# Ex.4.3 Q8 Linear Equations in Two Variables Solution - NCERT Maths Class 9

## Question

In countries like USA and Canada, the 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: \(\begin{align} F = \left( \frac { 9 } { 5 } \right) C + 32 \end{align}\)

(i) Draw the graph of the linear equation above using Celsius for \(x\)-axis and Fahrenheit for \(y\)-axis.

(ii) If the temperature is \(30^{\circ} C\), what is the temperature in Fahrenheit?

(iii) If the temperature is \(95^{\circ} F\), what is the temperature in Celsius?

(iv) If the temperature is \(0^{\circ} C\), what is the temperature in Fahrenheit and if the temperature is \(0^{\circ} F\), 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.

## Text Solution

**Steps:**

(i) **Given,**

\[\begin{align} F = \left( \frac { 9 } { 5 } \right) C + 32 \end{align} \dots \text{ Equation } (1)\]

Equation (1) represents a linear equation of the form \(ax + by + c = 0\) where \(C\) and \(F\) are the two variables.

By substituting different values for \(C\) in the Equation (1), we obtain different values for \(F\)

When \(C = 0,\)

\(\begin{align}F = \left( {\frac{9}{5}} \right)C + 32 = \left( {\frac{9}{5}} \right)0 + 32 = 32\end{align}\)

When \(C = \,– 40, \)

\(\begin{align}F& = \left( {\frac{9}{5}} \right) \times ( - 40) + 32 \\&= - 72 + 32 = - 40\end{align}\)

When \(C = 10, \)

\(\begin{align}F = \left( {\frac{9}{5}} \right) \times (10) + 32 = 18 + 32 = 50\end{align}\)

Thus, we have the following table with all the obtained solutions:

\(x\) | \(0\) | \(-40\) | \(10\) |

\(y\) | \(32\) | \(-40\) | \(50\) |

By Plotting the points \((– 40, – 40)\) and \((10, 50)\) on the graph point. On joining these points by line segment, we obtain the required graph

The graph of the line represented by the given equation as shown.

(ii) **Given:**

Temperature \(= 30 ^\circ C\)

To find: \(F=?\)

We know that,\(\begin{align}F = \left( {\frac{9}{5}} \right)C + 32\end{align}\)

By Substituting the value of \(C= 30 ^\circ C\) in the Equation above,

\[\begin{align}F &= \left( {\frac{9}{5}} \right)C + 32\\ &= \,\left( {\frac{9}{5}} \right)30 + 32\\ &= 54 + 32 \\&= 86 \end{align}\]

Therefore, the temperature in Fahrenheit is \(86^ {\circ} F\).

(iii) **Given:**

Temperature \(=\) \(95^ {\circ} F\)

To find: \(C= ?\)

We know that,\(\begin{align} F = \left( \frac { 9 } { 5 } \right) C + 32 \end{align}\)

By Substituting the value of temperature in the Equation above,

\[\begin{align}95 &= \left( {\frac{9}{5}} \right)C + 32\\95 - 32 &= \left( {\frac{9}{5}} \right)C\\63 &= \left( {\frac{9}{5}} \right)C\\\therefore C &= \frac{{63 \times 5}}{9} = 35\end{align}\]

Therefore, the temperature in Celsius is \(35^ {\circ} C\).

(iv) We know that, \(\begin{align} F = \left( \frac { 9 } { 5 } \right) C + 32 \end{align}\)

If \(C = 0^ {\circ}C\), then by Substituting the value in the above Equation,

\[\begin{align}F &= \left( {\frac{9}{5}} \right)0 + 32\\F &= 0 + 32\\F &= 32\end{align}\]

Therefore, if \(C = 0^ {\circ}C\), then \(F = 32^ {\circ}F\).

If \(F = 0^ {\circ}F\), then by Substituting the value in the above Equation,

\[\begin{align}0 &= \left( {\frac{9}{5}} \right)C + 32\\\left( {\frac{9}{5}} \right)C &= - 32\\C &= \frac{{ - 32 \times 5}}{9} = - 17.77\end{align}\]

Therefore, if \(F = 0^ {\circ}F\), then \(C = 17.8^ {\circ}C\).

(v) We know that,

\(\begin{align} F = \left( \frac { 9 } { 5 } \right) C + 32 \end{align}\)

Let us consider, \(F = C\)

By Substituting this value in the Equation above,

\[\begin{align}&F = \left( {\frac{9}{5}} \right)F + 32\\&\left( {\frac{9}{5} - 1} \right)F + 32 = 0\\&\left( {\frac{4}{5}} \right)F = - 32\\&F = \frac{{ - 32 \times 5}}{4}\end{align}\]

Hence, \(F = \,–40^ {\circ}F\)

Yes, there is a temperature, \(−40^ {\circ}F\), which is numerically the same in both Fahrenheit and Celsius.