, 2018

Simplifying, we get $\dfrac{12x^2}{3x}=4x$ for $x\neq0$.
Since $\dfrac{12x^2}{3x}=4x$, the value of the expression $\dfrac{12x^2}{3x}$ is equal to the value of the simplified expression $4x$ for all values of $x\neq0$.

So when $x=5$, the value of the expression $\dfrac{12x^2}{3x}$ is equal to $4(5)=20$.
Simplifying, we get $\dfrac{8mn}{3m^2}=\dfrac{8n}{3m}$ for $m\neq0$.

The value of the expression $\dfrac{8mn}{3m^2}$ is equal to the value of the simplified expression $\dfrac{8n}{3m}$ for all values of $m\neq0$.

Substituting $n=2m$ into $\dfrac{8n}{3m}$ and simplifying, we get $\dfrac{8(2m)}{3m}=\dfrac{16m}{3m}=\dfrac{16}{3}$.

When $n=2m$ and $m\neq0$, the value of the expression $\dfrac{8mn}{3m^2}$ is $\dfrac{16}{3}$.
Simplifying, we get $\dfrac{8p^2q}{5pq^2}=\dfrac{8p}{5q}$ for $p\neq0$, $q\neq0$.

When $q=6$, we get $\dfrac{8p}{5q}=\dfrac{8p}{5(6)}=\dfrac{8p}{30}=\dfrac{4p}{15}$.

That is, when $q=6$ (and $p\neq0$) the expression $\dfrac{8p^2q}{5pq^2}$ is equal to $\dfrac{4p}{15}$, and so the solution to $3\leq\dfrac{8p^2q}{5pq^2}\leq4$ is equivalent to the solution to $3\leq\dfrac{4p}{15}\leq4$.

Solving $3\leq\dfrac{4p}{15}\leq4$, we get $45\leq4p\leq60$ or $\dfrac{45}{4}\leq p\leq\dfrac{60}{4}$, and so $11.25\leq p\leq15$.

Since $p$ is a positive integer, then $p=12,13,14,15$.

Note: In each of the solutions to (b), (c) and (d), we chose to simplify the expression before substituting. Changing the order to substitution followed by simplification would also allow us to solve these problems.