Problem 26 - Algebra for College Students

Question

$$ \mathrm{C}=\frac{5}{9}(F-32) \quad \text { for } \mathrm{F} $$

Step-by-Step Solution

Answer

\( F = \frac{9 \mathrm{C} + 160}{5} \)

Step 1: Identify the Equation

The given equation to solve is \( \mathrm{C} = \frac{5}{9}(F-32) \). We need to solve this equation for \( F \).

Step 2: Clear the Fraction

To eliminate the fraction, multiply both sides of the equation by 9, resulting in \( 9 \mathrm{C} = 5(F - 32) \).

Step 3: Distribute the 5

Use the distributive property to expand the right side of the equation: \( 9 \mathrm{C} = 5F - 160 \).

Step 4: Isolate the Term with F

Add 160 to both sides to start isolating the \( F \) term: \( 9 \mathrm{C} + 160 = 5F \).

Step 5: Solve for F

Divide both sides by 5 to solve for \( F \): \( F = \frac{9 \mathrm{C} + 160}{5} \).

Key Concepts

Temperature Conversion

Temperature conversion is essential when dealing with different measurement units across the world. The equation \( \mathrm{C} = \frac{5}{9}(F-32) \) is commonly used to convert Fahrenheit to Celsius. In this formula:

• \( \mathrm{C} \) represents the temperature in Celsius.
• \( \mathrm{F} \) represents the temperature in Fahrenheit.

This conversion is vital for scientific calculations, travel, and cooking. To make it practical:
1. Identify the current temperature in Fahrenheit.
2. Subtract 32 from the Fahrenheit value.
3. Multiply the result by \( \frac{5}{9} \) to find the Celsius temperature.Understanding temperature conversion helps in grasping how different regions perceive weather and climate differently due to varying scales of measurement. It also sharpens calculation skills used in everyday contexts.

Isolating Variables

Isolating variables is a fundamental skill in algebra, helping us solve equations and find unknown values. When solving \( \mathrm{C} = \frac{5}{9}(F-32) \) for \( F \), transforming the equation involves isolating \( F \) to express it only in terms of \( C \):
1. Begin by removing fractions to simplify the problem, resulting in an equation without fractions.
2. Rearrange the equation by performing operations that will "get rid of" additional constants around the variable.

• In our example, start by multiplying both sides by 9 to clear the fraction \( \frac{5}{9} \).
• Then distribute the 5 across the terms within the parentheses.
• Finally, add or subtract terms as needed to isolate \( F \).

This step-by-step approach is not only helpful in solving equations but also provides insight into how equations work. It aids in understanding complex algebraic problems and is critical in scientific computations and engineering tasks.

Distributive Property

The distributive property is an important algebraic tool used to simplify and expand expressions. It involves multiplying a single term across terms inside a parenthesis. In our exercise, we have transformed \( 9 \mathrm{C} = 5(F - 32) \) by distributing 5:
1. Expand the expression \( 5(F - 32) \) into \( 5F - 160 \).

• Distribute the 5 to both \( F \) and -32 within the parentheses.

Applying the distributive property simplifies equations and helps in isolating variables by making expressions easier to handle. This method is widely used in algebra to deal with more complicated equations and is foundational in advanced mathematical problems. Developing a solid understanding of this property enhances one's ability to manipulate and solve diverse algebraic expressions efficiently.