Simplyfying a vectorial equation

[rugalas]

Hello all, was hoping i could get some help on the following problem.
I have the following equation:

t = (N1*V1 + N2*V2 + N3*V3 - b) / (N1*V1 + N2*V2 + N3*V3 - N1*P1 + N2*P2 + N3*P3)
Which I simplified as follows:
d = N1*V1 + N2*V2 + N3*V3
l = N1*P1 + N2*P2 + N3*P3
t = (d-b) / (d-l)

So the simpler question is can I further simplify
t = (d-b) / (d-l)
Perhaps somehow cancel out the d?

The harder question is can I simplify at a higher level...
t = (N1*V1 + N2*V2 + N3*V3 - b) / (N1*V1 + N2*V2 + N3*V3 - N1*P1 + N2*P2 + N3*P3)
Perhaps by somehow cancelling out N1,N2,N3?
I know I can do (N1*V1-P1) + (N2*V2-p2) + (N3*V3-P2) in the denominator but that doesn't simplify it because
you still have (N1*V1 + N2*V2 + N3*V3) in the numerator and so it is simpler to make it into t = (d-b) / (d-l).

Thanks

 

[Deleted member 4993]

Hello all, was hoping i could get some help on the following problem.
I have the following equation:

t = (N1*V1 + N2*V2 + N3*V3 - b) / (N1*V1 + N2*V2 + N3*V3 - N1*P1 + N2*P2 + N3*P3)
Which I simplified as follows:
d = N1*V1 + N2*V2 + N3*V3
l = N1*P1 + N2*P2 + N3*P3
t = (d-b) / (d-l)

So the simpler question is can I further simplify
t = (d-b) / (d-l)
Perhaps somehow cancel out the d?

The harder question is can I simplify at a higher level...
t = (N1*V1 + N2*V2 + N3*V3 - b) / (N1*V1 + N2*V2 + N3*V3 - N1*P1 + N2*P2 + N3*P3)
Perhaps by somehow cancelling out N1,N2,N3?
I know I can do (N1*V1-P1) + (N2*V2-p2) + (N3*V3-P2) in the denominator but that doesn't simplify it because
you still have (N1*V1 + N2*V2 + N3*V3) in the numerator and so it is simpler to make it into t = (d-b) / (d-l).

Thanks

What do you need to find?

Simplification is only useful - when you know that you need to find somethingelse.