SAT Question 63d03c0b: Linear Functions answer and explanation with Desmos solution

For engine speeds between $1,000$ revolutions per minute (rpm) and $6,000$ rpm, it’s given that the equation $f(x)=\frac{1}{7}(x-a)+433$ defines the linear function $f$ and gives the predicted power, in brake horsepower $(\text{bhp})$, at an engine speed of $x$ rpm, where $a$ is a constant. It’s also given that the car's predicted power is $433$ bhp at an engine speed of $3,331$ rpm. Substituting $3,331$ for $x$ and $433$ for $f(x)$ in the equation yields $433=\frac{1}{7}(3,331-a)+433$. Subtracting $433$ from both sides of this equation yields $0=\frac{1}{7}(3,331-a)$. Multiplying both sides of this equation by $7$ yields $0=3,331-a$. Adding $a$ to both sides of this equation yields $a=3,331$. Thus, the value of $a$ is $3,331$.