SOLUTION: A submarine left Hawaii travelling west four hours before an aircraft carrier. The aircraft carrier traveled in the opposite direction going 8 mph slower than the submarine for thr

Algebra.Com
Question 1135794: A submarine left Hawaii travelling west four hours before an aircraft carrier. The aircraft carrier traveled in the opposite direction going 8 mph slower than the submarine for three hours which time the ships were 206 miles apart. Find the submarine's speed.
Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.
4x + 3*(x + (x-8)) = 206       <<<---===  the sum of the two distances


4x + 3x + 3x - 24 = 206


10x = 206 + 24


10x = 230


x = 230/10 = 23 miles per hour is the submarine' speed.     ANSWER


RELATED QUESTIONS

A submarine let Hawaii two hours before an aircraft carrier. The vessels traveled in... (answered by ewatrrr)
An aircraft carrier left hawaii traveling west seven hours before a container ship. The... (answered by josmiceli)
An aircraft carrier left Hawaii traveling west seven hours before a container ship. The... (answered by ccs2011)
As an aircraft carrier left hawaii traveling west seven hours before a container ship.... (answered by Theo)
an aircraft carrier left port 53 and traveled west an an average speed of 10km/h. a... (answered by jorel1380)
A submarine left the Azores and traveled west. Three hours later an aircraft carrier... (answered by ankor@dixie-net.com)
An aircraft carrier left the pier traveling east eight hours before a fishing boat. The... (answered by stanbon)
an aircraft carrier left port 31 and traveled west at an avarage speed of 14 mph.some... (answered by ankor@dixie-net.com)
An aircraft carrier left Diego Garcia and traveled toward St. Vincent at an average speed (answered by mananth)