Michael is working two summer jobs, making \$19 per hour lifeguarding and making \$14 per hour clearing tables, In a given week, be can work at most 17 total hours and must earn a minimum of $\$ 280$. If $x$ represents the number of hours lifeguarding and $y$ represents the number of hours clearing tables, write and solve a system of inequalities graphically and determine one possible solution.
Answer
The feasible region is the area that satisfies both inequalities. We can find one possible solution by picking a point in this region. For example, if Michael works 10 hours lifeguarding (\(x = 10\)) and 7 hours clearing tables (\(y = 7\)), he will work a total of 17 hours and earn $310, which is above the minimum requirement. The final answer is \(\boxed{x = 10, y = 7}\).
Step 1 ：First, we can write the system of inequalities based on the problem. The total hours Michael can work is at most 17, so \(x + y \leq 17\). He must earn a minimum of $280, so \(19x + 14y \geq 280\).
Step 2 ：Next, we can graph these inequalities. The line \(x + y = 17\) is the boundary for the hours he can work, and the line \(19x + 14y = 280\) is the boundary for the money he must earn.
Step 3 ：The feasible region is the area that satisfies both inequalities. We can find one possible solution by picking a point in this region. For example, if Michael works 10 hours lifeguarding (\(x = 10\)) and 7 hours clearing tables (\(y = 7\)), he will work a total of 17 hours and earn $310, which is above the minimum requirement. The final answer is \(\boxed{x = 10, y = 7}\).
