Question

What interval on the Celsius scale corresponds to the temperature range$50\le F\le 95$?

Short answer

The interval on the Celsius scale is $10\le C\le 35$.

Step-by-step solution

Step 1. Apply the concept of Solving linear inequalities.

An inequality looks like an equation, except that in the place of the equal sign is one of the symbols, $<,>,\le ,\ge$.

Example of an inequality is $4x+7\le 19$.

An inequality is linear if each term is constant or a multiple of the variable. To solve a linear inequality, isolate the variable on one side of the inequality sign.

Step 2. Find the inequality.

The relationship between degrees Celsius (C)and degrees Fahrenheit (F) is given by the equation $C=\frac{5}{9}(F-32)$. Rewrite this relation such that Fis isolated.

Therefore,

$\begin{array}{c}C=\frac{5}{9}(F-32)\\ \Rightarrow \frac{9}{5}C=F-32\text{(Multiplyby}\frac{9}{5})\\ \Rightarrow \frac{9}{5}C+32=F-32+32\text{(add32)}\\ \Rightarrow \frac{9}{5}C+32=F\end{array}$

Hence by using this relation and the given data, the following inequality is obtained:

$\begin{array}{c}50\le F\le 95\\ \Rightarrow 50\le \frac{9}{5}C+32\le 95\end{array}$

is the required initial inequality.

Step 3. Simplify the inequality.

Simplify the inequality $50\le \frac{9}{5}C+32\le 95$. Therefore,

$\begin{array}{c}50\le \frac{9}{5}C+32\le 95\\ 50-32\le \frac{9}{5}C\le 95-32\left(\text{subtract32}\right)\\ 18\le \frac{9}{5}C\le 63\\ 18\cdot \frac{5}{9}\le C\le 63\cdot \frac{5}{9}\text{(Multiplyby}\frac{5}{9})\\ 10\le C\le 35\end{array}$