Question

december review assignment
algebra k
for each question, you must show work or write a sentence explaining why the answer you chose is
correct. you will earn 2 points for each correct answer with work or a justification, and 0 points if no
work or justification is shown regardless of whether your answer is correct.

1. to keep track of his profits, the owner of a carnival booth decided to model his ticket sales on a

graph. he found that his profits only declined when he sold between 10 and 40 tickets. which
graph could represent his profits?
(1)
(3)
(2)
(4)

1. what is the solution to the system of equations below?

y = 2x + 8
3(-2x + y) = 12
(1) no solution
(3) (-1,6)
(2) infinite solutions
(4) ( 0.5,9)

Explanation:

Question 1

Step1: Analyze profit trend condition

We need a graph where profit declines (the graph has a decreasing slope) only between 10 and 40 tickets sold.

Step2: Evaluate each graph

• Graph (1): Profit is constant between 10 - 40, not declining. Eliminate.
• Graph (2): Profit is always increasing. Eliminate.
• Graph (3): Profit decreases from 10 - 40 (slope negative) and increases otherwise. Matches the condition.
• Graph (4): Profit has two declines (around 10 - 30 and 30 - 50), not only 10 - 40. Eliminate.

Step1: Simplify the second equation

Start with \( 3(-2x + y)=12 \). Divide both sides by 3: \( -2x + y = 4 \), which can be rewritten as \( y = 2x + 4 \).

Step2: Compare with first equation

The first equation is \( y = 2x + 8 \). These are two linear equations with the same slope (\( m = 2 \)) but different y - intercepts (\( 8 \) vs \( 4 \)). Parallel lines (same slope, different intercepts) never intersect, so there's no solution.

Answer:

Graph (3)

Question 2