Challenge. The price of Stock A at 9 A. M. was $13.66. Since then, the price has been increasing at the rate of $0.05 each hour. At noon the price of Stock B was $14.16. It begins to decrease at the rate of $0.15 each hour. If the two rates continue, in how many hours will the prices of the two stocks be the same?

7/4

Step-by-step explanation:

We can write equations for increasing stock and decreasing stock and find our answer.

Let number of hours be "h" for both to be same

For first, we can write:

13.66 + 0.05h

But, second one starts from noon, so already 12pm - 9am = 3 hours gone, so stock A increased by 3 x 0.05 = 0.15, so we equation would be:  

13.81 + 0.05h

Now, Stock B's equation [from noon]:

14.16 - 0.15h

Equate and solve:

so we will do 13.81 + 0.05h = 14.16 - 0.15h

number with number

variables with variables

0.2h   = 0.35

divide 0.2 both sides

answer is

1.75 or if its in fraction form 7/4

1.75 hours

Step-by-step explanation:

(13.66 +3*0.05) + (0.05 * h) = 14.16 - (0.15 * h)

13.81 + 0.05h = 14.16 - 0.15h

0.05h + 0.15h = 14.16 - 13.81

0.20h = 0.35

h = 0.35/ 0.20

h = 7/4

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3