Problem 44 - Algebra and Trigonometry

Question

The given equation is either linear or equivalent to a linear equation. Solve the equation. \(\frac{1}{1-\frac{3}{2+w}}=60\)

Step-by-Step Solution

Answer

\(w = \frac{62}{59}\)

Step 1: Simplify the Denominator

Start with the inner part of the equation: \(1 - \frac{3}{2+w}\). To simplify, let the denominator be \(D = 2 + w\). Then the expression becomes \(1 - \frac{3}{D}\).

Step 2: Create a Common Denominator

To simplify \(1 - \frac{3}{D}\), convert the number 1 to \(\frac{D}{D}\). Hence, the expression becomes \(\frac{D - 3}{D}\).

Step 3: Substitute Back into the Original Equation

Replace \(1 - \frac{3}{2+w}\) with \(\frac{D - 3}{D}\). The equation now becomes \(\frac{1}{\frac{D - 3}{D}} = 60\).

Step 4: Simplify the Fraction

Simplify \(\frac{1}{\frac{D - 3}{D}}\) by taking the reciprocal: \(\frac{D}{D - 3}\). Thus, the equation is \(\frac{D}{D - 3} = 60\).

Step 5: Cross-Multiply to Solve for D

Cross-multiply to solve for \(D\): \(D = 60(D - 3)\). Expand this to get \(D = 60D - 180\).

Step 6: Rearrange to Isolate D

Subtract \(60D\) from both sides: \(D - 60D = -180\). Simplify to get \(-59D = -180\).

Step 7: Solve for D

Divide both sides by -59: \(D = \frac{180}{59}\).

Step 8: Return to the Original Expression for w

Recall that \(D = 2 + w\), so substitute back: \(2 + w = \frac{180}{59}\).

Step 9: Solve for w

Subtract 2 from both sides to isolate \(w\): \(w = \frac{180}{59} - 2\).

Step 10: Simplify the Final Expression

Calculate \(w = \frac{180 - 2 \times 59}{59} = \frac{180 - 118}{59} = \frac{62}{59}\). Therefore, \(w = \frac{62}{59}\).

Key Concepts

Cross-Multiplication

Cross-multiplication is a powerful tool for solving equations that involve fractions or ratios. It involves the process of multiplying across the diagonal of a fraction equation to eliminate the fractions.

This technique often simplifies the equation significantly, making it easier to solve for the unknown variable. Consider an equation like \( \frac{a}{b} = \frac{c}{d} \). To cross-multiply, you multiply the numerator of one fraction by the denominator of the other, and vice versa. This gives you an equation without fractions: \( a \times d = b \times c \).

In the original exercise, after simplifying the fractions, the equation \( \frac{D}{D-3} = 60 \) was solved using cross-multiplication, resulting in \( D = 60(D - 3) \). Cross-multiplication made it possible to clear out the fraction part of the problem entirely, streamlining the path to the solution.

Simplifying Fractions

Simplifying fractions is the process of making a fraction as simple as possible. This means the numerator and the denominator have no common factors other than 1.

There are several steps for simplifying a fraction, which help in making complex problems easier to handle:

• Identify any common factors of the numerator and the denominator.
• Divide both the numerator and the denominator by their greatest common factor (GCF).
• Repeat the process if further simplification is possible.

In our exercise, the fraction \( \frac{1}{\frac{D-3}{D}} \) was simplified by understanding that taking the reciprocal is a way to simplify, leading directly to \( \frac{D}{D-3} \). This step was crucial because it brought the equation to a simpler form that's easier to solve.

Common Denominator

Finding a common denominator is essential when dealing with expressions or equations that involve multiple fractions. A common denominator allows you to combine, compare, or subtract fractions by rewriting them with the same denominator.

• Identify the least common multiple (LCM) of the denominators involved.
• Adjust the numerators according to the new denominator.

In the example provided, we converted the number 1 to \( \frac{D}{D} \) to match the denominator of \( \frac{3}{D} \). By approaching the problem this way, we crafted a new, unified expression \( \frac{D - 3}{D} \), which was instrumental for further simplification and solving the equation.

Reciprocal

The concept of a reciprocal involves flipping a fraction over. In mathematical terms, the reciprocal of a fraction \( \frac{a}{b} \) is \( \frac{b}{a} \). This property is especially useful in solving equations because taking the reciprocal of both sides maintains the equality and often simplifies operations.

When dealing with a complex fraction, finding its reciprocal can be a key step in rearranging and solving the equation, as demonstrated in our exercise.

The expression \( \frac{1}{\frac{D-3}{D}} \) involved a fraction within a fraction, often referred to as a complex fraction. To simplify it, the reciprocal \( \frac{D}{D-3} \) was taken, effectively removing the fraction from the denominator and streamlining the calculation. Recognizing when to use the reciprocal is essential in turning tricky algebraic expressions into solvable equations.