, 2025

Solution 1:

Since $p + q = 5$ and $q + r = 9$, then $p + q + q + r = 5 + 9$ or $p + 2q + r = 14$. Since $p + 2q + r = 14$ and $p + q + r = 18$, then subtracting the two equations gives $$(p+2q+r) - (p+q+r) = 14 - 18$$ and so $q = -4$. Solution 2: Since $p + q + r = 18$ and $p + q = 5$, then $5 + r = 18$ and so $r = 13$.  Since $q + r = 9$ and $r = 13$, then $q + 13 = 9$ and so $q = -4$. To find the $x$-intercept of the line with equation $6x + y = 24$, we set $y = 0$ to obtain $6x = 24$ which gives $x = 4$. To find the $y$-intercept of the line with equation $6x + y = 24$, we set $x = 0$ to obtain $y = 24$. Since the parabola whose equation we want to determine has only one $x$-intercept (namely $x = 4$), we can write its equation as $y = a(x-4)^2$ for some real number $a$. Additionally, we know that the $y$-intercept of the parabola is $y = 24$, so it passes through the point $(0, 24)$. Substituting $(x,y) = (0, 24)$ into $y = a(x-4)^2$, we obtain $24 = a(0-4)^2$ which gives $24 = 16a$ and so $a = \frac{3}{2}$. Therefore, an equation of the parabola is $y
= $\frac{3}{2}(x-4)^2$$. Since$$\dfrac{1}{2w} = \dfrac{1}{3y} = \dfrac{1}{4z}$$and$$\dfrac{1}{2w} + \dfrac{1}{3y} + \dfrac{1}{4z} = \dfrac{1}{24}\$,$ then each of
$\dfrac{1}{2w}$ and $\dfrac{1}{3y}$ and $\dfrac{1}{4z}$ is equal to one-third of the total, which gives $\$\dfrac{1}{2w} = \dfrac{1}{3y} = \dfrac{1}{4z} = \dfrac{1}{3} \cdot \dfrac{1}{24} = \dfrac{1}{72}$$. Therefore,$2w = 3y = 4z = 72$and so$w = 36$and$y = 24$and$z = 18$.

Thus, $w + y + z = 36 + 24 + 18 = 78$.