times b = -(1 \times (b-2))\)
• Or simply rewrite as: \(5b = -b + 2\)

All fractions are now removed, allowing you to focus solely on solving the equation.

This technique helps to simplify potentially daunting fractions and reveal the true numbers beneath them.

Verifying Solutions

Verifying solutions is an essential step in solving equations, especially for rational equations. It ensures that the values we find are valid and satisfy the original problem.

After determining \(b = \frac{1}{3}\), substitute this value back into each term of the original equation to verify it remains true. Simplify each fraction carefully to check consistency across both sides of the equation:

• First term: \(\frac{8}{(\frac{1}{3} - 2)}\)
• Second term: \(-\frac{3}{(\frac{1}{3} - 2)}\)
• Third term: \(-\frac{1}{\frac{1}{3}}\)

If both sides equate, then \(b = \frac{1}{3}\) is an accurate answer. This verification is crucial because sometimes, an unexpected restriction from the equation's domain can give an incorrect solution. Always evaluate each part thoroughly.