Question

select the correct locations on the image.
sean decides to start a small business creating and selling outdoor yard games. it will cost sean $50 to make each game as well as an initial cost of $300 to purchase the needed equipment and supplies. he plans to sell each game for $85.
the system of equations below models the cost and revenue for seans outdoor yard games where x represents the number of games and y represents the amount, in dollars.
\\( y = 50x + 300 \\)
\\( y = 85x \\)
first, select the point on the graph that represents seans break - even point, which is the point where the costs to make the games will equal the revenue from selling them. notice that the revenue equation has already been graphed.
then, determine the least number of games sean will need to sell in order to make a profit. profit occurs when revenue is greater than the cost.
graph of outdoor yard games profit with x - axis as number of games (0 - 20) and y - axis as dollars (0 - 1000), and a blue line for revenue. also, a set of numbers 5, 8, 9, 10, 12 with the question to make a profit, sean needs to sell games.

Explanation:

Step1: Find break - even point (solve the system)

To find the break - even point, we set the cost equation equal to the revenue equation since at break - even, cost = revenue. So we solve \(50x + 300=85x\).
Subtract \(50x\) from both sides: \(300 = 85x-50x\), which simplifies to \(300 = 35x\). Wait, no, \(85x - 50x=35x\)? Wait, no, \(85 - 50 = 35\)? Wait, no, \(85-50 = 35\)? Wait, no, \(85 - 50=35\)? Wait, actually, \(85x-50x = 35x\), so \(35x=300\)? Wait, that can't be right. Wait, I made a mistake. Let's do it again. The cost function is \(y = 50x+300\) (fixed cost + variable cost) and revenue is \(y = 85x\). So set \(50x + 300=85x\). Subtract \(50x\) from both sides: \(300=35x\)? No, \(85x - 50x = 35x\)? Wait, \(85 - 50 = 35\), yes. Then \(x=\frac{300}{35}=\frac{60}{7}\approx8.57\). Wait, but maybe we should solve it correctly. Wait, \(50x+300 = 85x\). Subtract \(50x\): \(300=35x\)? No, \(85x-50x = 35x\), so \(x=\frac{300}{35}=\frac{60}{7}\approx8.57\). But since we can't sell a fraction of a game, we need to sell more than the break - even quantity. The break - even quantity is when \(x=\frac{300}{85 - 50}=\frac{300}{35}=\frac{60}{7}\approx8.57\). So the least number of games to make a profit is the smallest integer greater than \(8.57\), which is \(9\).

Step2: Determine the least number of games

Since the break - even point is at \(x\approx8.57\), we need to sell \(x = 9\) games (because we can't sell a fraction of a game, and selling \(8\) games would give revenue less than cost, selling \(9\) games gives revenue greater than cost).

Answer:

To make a profit, Sean needs to sell \(\boldsymbol{9}\) games.