(i) Taking Celsius on x-axis and Fahrenheit on y-axis, the linear equation is given by: y = (9/5)x + 32
For plotting the graph:
Putting x = 0, we have, y = (9/5) × 0 + 32 = 32
putting x = 5, we have, y = (9/5) × 5 + 32 = 41
Putting x = 10, we have, y = (9/5) × 10 + 32 = 50
Hence, A(0, 100), B(5, 41) and c(10, 50) are the solutions of the equation.

(ii) If the temperature is 30° C, then
F = (9/5) × 30 + 32 = 54 + 32 = 86
Hence, if the temperature is 30°C, the temperature in Fahrenheit is 86°F.

(iii) If the temperature is 95°F, then
95 = (9/5)C + 32
⇒ 95 – 32 = (9/5)C
⇒ 63 × 5/9 = C
⇒ C = 35°
If the temperature is 95°F, the temperature in Celsius is 35°C.

(iv) If temperature is 0°C, then
F = (9/5) × 0 + 32 = 0 + 32 = 32
If the temperature is 0°F, then
0 = (9/5)C + 32
⇒ -32 = (9/5)C
⇒ -32 × 5/9 = C
⇒ – 160/9 = C
⇒ C = -17.8°
If the temperature is 0°C, the temperature in Fahrenheit is 32°F and if the temperature is 0°F, the temperature in Celsius is -17.8°C.

(v) Let x° be the temperature which is numerically the same in both Fahrenheit and Celsius, then
x = (9/5)x + 32
⇒ x – 32 = (9/5)x
⇒ (x – 32) × 5 = 9x
⇒ 5x – 160 = 9x
⇒ 4x = -160
⇒ x – 40°
Hence, -40° is the temperature which is numerically the same in both Fahrenheit and Celsius.