SPEED TIME DISTANCE GOING r 300/r 300 BACK r+10 300/(r+10) 300 DIFF 1
Let x be the rate (in miles per hour) going to friend's home. Then the speed going back is (x+10) mph. The time going to there ishours. The time going back is hours. The "time" equation is - = 1 hour. The solution/(the answer) is just seen from here: x = 50 mph. To get formal algebra solution, multiply both sides by x*(x+10). You will get 300*(x+10) - 300x = x*(x+10) 300x + 3000 - 300x = x^2 + 10x x^2 + 10x - 3000 = 0 (x-50)*(x+60) = 0 and x= 50 is the only meaningful positive solution. ANSWER. The rate going to friend's home was 50 mph. CHECK. - = 6 - 5 = 1 hour. ! Correct !
Let speed on outbound leg be S
Then speed on return leg = S + 10
We then get the following TIME equation:
300(S + 10) = 300S + S(S + 10) ------ Multiplying by LCD, S(S + 10)
0 = (S - 50)(S + 60)
S - 50 = 0 OR S + 60 = 0
Speed on outbound leg, orOR S = - 60 (ignore)