Question

The following equation can be used to convert Fahrenheit temperature to Celsius temperature. $$ { }^{\circ} \mathrm{C}=\frac{\left({ }^{\circ} \mathrm{F}-32\right)}{1.8} $$ Use algebra to change the equation to convert Celsius temperature to Fahrenheit temperature.

Short answer

\({ }^{\circ} F = 1.8 \times { }^{\circ} C + 32\)

Step-by-step solution

Identify the Current Equation

The given equation is in the form \( { }^{\circ} C = \frac{({ }^{\circ} F - 32)}{1.8} \), which converts Fahrenheit to Celsius. Our task is to manipulate this formula to convert Celsius to Fahrenheit.

Isolate the Fahrenheit Variable

First, we will make \( { }^{\circ} F \) the subject of the formula. To do this, start by multiplying both sides of the equation by 1.8 to remove the denominator on the right side: \({ 1.8 \times { }^{\circ} C = { }^{\circ} F - 32 \).

Solve for Fahrenheit

To isolate \( { }^{\circ} F \) completely, add 32 to both sides of the equation to get \( { }^{\circ} F \) on one side and the Celsius temperature on the other side: \({ 1.8 \times { }^{\circ} C + 32 = { }^{\circ} F \).

Key concepts

Temperature Conversion

Understanding temperature conversion is essential for comparing weather data, cooking recipes, or scientific measurements across different countries and systems. The most common conversions are between the Fahrenheit and Celsius scales. In simple terms, Celsius is the most used scale across the world and is based on 0°C as the freezing point of water and 100°C as its boiling point at sea level. On the other hand, the Fahrenheit scale uses 32°F as the freezing point of water and 212°F as the boiling point.

To convert between these two, we use the formula: \[ { }^{\text{°C}} = \frac{\big({ }^{\text{°F}}-32\big)}{1.8} \]. But what happens when we need to switch it up and instead convert Celsius back to Fahrenheit? This is where algebraic manipulation is employed to transform the equation into a format where the conversion scales work the other way around. The adjusted formula still respects the relationship between the two temperature scales, allowing for accurate and quick conversion.

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in mathematics that involves rearranging equations and formulas to solve for different variables. The process often includes a series of procedures such as adding, subtracting, multiplying, and dividing terms in an equation. For temperature conversion, algebraic manipulation helps to transition between the Fahrenheit to Celsius conversion equation and vice versa.

For instance, starting with the original equation for converting Fahrenheit to Celsius, \[ { }^{\text{°C}} = \frac{\big({ }^{\text{°F}}-32\big)}{1.8} \], algebra is used to reconfigure this formula to solve for Fahrenheit instead. Through strategic manipulations, which include multiplying both sides to eliminate fractions and adding or subtracting constants, the equation is rewritten in the desired form. This skill is broadly applicable beyond just temperature conversion, helping to solve for unknowns in a variety of scientific and mathematical contexts.

Isolating Variables

When faced with an equation, isolating variables is the process of manipulating the equation to get one variable alone on one side—typically the variable for which we want to solve. It is an integral part of solving algebraic equations and is central to clearly understanding how to express certain quantities in terms of others.

In the context of converting Celsius temperatures to Fahrenheit, we start with the known formula for converting from Fahrenheit to Celsius. The goal is to rearrange the equation so that the Fahrenheit temperature \( { }^{\text{°F}} \) is isolated on one side, making it the subject of the formula. This is achieved through a series of steps that systematically move numbers and variables around without changing the overall equality—eventually leading to the formula \( { }^{\text{°F}} = {1.8 \times { }^{\text{°C}} + 32} \). This skill is crucial for students studying algebra, enabling them to solve equations, understand direct and inverse relationships, and carry out conversions like the temperature example given.