To find when the two machines will have the same amount of ice cream left, we can express the amount of ice cream left in each machine after \( x \) days.
For the vanilla ice cream machine:
- Initial amount: 180 cups
- Sales per day: 20 cups
- Amount left after \( x \) days: \( 180 - 20x \)
For the chocolate ice cream machine:
- Initial amount: 140 cups
- Sales per day: 16 cups
- Amount left after \( x \) days: \( 140 - 16x \)
We want to find the point in time (after \( x \) days) when both machines have the same amount left, so we set the two expressions equal to each other:
\[ 180 - 20x = 140 - 16x \]
This matches option d from your list:
\[ 180 - 20x = 140 - 16x \]
Now, let's solve for \( x \):
- Rearranging gives: \[ 180 - 140 = 20x - 16x \]
- Simplifying further: \[ 40 = 4x \]
- Therefore: \[ x = 10 \]
So, the correct choice is d: \( 180 - 20x = 140 - 16x; x = 10 \) (though there is a correction needed in labeling the choice; it indeed leads to \( x = 10 \)).