Wanda Kane

2022-01-04

Write an equation to solve each problem.
Two planes left an airport at noon. One flew east and the other flew west at twice the speed. After 3 hours the planes were 2700 mi apart. How fast was each plane flying?

Jeremy Merritt

Let x be the speed of the slower plane going east in mi/hr
2x is the speed of the plane going west.
$speed\cdot time=dis\mathrm{tan}ce$
If they are 2700 mi apart in 3 hr then
$x\cdot 3+2x\cdot 3=2700$
$3x+6x=2700$
$9x=2700$
$x=300m\frac{i}{h}r$
So the faster plane flew at 60 mi/hr.
East at 300mi/hr, west at 600 mi/hr

Lynne Trussell

If we call the speed of the first plane v then the other plane has a speed of $2\cdot v$
Explanation:
So the distance between the planes will get greater by
$v+2\cdot v=3\cdot v$ every hour
So in three hours their distance will be:
$3\cdot 3\cdot v$ which is equal to 2700 mi.
So $9\cdot v=2700⇒v=\frac{2700}{9}=300\mp h$
And the other plane had twice that speed: 600mph

karton

x= speed of the first plane. (east)
2x= speed of the second plane (west).
we can suggest this equation:
3(x+2x)=2700
3(3x)=2700
9x=2700
x=$\frac{2700}{9}$
x=300
speed of the first plane =300 miles/h
speed of the second plane =2x=2(300 miles/h)=600 miles/h
Answer: the speed of the first plane was 300 miles/h, and the speed of the second plane was 600 miles/h.

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