, 2024

For every $x \neq 0$, we note
that $\dfrac{x^4 + 3x^2}{x^2} = x^2 + 3$.

Therefore, when $x = 2$, we have
$\dfrac{x^4 + 3x^2}{x^2} = x^2 + 3 = 2^2 + 3 = 7$.

Alternatively, when $x = 2$, we have
$\dfrac{x^4 + 3x^2}{x^2} = \dfrac{2^4 + 3 \cdot 2^2}{2^2} = \dfrac{28}{4} = 7$.
By the Pythagorean Theorem in $\triangle ABC$, we have $$\begin{align*} AC^2 & = AB^2 + BC^2 \\ (t+1)^2 & = 10^2 + (t-1)^2 \\ t^2 + 2t + 1 & = 100 + t^2 - 2t + 1 \\ 4t & = 100\end{align*}$$ and so $t = 25$.

Alternatively, we could remember the Pythagorean triple $5$-$12$-$13$ and scale this triple by a factor of
$2$ to obtain the Pythagorean triple
$10$-$24$-$26$, noting that the difference between
$t+1$ and $t-1$ is $2$ as is the difference between $26$ and $24$, which gives $t + 1 = 26$ and so $t = 25$.
Since $\dfrac{2}{y} + \dfrac{3}{2y} = 14$, then $\dfrac{4}{2y} + \dfrac{3}{2y}=14$ or $\dfrac{7}{2y} = 14$.

Therefore, $2y = \dfrac{7}{14} = \dfrac{1}{2}$ and so $y = \dfrac{1}{4}$.