SAT Question 0bd33265: Linear Functions answer and explanation with Desmos solution

It’s given that the equation $h=\frac{9(v-273.15)}{5}+32$ gives the corresponding temperature $h$, in degrees Fahrenheit, of any substance that has a temperature of $v$ kelvins, where $v>0$. Substituting $467.33$ for $h$ in the given equation yields $467.33=\frac{9(v-273.15)}{5}+32$. Subtracting $32$ from both sides of this equation yields $435.33=\frac{9(v-273.15)}{5}$. Multiplying both sides of this equation by $5$ yields $2,176.65=9(v-273.15)$. Dividing both sides of this equation by $9$ yields $241.85=v-273.15$. Adding $273.15$ to both sides of this equation yields $515=v$. Therefore, if a substance has a temperature of $467.33$ degrees Fahrenheit, the corresponding temperature, in kelvins, of this substance is $515$.