Is this conversion valid? Y = u + v − v *t* Y + I + G to Y (1 − v + v* t) = u + I + G

jangetzen

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Is this conversion valid? Y = u + v − v *t* Y + I + G to Y (1 − v + v* t) = u + I + G

I think this equation/conversion is not correct but I'm not entirely sure:
Y = u + v − v *t* Y + I + G
Y (1 − v + v* t) = u + I + G

As I understand it the applied operations are -v and +v*t*Y but then I think the factorising on the second line is wrong?
Thanks for any help
 
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I think this equation/conversion is not correct but I'm not entirely sure:
Y = Z = c 0 + c 1 − c 1 *t* Y + I + G
Y (1 − c 1 + c 1* t) = c 0 + I + G

As I understand it the applied operations are -c1 and +c1*t*Y but then I think the factorising on the second line is wrong?
Thanks for any help
Your thinking is correct!

Did you de-factorize the second line and see where could be the problem?
 
Did you de-factorize the second line and see where could be the problem?

Yes, defactorizing the first part would be equal to Y-Y*v but (leaving the other aside) the operation -v would lead to : Y-v instead which would be correct.
I just wanted to make sure that I'm not crazy because this is taken out of 6 semester Economic bachelor lecture slides :mrgreen:
 
Where did the Z go?!
Plus you're showing 2 equal signs...

Please repost the original equation PROPERLY.
Also, use u instead of c 0 and v instead of c 1 : less confusing!

Sorry, I now edited the post to reduce confusion. The Z is just another symbol that in the thematic context is always equal to Y - but it is not relevant for this question so I removed it as well
 
Sorry, I now edited the post to reduce confusion. The Z is just another symbol that in the thematic context is always equal to Y - but it is not relevant for this question so I removed it as well

The next thing to do is to show the steps of your work. Don't factorize until you have first written the result of the subtraction and addition you described; skipping that step is a common cause of errors.

Then, check that your final factorized form agrees with the step before, by multiplying. This catches most errors.
 
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