Question

What was David's mistake? I have the rest of the question solved I just can't figure this out.

Answer 1

I attached the question here

Answer 2

I don't need a very in depth answer just an explanation of David's basic mistake.

Answer 3

\(d=rt\)

His solving:
\(d=3(t+7)\)
\(d=4(t-5)\)

\(3(t+7)=4(t-5)\)
\(3t+21=4t-20\)
\(t=41\)
\(d=rt\), so \(d=3(41)=123\)

Let's find his mistake:
\(3(t+7)=4(t-5)\) this is correct because the question stated they were the same.
\(3t+21=4t-20\) this is correct because they just distributed.
\(t=41\) skipped a couple of steps, but still is correct.
\(d=3(41)=123\) this is where it is wrong.

Answer 4

i got 144miles

Answer 5

Wht grade u in?

Answer 6

9th, and wht are you referring to as this "144"?

Answer 7

here

Answer 8

You made a mistake, you added 21 back into the d=rt equation.

Answer 9

You found t for `both` of the equations.

Answer 10

But you have to find the distance for `both` equations not just one.

Answer 11

okay

Answer 12

d+3(41) `+21` = 144 miles , Where did you get 21 from,
3(t+7) you got it from here, 3(t) + 3(7)
but you need to solve for the distance from wherever she is to the train station, so you would subtract:
\(D_\text{early speed}~-~D_\text{late speed}\)

Answer 13

I'm kind of lost

Answer 14

So, you messed up when you added 21 into your last equation.

Answer 15

okay. So it should just be 80? Is that what you're saying?

Answer 16

No no, on the question it gives you these two equations
\(d=3(t+7)\)
and
\(d=4(t-5)\)

You must find the distance for BOTH of them, not just d=3(t+7)

Answer 17

But, because they are the same, you must find the value of t, and then PLUG IN t into your equation d=rt TWICE, once for 3mph and once for 4mph.

Answer 18

okay