You can
put this solution on YOUR website!Starting with the expression

to represent the distance from home to the metro is spot on IF you meant that x is the distance travelled on the metro.
But then you went off and created the expression x + .35 hour for time on the metro. You can't use the same variable for two different quantities, distance on the one hand, and time on the other.
The relationships you need to create must be based on the basic formula for distance given rate and time:
Since the relationships we will be able to establish to compare the two parts of the trip are based on time, we need to solve the basic equation for time:
Now, let's build a relationship for the trip from home to the metro. She travels 33-x miles at 7 miles per hour. So:

.
The trip on the metro is x miles at 39 mph. So:

.
So far, so good. Now we also know that the time on the bike is .35 hour less than the time on the metro, so we can say

. Now all we have to do is replace the two time variables in this statement with the expressions we developed to represent them and solve for x.

Lowest Common Denominator

Calculate

Cross-multiply

Distributive property

Add -1287 and -7x to both sides

Combine terms

Divide by -46

Round off to nearest whole mile
Since total trip was 33 miles, home to metro is 33 - 30, or 3 miles.
You had the right answer, I just couldn't figure how you got there. I hope what I did makes sense.
John
You can
put this solution on YOUR website!you almost got it.
st=d t=time (hrs) s=speed d=distance
.
Let t be time on the metro.
7(t+.35)+39t=33
7t+2.45+39t=33
46t=33-2.45
46t=30.55
t=.66413 hrs time on metro
t+.35=1.01413 hrs on the bike
1.01413*7=7 mi from her house to the metro station.
.
Check:
.66413*39=26
26+7=33
.
Ed