SOLUTION: Hi, I have some sheets on relativity and i was following the maths used in creating the Lorentz factor. It's a factoring question i have, i can factorise equations but this one

Question 87755: Hi,
I have some sheets on relativity and i was following the maths used in creating the Lorentz factor.
It's a factoring question i have, i can factorise equations but this one i don't understand how they did it and would appreciate it if you could explain it to me please.
The equation starts as:
t'^2=(v^2*t'^2+I^2)/c^2
They factorise the equation to be:
t'^2(1-[v^2/c^2])=I^2/c^2
How exactly did they get from that top equation to the second one?
Many thanks!
Steve
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
t'^2=(v^2*t'^2+I^2)/c^2
Multiply both sides by c^2 to get:
c^2t'^2 = v^2t'^2 + I^2
---------------------
You need to get the two terms which contain the factor t'^2 on the
same side of the equal sign. you get
[c^2t'^2 - v^2t'^2] = I^2
-----------------------
Now factor out the t'^2 to get:
t'^2[c^2-v^2] = I^2
-----------
Isolate t'^2 to get:
t'^2 = I^2/[c^2-v^2]
----------------------------
Let's see if their result is the same as ours:

t'^2(1-[v^2/c^2])=I^2/c^2
t'^2 = [I^2/c^2]/[1 - (v^2/c^2)]
t'^2 = [I^2/c^2]/[(c^2-v^2)/c^2)
Cancel the c^2 to get:
t'^2 = I^2/[c^2-v^2]
=======================
Cheers,
Stan H.
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