Hello!
Let's write some information:
Plane A:
• altitude: 2667 feet
,
• each second: +65.5 feet
Plane B:
• altitude: 4500 feet
,
• each second: +30.25 feet
Knowing it, we can write an equation for each airplane:
• A: 2667 + 65.5s
,
• B: 4500 + 30.25s
Now, let's solve the exercise:
How many seconds will pass before the plans are at the same altitude?
To solve this exercise we just have to equal the two equations that I create above and solve it, look:
[tex]\begin{gathered} 2667+65.5s=4500+30.25s \\ 65.5s-30.25s=4500-2667 \\ 35.25s=1833 \\ s=\frac{1833}{35.25} \\ \\ s=52\text{ seconds} \end{gathered}[/tex]
What will their altitude be when they're at the same altitude?
Now we will use the same equations again, but replace where's "s" by 52:
[tex]\begin{gathered} 2667+65.5s=4500+30.25s \\ 2667+(65.5\cdot52)=4500+(30.25\cdot52) \\ 2667+3406=4500+1573 \\ 6073=6073 \end{gathered}[/tex]
The altitude will be 6073 feet.
