A collection of 5-unit coins and 10-unit coins contains 60 coins altogether and is worth 450 in total. How many coins of each kind are there?

Step-by-step solution

1. 1

   Use one variable, not two. There are two unknowns but also two facts, so you can eliminate one unknown immediately using the count fact. Let

   $x=\text{number of }5\text{-unit coins}\quad\Longrightarrow\quad 60-x=\text{number of }10\text{-unit coins}$

2. 2

   Convert counts to value. A count is not a value: multiply each count by what one of those coins is worth.

   $\text{value of the }5\text{s}=5x,\qquad \text{value of the }10\text{s}=10(60-x)$

3. 3

   Set the total value equal to 450.

   $5x+10(60-x)=450$

4. 4

   Expand and solve. Watch the sign when $10$ multiplies $-x$:

   $5x+600-10x=450$

   $600-5x=450$

   $-5x=-150$

   $x=30$

5. 5

   Recover both counts and check both conditions. There are $30$ five-unit coins and $60-30=30$ ten-unit coins. Count: $30+30=60$ $\checkmark$. Value:

   $5(30)+10(30)=150+300=450\quad\checkmark$

   Both conditions hold, so the split is $30$ and $30$.