Question 767533
Distance = rate x time


Sheila's rate = 25 mph
Sheila's time = t
Sheila's distance = 25t {distance = rate x time}


Allison's rate = 50mph
Allison's time = t - 2 {she started 2 hours later or lost two hours off of the starting time}
Allison's distance = 50(t - 2) = 50t - 100 {distance = rate x time}


25t = 50(t - 2) {when Allison catches up with Sheila the distances are equal}
25t = 50t - 100 {used distributive property}
-25t = -100 {subtracted 50t from each side}
t = 4 {divided each side by -25}
{t corresponds with Sheila's time, which is what the question is asking for}


In 4 hours Allison will catch up with Sheila
{Your mistake was probably on Allison's time. You probably had t + 2.}
<br>For more help from me, visit:<br><a href = "http://www.algebrahouse.com/" target = "_blank">www.algebrahouse.com</a><br><br>