problem: All triangles have the same value, and all circles have the same value. What is the sum of two circles? \begin{align*}
  \Delta + \bigcirc + \Delta + \bigcirc + \Delta + \bigcirc + \Delta + \Delta&= 21\\
  \bigcirc + \Delta+\bigcirc+\Delta+\bigcirc + \Delta + \bigcirc + \bigcirc &= 16\\
  \bigcirc + \bigcirc &= \ \, ?
  \end{align*}
solution: Replace a triangle with the letter $a$ and a circle with the letter $b.$ The two given equations become \begin{align*}
  5a+3b&=21\\
  3a+5b&=16
  \end{align*}Multiplying the first equation by $3,$ we get $15a+9b=63.$ Multiplying the second equation by $5,$ we get $15a+25b=80.$ Subtracting this last equation from the second-to-last equation to eliminate $a,$ we have $16b=17.$ Multiplying both sides by $\frac{2}{16},$ we get $$\frac{2}{16}\cdot 16b = \frac{2}{16} \cdot 17 \implies 2b=\frac{17}{8}.$$Thus, two circles equals $\boxed{\frac{17}{8}}.$