Problem 68 - Algebra 2

Question

Solve each equation or formula for the specified variable. $$ C=\frac{5}{9}(F-32), \text { for } F $$

Step-by-Step Solution

Answer

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

Step 1: Understand the given equation

We are given the equation for temperature conversion from Fahrenheit to Celsius: \( C = \frac{5}{9}(F-32) \). We need to solve this equation for the variable \( F \) (Fahrenheit).

Step 2: Isolate the parenthesis term

To eliminate the fraction, multiply both sides of the equation by \( 9 \) to get rid of the denominator: \( 9C = 5(F - 32) \).

Step 3: Distribute the 5

Next, distribute the \(5\) on the right-hand side: \( 9C = 5F - 160 \).

Step 4: Isolate the F term

To get \( F \) by itself, add \( 160 \) to both sides of the equation: \( 9C + 160 = 5F \).

Step 5: Solve for F

Finally, divide both sides of the equation by \( 5 \) to solve for \( F \): \( F = \frac{9C + 160}{5} \).

Key Concepts

Temperature Conversion

When we talk about temperature conversion in math, we often need to convert from Fahrenheit to Celsius or vice versa. This process is helpful for science subjects, weather forecasts, or understanding temperature in different countries.

The formula to convert Fahrenheit to Celsius is \( C = \frac{5}{9}(F - 32) \). Here, \( C \) stands for Celsius, and \( F \) represents Fahrenheit.

• This formula subtracts 32 from the Fahrenheit temperature because 32°F is the freezing point of water, equal to 0°C.
• The fraction \( \frac{5}{9} \) is used to adjust for the different temperature scales between Fahrenheit and Celsius.

By rearranging the equation, one can easily solve for Fahrenheit if the Celsius temperature is given.

Isolate Variable

Isolating a variable is an essential part of solving equations. It involves getting the variable by itself on one side of the equation.

In the context of the given equation, we want to solve for \( F \). That means we need to rearrange the formula so \( F \) stands alone on one side. To do that, we perform operations that "undo" other terms added or multiplied to \( F \).

• First, any terms or coefficients on the same side as \( F \) should be moved or adjusted to the other side of the equation.
• Addition, subtraction, multiplication, and division are used to isolate the variable.
• Think of it like peeling the layers of an onion until you reach the core, which is your variable.

By isolating the variable correctly, you ensure you have accurately solved the equation.

Fraction Elimination

Fractions can often complicate equations, but there is a strategic way to handle them - fraction elimination. It's a technique used to simplify equations by removing fractions.

In our example, \( C = \frac{5}{9}(F - 32) \), the fraction \( \frac{5}{9} \) is a coefficient of the expression. To eliminate it, we multiply everything by its denominator, \( 9 \), to clear the fraction:

• Multiply both sides by \( 9 \) to get: \( 9C = 5(F - 32) \).
• This step helps us avoid dealing with fractions, making further steps straightforward.
• Once the fraction is eliminated, moving forward with calculations becomes simpler.

Removing fractions reduces complexity and leads to clearer, more manageable expressions.

Distributive Property

The distributive property is a fundamental concept in algebra. It lets us simplify expressions by distributing multiplication over addition or subtraction.

After eliminating the fraction in the equation \( 9C = 5(F - 32) \), we use the distributive property. We need to multiply \( 5 \) by each term in the parentheses:

• The distributive property states: \( a(b + c) = ab + ac \).
• Apply this rule: \( 9C = 5F - 160 \) after multiplication, splitting into individual operations.
• Distributing helps to break down expressions into parts, making it easier to isolate and solve for the variable.

By using the distributive property correctly, we transition smoothly from one stage of the equation to the next.