If countries like the USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is ${30^ \circ }{\text{C}}$, what is the temperature in Fahrenheit?
(iii) If the temperature is \[{95^ \circ }{\text{F}}\], what is the temperature in Celsius?
(iv) If the temperature is ${0^ \circ }{\text{C}}$, what is the temperature in Fahrenheit and if the temperature is \[{0^ \circ }{\text{F}}\], what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Hint: We will first find coordinates that we will plot on the graph using the equation ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$ Next, draw a continuous line joining the points to obtain the graph of the given equation. Then, we will substitute the given values of C and F to find the required values.

Complete step-by-step answer:
For the first part, we will begin by finding values $\left( {{\text{C,F}}} \right)$ for the equation ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$
Thus, when, ${\text{C = 0}}$, then, the value of ${\text{F}} = \left( {\dfrac{9}{5}} \right)\left( 0 \right) + 32 = 32$
Similarly, when ${\text{F = 0}}$, then, the value of C is :
 $
  0 = \left( {\dfrac{9}{5}} \right){\text{C + }}32 \\
   \Rightarrow {\text{C}} = - \dfrac{{160}}{9} \\
$
We will plot the points $\left( {0,32} \right)$ and $\left( { - \dfrac{{160}}{9},0} \right)$ .
We will now join the points through a line.

For part (ii), we are given that the value of C is ${30^ \circ }{\text{C}}$, we have to find the corresponding values of F.
We will substitute 30 for C in the equation ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$
$
  {\text{F}} = \left( {\dfrac{9}{5}} \right)\left( {30} \right) + 32 \\
   \Rightarrow {\text{F}} = 54 + 32 \\
   \Rightarrow {\text{F}} = 86 \\
$
Thus, the temperature is ${86^ \circ }{\text{F}}$ when it is ${30^ \circ }{\text{C}}$

For part (iii), we are given that the temperature is \[{95^ \circ }{\text{F}}\], we will find the corresponding value in Celsius by substituting 95 for F in equation ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$
$
  {\text{95}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32 \\
   \Rightarrow 95 - 32 = \left( {\dfrac{9}{5}} \right){\text{C}} \\
   \Rightarrow \left( {\dfrac{9}{5}} \right){\text{C}} = 63 \\
   \Rightarrow {\text{C = }}\dfrac{{63\left( 5 \right)}}{9} \\
   \Rightarrow {\text{C = 35}} \\
$
Hence, the temperature is ${35^ \circ }$

Now, in part (iv), first let 0 for C and find the value of F.
${\text{F}} = \left( {\dfrac{9}{5}} \right)\left( 0 \right) + 32 = 32$
${0^ \circ }{\text{C}}$ is equivalent to ${32^ \circ }{\text{F}}$
Now, substitute 0 for F and find the value of C
$
  0 = \left( {\dfrac{9}{5}} \right){\text{C + }}32 \\
   \Rightarrow {\text{C}} = - \dfrac{{160}}{9} \\
$
Hence, \[{0^ \circ }{\text{F}}\] is equivalent to $ - {\dfrac{{160}}{9}^ \circ }{\text{C}}$
In part (v), we have to find the value such that it is the same for both F and C.
Let ${\text{F = C}}$ and substitute in ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$
$
  {\text{F}} = \left( {\dfrac{9}{5}} \right){\text{F}} + 32 \\
   \Rightarrow {\text{F}} - \left( {\dfrac{9}{5}} \right){\text{F = }}32 \\
   \Rightarrow \dfrac{{5{\text{F}} - 9{\text{F}}}}{5} = 32 \\
   \Rightarrow \dfrac{{ - 4{\text{F}}}}{5} = 32 \\
   \Rightarrow - 4{\text{F}} = 160 \\
   \Rightarrow {\text{F}} = - 40 \\
 $

Hence, at $ - {40^ \circ }$ temperature is numerically the same in both Fahrenheit and Celsius.

Note: Substitute the values correctly in the given equation. Avoid calculation mistakes. Also, the given equation ${\text{F}} = \left( {\dfrac{9}{5}} \right){\text{C}} + 32$ is a linear equation, that is , maximum degree is 1 of the equation. The equation is linear, hence, the graph of the equation is a straight line.