A collection of 5-unit coins and 10-unit coins contains 80 coins altogether and is worth 650 in total. How many coins of each kind are there?

Step-by-step solution

1. 1

   Absorb the count condition into the variable. If $x$ is the number of $5$-unit coins, the rest of the $80$ coins must be $10$-unit coins:

   $x\ \text{and}\ 80-x$

   This is why the problem needs only one equation rather than a system.

2. 2

   Write each group's value.

   $5x\quad\text{from the }5\text{s},\qquad 10(80-x)\quad\text{from the }10\text{s}$

3. 3

   Set up the value equation.

   $5x+10(80-x)=650$

4. 4

   Expand, combine, solve.

   $5x+800-10x=650$

   $800-5x=650$

   $-5x=-150$

   $x=30$

5. 5

   Read off both counts and check. There are $30$ five-unit coins and $80-30=50$ ten-unit coins. Value:

   $5(30)+10(50)=150+500=650\quad\checkmark$

   A useful sanity check: if all $80$ coins were $5$s the total would be $400$, and every swap of a $5$ for a $10$ adds $5$. Needing $650-400=250$ extra means $250/5=50$ tens — the same answer by a different route.