Medium
If the temperature of a liquid can be measured in Kelvin units as $x^{\circ} K$ or in Fahrenheit units as $y^{\circ} F$,the relation between the two systems of measurement of temperature is given by the linear equation
$y = \frac{9}{5}(x - 273) + 32$
$(i)$ Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313^{\circ} K$.
$(ii)$ If the temperature is $158^{\circ} F$,then find the temperature in Kelvin.
$y = \frac{9}{5}(x - 273) + 32$
$(i)$ Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313^{\circ} K$.
$(ii)$ If the temperature is $158^{\circ} F$,then find the temperature in Kelvin.
(N/A) The given linear equation is $y = \frac{9}{5}(x - 273) + 32$.
$(i)$ Given $x = 313^{\circ} K$:
$y = \frac{9}{5}(313 - 273) + 32$
$y = \frac{9}{5}(40) + 32$
$y = 9 \times 8 + 32 = 72 + 32 = 104^{\circ} F$.
$(ii)$ Given $y = 158^{\circ} F$:
$158 = \frac{9}{5}(x - 273) + 32$
$158 - 32 = \frac{9}{5}(x - 273)$
$126 = \frac{9}{5}(x - 273)$
$x - 273 = 126 \times \frac{5}{9}$
$x - 273 = 14 \times 5 = 70$
$x = 273 + 70 = 343^{\circ} K$.
$(i)$ Given $x = 313^{\circ} K$:
$y = \frac{9}{5}(313 - 273) + 32$
$y = \frac{9}{5}(40) + 32$
$y = 9 \times 8 + 32 = 72 + 32 = 104^{\circ} F$.
$(ii)$ Given $y = 158^{\circ} F$:
$158 = \frac{9}{5}(x - 273) + 32$
$158 - 32 = \frac{9}{5}(x - 273)$
$126 = \frac{9}{5}(x - 273)$
$x - 273 = 126 \times \frac{5}{9}$
$x - 273 = 14 \times 5 = 70$
$x = 273 + 70 = 343^{\circ} K$.
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