Two planes left the same airport traveling in opposite directions. the first plane left at 9: 00 a. m. and 2.25 hours later, the two planes were 1825 miles apart. the second plane left at 10: 00 a. m. and its average rate was 108 miles per hour slower than the first plane's average rate. let x represent the first plane's average rate. what was the first plane's average rate? enter an equation that can be used to solve this problem in the first box. solve for x and enter the first plane's average rate in the second box.

(a) 1825 = 2.25x + (2.25-1)(x -108)

(b) 560 mi/h

Step-by-step explanation:

(a) distance = speed·time

The first plane's speed is x. The distance it travels in 2.25 hours is 2.25x.

The second plane's speed is x-108. It travels only 1.25 hours (since it started an hour later). The distance it travels is then (2.25 -1)(x -108).

The problem statement tells us the total of the distances traveled by the two planes is 1825 miles, so we can write the equation ...

... 1825 = 2.25x + (2.25 -1)(x -108)

(b) Simplifying the equation gives ...

... 1825 = 3.50x -135

To solve this 2-step equation, we add 135, then divide by 3.50.

.. 1960 = 3.50x

... 1960/3.50 = x = 560

The first airplane's speed is 560 mph.

Check

In 2.25 hours, the first plane travels (560 mi/h)·(2.25 h) = 1260 mi.

In 1.25 hours, the second plane travels (452 mi/h)·(1.25 h) = 565 mi.

Then 2.25 hours after the first plane leaves, the planes are 1260 +565 = 1825 miles apart, as given in the problem statement.

the solution is h=10 because you add 9 to each side of the equation.

step-by-step explanation: