Threse are two questions and two answers:

Question 1. The school band sells carnations on Valentine’s Day for \$2 each. They buy the carnations from a florist for \$0.50 each, plus a \$16 delivery charge.

a. Write a system of equations to describe the situation.

1) Sale:

Each carnation at \$2 ⇒ y = 2x

2) Cost:

Fixed cost + variable cost = deliver charge + unit cost × number of carnations ⇒

y = 0.5x + 16

∴ System of equations:

y = 2x; y = 0.50x + 16. ← answer

b. Graph the system. What does the solution represent?

(Which x-values should I choose to graph the equations above so that they intersect?)

The solution represents the number of carnations for which the cost and the sale are equals (zero profit).

You can use the values of this table

x —– Sale = 2x ——– 0.5x + 16

0 ——- 0 ————— 16

2 —— 4 —————- 1 + 16 = 17

4 —— 8 ————— 2 + 16 = 18

6 —— 12 ————– 3 + 16 = 19

8 —— 16 ————– 4 + 16 = 20

10 —— 20 ———— 5 + 16 = 21

12 —– 24 ———— 6 + 16 = 22

Then, the solution is between 10 and 12 carnations.

Indeed it is: 2x = 0.5x + 16 ⇒ 1.5x = 16 ⇒ x = 16 / 1.5 = 10.67

Therefore, choose values of x that include the interval [0, 12].

c. Explain whether the solution shown on the graph makes sense in this situation. If not, give a reasonable solution.

The exact solution is x = 10.67, which is not a real solution, since the number of carnations must be integer.

The most reasonable solution is the next integer, i.e. 11.

Question 2. 6(x + 2) = -2(x + 10)

1) Expand using distributive property:

6x + 12 = -2x – 20

2) Addition and subtraction property of equality:

6x + 2x = – 20 – 12

3) Combine like terms:

8x = – 32

4) Division property of equality:

x = – 32 / 8 = – 4 ← answer,