Inverse lorentz transformation question

[Benzoate]

Homework Statement

Suppose that an event occurs in inertial frame S with cooridinates x=75 m y=18 m ,z=4.0 m and t=2.0*10^-5 seconds . The inertial frame S' moves in the +x direction with v=c*.85 . The origins of S and S' coincided at t=t'=0. a)what are the coordinates of the event in S' and b) Use the inverse transformation on the results of part(a) to obtain the original coordinates

Homework Equations

x=gamma*(x'+vt') , t=gamma*(t'+v*x'/c^2) , t'=gamma*(t'-v*x'/c^2) , x'=gamma*(x-vt)
gamma=1/sqrt(1-B^2) ; B=v/c)

The Attempt at a Solution

I didn't have a problem calculating the coordinates in part a, but I will display my results from part a nevertheless

a) gamma =1/sqrt(1-(.85)^2= 1.9
y'=y=18 m
z'=z=4 m
x'=1.9*(75 m-(.85c)(2.0e-5 sec))=-9547.00 m
t'=1.9*(2.0e-5 - (.85c)(75)/(c^2))= .00004 secondsb) y=y'=18 m
z=z'= 4 m
x= 1.9*(-9547 m + (.85c)(.00004 sec)) = 1240.700
t=1.9*(.00004 seconds +(.85c)(-9547 m)/(c^2))= .00002

In part b,I don't understand why my calculations are incorrect for x , but not for t , z, or y

 

[learningphysics]

You're not carrying enough decimal places... for t' and gamma you need more decimal places.

 

[Benzoate]

sorry, t is suppose to be equal to .00002 seconds, not .000002 seconds .gamma's actually value is 1.898 , but that still shouldn't effect the result of my other space coordinates if I round gamma to 1.9. no, t' is correct because when I plug t' into the equation for t, I get the correct result for t.

 

[learningphysics]

when I plug in t' = 3.759625*10^-5 into the inverse formula for x, I get back the correct x.

 

[Benzoate]

oh, so you didn't round t'?

 

[learningphysics]

Yeah, I didn't round... but seems I still wasn't accurate enough.

Also, I'm getting x = -9537.45... not x = -9547...

It seems to work out when x = -9537.45 and t = 3.75567*10^-5 s