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Please help me solve this algebra problem?
Wilma drove at an average speed of 35 mph from her home in City A to visit her sister in City B.
She stayed in City B hours, and on the trip back averaged 50 mph. She returned home 43 hours after leaving.
1 ) We have to determine how many miles "X" is it from City A to City B.
In terms of , express Wilma's driving time from A to B
It is:____________
2.) In terms of "X", her return driving time is:__________
3.) Her TOTAL driving time is: __________ (trip time minus visiting time)
So answer from 1. PLUS 2. set equal to 3. is the equation:
__________=__________ .
which is the equation we must use to solve for "X". Your answer for "X" is : _________
You don't need this "hint", but many are making a simple mistake. Distance equals Speed times Time; Time does not equal Distance times Speed.
The teacher provided a help video of how to do it but the link he gave was broken. I know D=R*T but i keep getting everything mixed up.
grr stayed in City B 15 hours
2 Answers
- 10 years agoFavorite Answer
you say "She stayed in City B hours," but you don't have a number of hours and that is needed to find the answers.
1) X= 35*T
T[AB]=X/35
2) X= 50*T
T[BA]= X/50
3) T[total]=43-T[stay]
T[total]=T[AB]+T[BA]
43-T[stay]=X/35+X/50
you still need the stay time to get an actual time or distance.
- 10 years ago
1) rate*time=distance.
35mph*ta=x
2) 50*tb=x
3)
T1+T2+Vt=43
T1=x/35 T2=x/50
x/35+x/50=43-Vt
distance over the rate gives you the time. time one plus time two plus visiting time equals 43 hrs.
with a little rearranging you get this equation. x=distance. Vt is visiting time.
