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Chapter 4: Problem 5

For Problems \(1-10\), solve for the specified variable using the given facts. (Objective 1) $$ \text { Solve } \mathrm{F}=\frac{9}{5} \mathrm{C}+32 \text { for } \mathrm{C} \text { if } \mathrm{F}=68 $$

Short Answer

Expert verified
C = 20

Step by step solution

01

Start with the given equation

We begin with the equation given in the problem: \( F = \frac{9}{5}C + 32 \).
02

Substitute the given value of F

Replace \( F \) with \( 68 \) in the equation, as specified in the problem: \( 68 = \frac{9}{5}C + 32 \).
03

Isolate the term with C

Subtract 32 from both sides of the equation to move the constant to the left side: \( 68 - 32 = \frac{9}{5}C \). This simplifies to \( 36 = \frac{9}{5}C \).
04

Solve for C by eliminating the fraction

Multiply both sides by the reciprocal of \( \frac{9}{5} \), which is \( \frac{5}{9} \), to solve for \( C \): \[ C = 36 \times \frac{5}{9} \]
05

Simplify the expression

Calculate \( 36 \times \frac{5}{9} \). Start by simplifying \( \frac{5}{9} \times 36 \) as follows: \( C = \frac{5 \times 36}{9} \). Since 36 divided by 9 equals 4, we have: \( C = 5 \times 4 \).
06

Compute the final solution

Multiply \( 5 \times 4 \) to find the value of \( C \): \[ C = 20 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
Solving equations is a fundamental skill in algebra that involves finding the value of a variable that makes an equation true. In this exercise, the equation is given as \( F = \frac{9}{5}C + 32 \), and we are tasked with finding the value of \( C \) when \( F = 68 \). The goal is to determine what number \( C \) must be so that when inserted into the equation, the left and right sides are equal.

To solve an equation, it is essential to follow a clear set of steps:
  • Identify the variable you need to solve for (in this case, \( C \)).
  • Apply necessary arithmetic operations to isolate this variable.
  • Check your solution by substituting it back into the original equation to verify correctness.
These steps ensure that the solution you find will make the initial equation hold true.
Substitution Method
The substitution method is a useful technique in solving equations, especially when specific values for variables are provided. In this exercise, the problem supplies the value for \( F \), which is 68. The substitution process involves:
  • Re-writing the original equation to insert the given value (e.g., replacing \( F \) with 68).
  • Adjusting the equation accordingly, resulting in a simpler form that is easier to solve.
By substituting \( F = 68 \) into the initial equation, \( 68 = \frac{9}{5}C + 32 \) is established. This lets us focus on solving for the unknown variable \( C \).

It is important to understand substitution as a powerful tool to transition from known values to solving for unknowns, enhancing problem-solving efficiency.
Isolation of Variables
Isolating a variable means arranging an equation such that the variable stands alone on one side, typically the left side. In our case, we aimed to isolate \( C \). The process of isolation involves moving all other terms away from the variable.

Here鈥檚 how we proceed:
  • Subtract 32 from both sides of the equation: \( 68 - 32 = \frac{9}{5}C \), simplifying to \( 36 = \frac{9}{5}C \).
  • To eliminate the fraction, multiply both sides by the reciprocal of \( \frac{9}{5} \), which is \( \frac{5}{9} \). This effectively "clears" the fraction, resulting in \( C = 36 \times \frac{5}{9} \).
This step-by-step process clears away distractions, leaving the variable of interest cleanly isolated for a clear, algebraic solution. Ultimately, this process resulted in finding that \( C = 20 \), demonstrating the effectiveness of isolation in solving algebraic equations.

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Most popular questions from this chapter

For Problems 1-12, solve each equation. You will be using these types of equations in Problems \(13-41\). $$ 30 x+14\left(3 \frac{1}{2}-x\right)=97 $$

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