SAT Question e3cf671f: Linear Functions answer and explanation with Desmos solution

It's given that $f\left(5\right)=32$. Therefore, for the given function $f$, when $x=5$, $f\left(x\right)=32$. Substituting $5$ for $x$ and $32$ for $f\left(x\right)$ in the given function $f\left(x\right)=4x+k\left(x-1\right)$ yields $32=4\left(5\right)+k\left(5-1\right)$, or $32=20+4k$. Subtracting $20$ from each side of this equation yields $12=4k$. Dividing each side of this equation by $4$ yields $k=3$. Substituting $3$ for $k$ in the given function $f\left(x\right)=4x+k\left(x-1\right)$ yields $f\left(x\right)=4x+3\left(x-1\right)$, which is equivalent to $f\left(x\right)=4x+3x-3$, or $f\left(x\right)=7x-3$. Substituting $10$ for $x$ into this equation yields $f\left(10\right)=7\left(10\right)-3$, or $f\left(10\right)=67$. Therefore, the value of $f\left(10\right)$ is $67$.