iam having a doubt on Chapter 6 (Limiting Factors ). in the example we have taken the equation as follows
MATERIAL – 2S+4E=80 KG
LABOUR – 5S+6E=180 HOURS
In this case we are trying to solve the equation by multiplying the material by 2.5. how can we solve this equation since the first one in KG and second one in Labour hours. Can we do in this way? iam confused. please explain. iam sorry if i asked you a stupid question
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Limiting factors
9y ago#1S and E are both numbers of units - they are neither Kg or hours. (It is only when, for example, S is multiplied by 2 in the first equation that we then have kg.)
sorry for replyng you back, iam still confused. the first equation is in kg ( 2S+4E=80KG ), and second equation in Hours ( 5S+6E=180 Hours ). we are solving this equation by multiplying by 2.5. i understood that part. but first equation is in KG ( 80KG) second equation is in Hours (180)Hours. when we deduct after multiplication it is showing 200 kg-180 hours, and we arrived to the results of 4E = 20. is this correct to deduct 200 from 180 since both are in different category ( KG & Hours )?
9y ago#3It is perfectly valid.
This is basic arithmetic - the equations are setting limits on the number of units of S and E and it is completely irrelevant whether the limits are caused by hours or kgs or whatever.
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