When both conditions must hold, solve them separately and keep their overlap. Translate “at most” as <= and “at least” as >=. Counts must be whole numbers, so interpret the bound before choosing a value. Example From 2x+1>5, x>2. From 3x-2 10, x 4. Combining x>2 and x 4, the top row below shows the first condition and the lower row shows the second. Their common range is 2<x 4; the integer solutions are 3 and 4. A hire fee is 12 dollars plus 7 dollars per item, with a budget of 50 dollars. For n items, 12+7n 50, so n <= 387. The greatest possible count is 5: five cost 47 dollars, while six cost 54 dollars.

When both conditions must hold, solve them separately and keep their overlap. Translate “at most” as \(\le\) and “at least” as \(\ge\). Counts must be whole numbers, so interpret the bound before choosing a value.

From \(2x+1>5\), \(x>2\). From \(3x-2\le10\), \(x\le4\). Combining \(x>2\) and \(x\le4\), the top row below shows the first condition and the lower row shows the second. Their common range is \(2<x\le4\); the integer solutions are \(3\) and \(4\).

A hire fee is \(12\) dollars plus \(7\) dollars per item, with a budget of \(50\) dollars. For \(n\) items, \(12+7n\le50\), so \(n\le\dfrac{38}{7}\). The greatest possible count is \(5\): five cost \(47\) dollars, while six cost \(54\) dollars.

Common bounds and feasible counts

Common bounds and feasible counts