Daily math problems for June 22, 2026 — Combinatorics, Algebra

$\binom{10}{3} = \dfrac{10 \cdot 9 \cdot 8}{3!} = \dfrac{720}{6} = 120$.

$\bullet$ Take reciprocals of each equation, using $\tfrac{1}{x}+\tfrac{1}{y}=\tfrac{x+y}{xy}$:

$$\tfrac{1}{x}+\tfrac{1}{y}=\tfrac{3}{5},\quad \tfrac{1}{x}+\tfrac{1}{z}=\tfrac{2}{3},\quad \tfrac{1}{y}+\tfrac{1}{z}=\tfrac{1}{4}$$

$\bullet$ Let $a=1/x$, $b=1/y$, $c=1/z$. Sum all three:

$$2(a+b+c)=\tfrac{3}{5}+\tfrac{2}{3}+\tfrac{1}{4}=\tfrac{91}{60}$$

$$a+b+c=\tfrac{91}{120}$$

$\bullet$ Subtract each pair-sum from $a+b+c$:

$$a=\tfrac{61}{120},\ b=\tfrac{11}{120},\ c=\tfrac{19}{120}$$

$\bullet$ Recover $x,y,z$ and add:

$$x+y+z=\tfrac{120}{61}+\tfrac{120}{11}+\tfrac{120}{19}$$