, 2021

Substituting $a=5$ and $b=1$, we get $5 \triangle 1 = 5(2\times1 +4)=5(6)=30$. If $k \triangle 2 = 24$, then $k(2\times2 +4)=24$ or $8k=24$, and so $k=3$. Solving the given equation for $p$, we get $$\begin{aligned} p \triangle 3 & = 3 \triangle p \\ p(2\times3+4) & = 3(2p+4)\\ p(10) & = 6p+12\\ 10p-6p & = 12\\ 4p & = 12\\ p & = 3\end{aligned}$$ The only value of $p$ for which $p \triangle 3 = 3 \triangle p$ is $p=3$. Simplifying the given equation, we get $$\begin{aligned} m \triangle (m+1) & = 0 \\ m(2(m+1)+4) & = 0\\ m(2m+2+4) & = 0\\ m(2m+6) & = 0\end{aligned}$$ Thus, $m=0$ or $2m+6=0$ which gives $m=-3$. The values of $m$ for which $m \triangle (m+1) = 0$ are $m=0$ and $m=-3$. (Substituting each of these values of $m$, we may check that $0 \triangle 1 = 0(2\times1 +4)=0(6)=0$, and that $(-3)\triangle(-2) = -3(2\times(-2) +4)=-3(0)=0$.)