A merchant could sell one model of digital cameras at lisice and receive $224 for .....

Question

A merchant could sell one model of digital cameras at list price and receive $224 for all of them. If he had eight more cameras, he could sell each one for $14 less and still receive $224. Find the list price of each camera.

Answer 1

So we don't know how many cameras he has at first.

let's say he has N cameras.

Then...

They each cost 224/N since N * cost = 224

If instead he had 8 more and they cost 14 less then

He would have N + 8 cameras

and they would cost (224/N - 14)

the cost of all cameras now should be 224 again.

This can be written as (N + 8) (224/N - 14) = 224

FOIL to find N

224 + 1792/N - 14N - 112 = 224

1792/N -14N - 112 = 0

Multiply everything by N

1792 - 14N² - 112N = 0

Rearange

14N² + 112N - 1792 = 0

Solve with quadratic formula.

only positive answer is

8 cameras.

He started with 8 cameras for $28 each

With 16 cameras he sold them for $14 each.

Answer 2

X cameras times list Price = 224

XP = 224

(X+8)(P-14)=224

X=224/P

(224/P+8)(P-14)=224

expand and solve for P

224 +8P-14(224)/P - 8(14) = 224

8P - 14(224)/P = 8(14)

multiply both sides by P

8P²- 14(224) = 8(14)P

P² - 14(28) = 14P

P²-14P - 14(28) = 0

solve for P, using either quadratic formula, completing the square or factor

(P-28)(P+14) = 0

P=$28 per camera

XP=224

X(28)=224

X=224/28=8 cameras

add 8 more to get 16 cameras sold at $14 per camera to get total 16x14=$224