Two small planes start from the same point and fly in opposite directions. The first plane is flying 35 mph slower than the second plane. In 3 h, the planes are 795 mi apart. Find the rate of each plane.

Solution

Tw o s ma ll pl an es s t a r t f ro m t h e s am e p o in t an d f l y in o pp os i t e d i rec t i o n s . T h e f i rs t pl an e i s f l y in g 35 m p h s l o w er t han t h e seco n d pl an e . I n 3 h, t h e pl an es a re 795 mi a p a r t . F in d t h e r a t e o f e a c h pl an e .

first plan e^{'} s speed : 115

second plan e^{'} s speed : 150

Solution steps

Translate the problem into an equation:

x \cdot 3 + (x + 35) \cdot 3 = 795

Solve for

x, x \cdot 3 + (x + 35) \cdot 3 = 795 : x = 115

Find speed

first plan e^{'} s speed : 115

second plan e^{'} s speed : 150