Two jets leave an air base at the same time and travel in opposite directions. One jet travels 74 mph faster than the other. If the two jets are 9492 miles apart after 7 hours, what is the rate of each jet?

Solution

Tw o j e t s l e a v e an ai r ba se a t t h e s am e t im e an d t r a v e l in o pp os i t e d i rec t i o n s . O n e j e t t r a v e l s 74 m p h f a s t er t han t h e o t h er . I f t h e tw o j e t s a re 9492 mi l es a p a r t a f t er 7 h o u rs, w ha t i s t h e r a t e o f e a c h j e t ?

first je t^{'} s speed : 715

second je t^{'} s speed : 641

étapes des solutions

Translate the problem into an equation:

x \cdot 7 + (x - 74) \cdot 7 = 9492

Solutionner pour

x, x \cdot 7 + (x - 74) \cdot 7 = 9492 : x = 715

Trouver speed

first je t^{'} s speed : 715

second je t^{'} s speed : 641