Problem 40 - Algebra and Trigonometry

Question

\(9- 46\) 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

The solution is \( w = \frac{62}{59} \).

Step 1: Simplify the Denominator

We start with the equation \( \frac{1}{1 - \frac{3}{2+w}} = 60 \). To simplify the denominator, we recognize that \( 1 - \frac{3}{2+w} \) can be rewritten with a common denominator, which is \( 2+w \). So, we have \( \frac{2+w}{2+w} - \frac{3}{2+w} = \frac{(2+w)-3}{2+w} = \frac{w-1}{2+w} \).

Step 2: Set Up the Equation

After simplifying, our equation becomes \( \frac{1}{\frac{w-1}{2+w}} = 60 \). Using the property of reciprocals, this is equivalent to \( \frac{2+w}{w-1} = 60 \).

Step 3: Cross-Multiply

To eliminate the fraction, we cross-multiply: \( 2+w = 60(w-1) \). This results in the equation \( 2+w = 60w - 60 \).

Step 4: Rearrange the Equation

Next, we move all terms involving \( w \) to one side of the equation: \( 2+w = 60w - 60 \). Subtract \( w \) from both sides to get \( 2 = 59w - 60 \).

Step 5: Solve for \( w \)

Add 60 to both sides to get \( 62 = 59w \). Then, divide both sides by 59 to solve for \( w \): \( w = \frac{62}{59} \).

Key Concepts

Reciprocals

In mathematics, the reciprocal of a number is essentially "one divided by" that number. For instance, the reciprocal of 5 is \( \frac{1}{5} \). Understanding reciprocals is key when dealing with fractions in equations.

In the original exercise, the equation \( \frac{1}{\frac{w-1}{2+w}} = 60 \) makes use of the reciprocal concept. By taking the reciprocal of a fraction, \( \frac{w-1}{2+w} \), we flip it, getting \( \frac{2+w}{w-1} \).

This simplification transforms the original complex equation into something that is straightforward to solve, thereby helping us deal with and solve for the variable \( w \). Remember that using reciprocals is a powerful tool for simplifying complex fractional equations.

Cross-Multiplication

Cross-multiplication is an efficient process used to solve equations involving fractions. It involves multiplying across the "equal" sign, considering terms across the diagonal of fractions.

When we reached the equation \( \frac{2+w}{w-1} = 60 \) in the solution, cross-multiplication was applied to eliminate the fraction. Here's how it works:

• Multiply \( 2+w \) by 1 (the denominator of 60).
• Multiply 60 by \( w-1 \) (the denominator of the fraction).

This gives the new equation: \( 2+w = 60(w-1) \).

This step effectively removes the fractions from our equation, simplifying it to a manageable linear equation that we can solve using basic algebraic techniques.

Common Denominators

Finding common denominators is crucial when subtracting or adding fractions. It ensures that all components are in a comparable form, eliminating complex fractions.

In the exercise, the denominator \( 1 - \frac{3}{2+w} \) was simplified by expressing both terms with a common denominator.

• Convert 1 to \( \frac{2+w}{2+w} \).
• Subtract \( \frac{3}{2+w} \) from \( \frac{2+w}{2+w} \).

This gave us \( \frac{(2+w)-3}{2+w} \), which simplifies to \( \frac{w-1}{2+w} \).

Using common denominators simplifies the fractional expression and allows us to proceed with further operations like taking the reciprocal or cross-multiplying, as seen in the solution process. This makes complex equations more approachable and easier to solve.