Question

A hot air balloon is 10 meters above the ground and rising at a rate of 15 meters per minute. Another balloon is 150 meters above the ground and descending at a rate of 20 meters per minute. What is the solution of the system of equations? Make sure you write the answer as an ordered pair.

Answer 1

(4, 70)

Explanation:

Let's call

x: time, in minutes

y: height of the balloon, in meters

The height of the rising balloon is modeled as follows:

y = 15x + 10 (eq. 1)

The height of the descending balloon is modeled as follows:

y = -20x + 150 (eq. 2)

Combining equations 1 and 2:

15x + 10 = -20x + 150

15x + 20x = 150 - 10

35x = 140

x = 140/35 = 4

Replacing it in equation 1:

y = 15(4) + 10 = 70

So, the solution is (4, 70), that is, after 4 minutes both balloons will have the same height, which is 70 meters.

Answer 2

(x,y,t) = (70,70,4)

Explanation:

For the first balloon,

Initial height = 10 m.

Rate at which it is going up = 15 m/min.

So, its height x at any time can be given by,

x = 10 + 15t -(1) , where t is in minutes.

For the second balloon,

Initial height = 150 m.

Rate at which it is going downwards = 20 m/min.

so, its height y at any time is given by,

y = 150 - 20t -(2), where t is any minutes

Thus, when both of them will be at same height , x= y.

So, 10 + 15t = 150 - 20t

35t = 140.

t = 4 min.

x = y = 10 + 60 = 70 m.

So, Solution is (x,y,t) = (70 , 70 , 4)