Aircraft are landing and taking off on parallel runways at a busy airport. On its approach, one aircraft descends from an altitude of 1200m to an altitude of 500m in 35s. During the same time, a departing aircraft climbs from an altitude of 200 m to an altitudes of 1250m.

A) Write a system of linear equations to model the altitudes of the aircraft

B)When are the aircraft at the same altitude? What is that altitude?

I really don't get how to to this, Please help:(

1200-500 = 700

so
x = 1200 - (700/35)t

1250 - 200 = 1050
so
y = 200 +(1050/35) t

when does x equal y ?

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Sure, I can help you with this problem. Let's break it down step by step:

A) To model the altitudes of the aircraft, we can use the equation of a straight line in slope-intercept form: y = mx + b, where y represents the altitude and x represents the time.

For the descending aircraft:
Altitude (y) = -35x + 1200

For the departing aircraft:
Altitude (y) = 37.14x + 200

Note that the slope for the descending aircraft is negative (-35) because it is descending, while the slope for the departing aircraft is positive (37.14) because it is climbing.

B) To find when the aircraft are at the same altitude, we need to set the two equations equal to each other and solve for x:

-35x + 1200 = 37.14x + 200

Now, let's solve for x.

First, we can simplify the equation by combining like terms:

-35x - 37.14x = 200 - 1200

-72.14x = -1000

Next, divide both sides of the equation by -72.14 to solve for x:

x = -1000 / -72.14

Calculating this, we find:

x ≈ 13.88 seconds

Now, to find the altitude at this time, substitute x = 13.88 into either of the equations. Let's use the equation for the descending aircraft:

Altitude (y) = -35(13.88) + 1200

Simplifying this, we have:

y ≈ 500.8 meters

Therefore, the aircraft are at the same altitude of approximately 500.8 meters after approximately 13.88 seconds.

I hope this explanation helps you understand how to approach this problem. Let me know if you have any further questions!