Doomtown is 300 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9 a. m. Mr. Archer leaves Sagebrush for Joshua. At 1 p. m. Mr. Sassoon leaves Doomtown for Joshua. If Mr. Archer travels at an average speed 20 mph faster than Mr. Sassoon and they each reach Joshua at 4 p. m., how fast is each traveling?

This is an example of distance problem in algebra.

Ley x be the rate of mr. sassoon

X + 20 be the rate of mr. archer

Y = distance from doomtown to joshua

So the distance travelled by mr. sassoon:

Y = 4x

The distance travelled by mr. archer:

(300 + y) = (x + 20) * 7

Substitute y = 4x to the second eqation

300 + 4x = (x + 20) * 7

Solving for x

X = 53.33 mph is the rate of mr sassoon

X + 20 = 73.33 mph the rate of mr. archer