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How do you solve for $K$ in $F = \dfrac{9}{5}\left( {K - 273} \right) + 32$?

seo-qna
Last updated date: 25th Jun 2024
Total views: 374.4k
Views today: 8.74k
Answer
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Hint:This question is related to linear equation concept. An equation for a straight line is known as linear equations. The term which is involved in a linear equation is either a constant or a single variable or product of a constant. The two variables can never be multiplied. All linear equations have a line graph. Linear equations are the same as linear function. The general form of writing a linear equation is $y = mx + c$ and $m$ is not equal to zero, where $m$ is the slope and $c$ is the point on which it cuts the y-axis. $y = mx + c$ is also known as the equation of the line in slope-intercept form. This given question deals with a specific type of linear equation and that is, formulas for problem solving.

Complete step by step solution:
Given is $F = \dfrac{9}{5}\left( {K - 273} \right) + 32$
We have to solve the given equation in order to find the value of $K$ for which left-hand side is equal to the right-hand side of the equation.
Let us simply start by simplifying the given equation by subtracting $32$ from both sides of the equation.
$
\Rightarrow F - 32 = \dfrac{9}{5}\left( {K - 273} \right) + 32 - 32 \\
\Rightarrow F - 32 = \dfrac{9}{5}\left( {K - 273} \right) \\
$
Next, let us multiply $\dfrac{5}{9}$ on both the sides of the equation and we get,
$
\Rightarrow \dfrac{5}{9} \times \left( {F - 32} \right) = \left( {\dfrac{5}{9} \times \dfrac{9}{5}}
\right)\left( {K - 273} \right) \\
\Rightarrow \dfrac{5}{9} \times \left( {F - 32} \right) = 1 \times \left( {K - 273} \right) \\
\Rightarrow \dfrac{5}{9} \times \left( {F - 32} \right) = \left( {K - 273} \right) \\
$
Now, we add $273$ to both the sides of the equation and we get,
$
\Rightarrow \dfrac{5}{9} \times \left( {F - 32} \right) + 273 = \left( {K - 273} \right) + 273 \\
\Rightarrow \dfrac{5}{9} \times \left( {F - 32} \right) + 273 = K \\
$
Therefore, the value of $K$ is $\dfrac{5}{9} \times \left( {F - 32} \right) + 273$.
Note: Now that we know the value of $K$ is $\dfrac{5}{9} \times \left( {F - 32} \right) + 273$, there is a way to double check our answer. In order to double check the solution we are supposed to substitute the value of $K$ which we got as $\dfrac{5}{9} \times \left( {F - 32} \right) + 273$ in the given equation, $F = \dfrac{9}{5}\left( {K - 273} \right) + 32$
$
\Rightarrow F = \dfrac{9}{5}\left( {K - 273} \right) + 32 \\
\Rightarrow F = \dfrac{9}{5}\left( {\dfrac{5}{9}\left( {F - 32} \right) + 273 - 273} \right) + 32
\\
\Rightarrow F = \dfrac{9}{5}\left( {\dfrac{5}{9}\left( {F - 32} \right)} \right) + 32 \\
\Rightarrow F = F - 32 + 32 \\
\Rightarrow F = F \\
$$$$$
Now, the left-hand side is equal to the right-hand side of the equation. So, we can conclude that our solution or the value of $K$ which we calculated was correct.