Doppler Effect and the speed of sound

[Potatochip911]

Homework Statement

You are moving at a speed of 35m/s and hear a siren coming from behind you and observe the frequency to be 1370 Hz. The siren goes past you and the new frequency heard is 1330 Hz. What is the speed of the siren traveling at? The speed of sound in air is 340m/s.
f₁=1370 Hz
f₂= 1330 Hz
D=35m/s
V=340m/s

Homework Equations

f^l=f_o*((V+D)/(V+S))

The Attempt at a Solution

Since the original frequency is not given I decided to make 2 equations with the different frequencies then just divide to eliminate the original frequency.
f₁=f_o*((V+D)/(V-S)) <-Since the siren is approaching
f₁*(V-S)=f_o*(V+D)
f₂=f_o*((V-D)/(V+S)) <-Since the siren is moving away
f₂*(V+S)=f_o*(V-D)
Dividing the 2nd equation by the first,
[f₂*(V+S)]/[f₁*(V-S)]=(V-D)/(V+D)
Letting C=f₂/f₁
C(V+S)(V+D)=(V-D)(V-S)
C(V²+SV+DV+SD)=(V²-DV-SV+SD)
CV²+CSV+CDV+CSD=V²-DV-SV+SD
CV²+CDV+DV-V²=-SV+SD-CSD-CSV
CV(V+D)+V(D-V)=S(-V+D-CD-CV)
S={V[C(V+D)+(D-V)]/[V(-1-C)+D(1-C)]}
Which gives me S=-40m/s which doesn't make any sense, I'm not sure where my mistake is but the correct answer is +40m/s.

 

[mfb]

Check the signs in the numerators of the equations for f_i.

When the siren is approaching, you are moving away from the siren (at least relative to the air).
When the siren is moving away, you are approaching the siren (at least relative to the air).

 

[Potatochip911]

Check the signs in the numerators of the equations for f_i.

When the siren is approaching, you are moving away from the siren (at least relative to the air).
When the siren is moving away, you are approaching the siren (at least relative to the air).

So in f₁ are both of the signs negative and in f₂ both signs are positive?

 

[mfb]

Right.
In both cases, the velocity of the ambulance and your velocity have opposite effects, in one case you have larger numerator and denominator and in the other case both get reduced. Which one is which depends on your choice of the coordinate system.