Question

If it is given that $F=\dfrac{9}{5}C+32$, then the expression for C is
[a] $C=\dfrac{5}{9}\left( F+32 \right)$
[b] $C=\dfrac{9}{5}\left( F-32 \right)$
[c] $C=\dfrac{5}{9}\left( F-32 \right)$
[d] $C=\dfrac{9}{5}\left( F+32 \right)$

Answer 1

Hint: Use the fact that addition and subtraction of equal things on both sides of an equation does not change the solutions of the equation. Similarly, multiplication and division of non-zero things on both sides of an equation does not change the solution of the equation. Subtract 32 from both sides of the equation, then multiply by 5 on both sides and finally, divide by 9 on both sides of the equation and hence express C in terms of F.

Complete step-by-step answer:
We have $F=\dfrac{9}{5}C+32$
We know that subtraction of equal things on both sides of an equation does not change solutions to an equation.
Hence, subtracting 32 from both sides of the equation, we get
$\begin{align}
  & F-32=\dfrac{9}{5}C+32-32 \\
 & \Rightarrow F-32=\dfrac{9}{5}C \\
\end{align}$
We know that the multiplication by equal non-zero things on both sides of an equation does not change the solutions of the equation.
Hence multiplying both sides of the equation by 5, we get
$\begin{align}
  & 5\left( F-32 \right)=\dfrac{9}{5}C\times 5 \\
 & \Rightarrow 5\left( F-32 \right)=9C \\
\end{align}$
We know that the division by equal non-zero things on both sides of an equation does not change the solutions of the equation.
Hence dividing by 9 on both sides of the equation, we get
$\begin{align}
  & \dfrac{5}{9}\left( F-32 \right)=\dfrac{9C}{9} \\
 & \Rightarrow C=\dfrac{5}{9}\left( F-32 \right) \\
\end{align}$
Hence option [c] is correct.

Note: Verification:
Same order pairs of (C, F) should satisfy both the equations.
Consider the ordered pair (0,32)
For the equation $F=\dfrac{9}{5}C+32$, we have
$\begin{align}
  & 32=\dfrac{9}{5}\left( 0 \right)+32 \\
 & \Rightarrow 32=32 \\
\end{align}$
which is true.
Hence (0,32) satisfies the equation.
 For equation $C=\dfrac{5}{9}\left( F-32 \right)$, we have
$\begin{align}
  & 0=\dfrac{5}{9}\left( 32-32 \right) \\
 & \Rightarrow 0=0 \\
\end{align}$
which is true.
Hence (0,32) satisfies the equation.
Hence (0,32) satisfies both the equation
Consider the ordered pair (-20,-4)
For equation $F=\dfrac{9}{5}C+32$, we have
$\begin{align}
  & -4=\dfrac{9}{5}\left( -20 \right)+32 \\
 & \Rightarrow -4=-36+32 \\
 & \Rightarrow -4=-4 \\
\end{align}$
which is true
Hence (-20,-4) satisfies the equation
For the equation $C=\dfrac{5}{9}\left( F-32 \right)$, we have
$\begin{align}
  & -20=\dfrac{5}{9}\left( -4-32 \right) \\
 & \Rightarrow -20=-20 \\
\end{align}$
which is true
Hence, (-20,-4) satisfies the equation.
Hence (-20,-4) satisfies both the equations.
Hence, our answer is verified to be correct.