Hey i have a question but the equation is to long so i will put the problem in the description box

Question

Two girls Emily and kristan, go for a run during track practice. Emily begins walking 30 seconds earlier than kristan. emily ran at a pace of 8 feet per second, and kristan ran at a pace of 10 feet per second. how many seconds had emily been running at the moment when the two have ran exactly the same distance?

Answer 1

Emily : 8 * (t+30)

Kristen: 10 * t

8(t+30) = 10t

8t + 240 = 10t

240 = 2t

t = 120

t = 120 seconds = 2 minutes

At that time Emily (120+30) * 8 = 150*8 = 1200 feet <--- 30 second head start

Kristen 120*10 = 1200 feet

they meet at 150 seconds

Answer 2

Two girls Emily and kristan, go for a run during track practice. Emily begins walking 30 seconds earlier than kristan. emily ran at a pace of 8 feet per second, and kristan ran at a pace of 10 feet per second. how many seconds had emily been running at the moment when the two have ran exactly the same distance?

This type of problem is very common throughout the Algebra world.

It is the classic Rate (R) x Time (T) = Distance (D) formula.

In this case....

Emily's Pace.........8 f/s x (T+30) = D.....time here is added because she starts earlier

Kristan's Pace....10f/s x T = D

You now have 2 equations and 2 unknowns...common algebra exercise....

8f/s x (Ts + 30s) = 8f/sTs + 240ft = D or simply 8T + 240 = D

10 f/s x Ts = D....or simply 10T = D

Next substitute D from the 2nd equation into the first

like so.... 8T + 240 = 10T........solve for T

2T = 240

T = 120 seconds ,,,,Kristan's Time

----check ----

8f/s x (120s + 30s) = 8 x 150 = 1200 ft

10f/s x 120s = 1200 ft

....so in 120s Kristan reaches 1200 ft..

however the question ASKS....

"how many seconds had emily been running at the moment when the two have ran exactly the same distance?"

Which is 1200ft / 8f/s = 150s or 120s + 30s = 150s....so 150s is the correct answer