Need help with simple math problem
#1
N = R + ( (R/3) * (P/2) )

I want to move P to N ='s place and N somewhere inside the formula.

How do I do this? I'm doing a formula in Excel and need to figure out something backwards. I'm sure the answer is beyond simple, but the arithmetic escapes me atm.
"The true value of a human being is determined primarily by the measure and the sense in which he has attained liberation from the self." -Albert Einsetin
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#2
(08-31-2013, 06:51 AM)Taem Wrote: N = R + ( (R/3) * (P/2) )

I want to move P to N ='s place and N somewhere inside the formula.

For R not equal to zero:

N = R + ( (R/3) * (P/2) )
R + ( (R/3) * (P/2) ) = N
(R/3) * (P/2) = N - R
P/2 = (3/R) * (N - R)
P = (6/R) * (N - R)
P = (6N/R) - 6

Hopefully I haven't made any goofs. Algebra 1 was a long time ago.
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#3
(08-31-2013, 06:51 AM)Taem Wrote: N = R + ( (R/3) * (P/2) )

I want to move P to N ='s place and N somewhere inside the formula.

How do I do this? I'm doing a formula in Excel and need to figure out something backwards. I'm sure the answer is beyond simple, but the arithmetic escapes me atm.

The answer is 2.
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#4
(08-31-2013, 08:08 AM)Nystul Wrote:
(08-31-2013, 06:51 AM)Taem Wrote: N = R + ( (R/3) * (P/2) )

I want to move P to N ='s place and N somewhere inside the formula.

For R not equal to zero:

N = R + ( (R/3) * (P/2) )
R + ( (R/3) * (P/2) ) = N
(R/3) * (P/2) = N - R
P/2 = (3/R) * (N - R)
P = (6/R) * (N - R)
P = (6N/R) - 6

Hopefully I haven't made any goofs. Algebra 1 was a long time ago.

Works perfectly! If this were a help forum, I'd add +1 to your reputation Big Grin ! Thank you for the help.
"The true value of a human being is determined primarily by the measure and the sense in which he has attained liberation from the self." -Albert Einsetin
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