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Simon Jowitt[_2_]

Solving multiple equations
 
Hi, I have an equation in the following form: x = (y*(z+1))/(z+a) which I
have the values for x, y and a, but need to find a value for z. I can do this
one at a time using the SOLVER add-in - but is there another way around this?
or can you run multiple versions of solver e.g. to determine 100+ values of z
for 100+ different equations? even 20+ would do!

Sheeloo[_3_]

Solving multiple equations
 
if you just need to calculate z for each set of x,y and a then use a formula...
enter x values in Col A, y in B and a in C then enter this in D1 and copy down
=(B1/A1-1)/(1-B1/A1)
[check my solution and correct if wrong, but you get the idea, right?]
You can enter as many sets as you want, one set per row... and copy down the
formula in D1



--
Always provide your feedback so that others know whether the solution worked
or problem still persists ...


"Simon Jowitt" wrote:

Hi, I have an equation in the following form: x = (y*(z+1))/(z+a) which I
have the values for x, y and a, but need to find a value for z. I can do this
one at a time using the SOLVER add-in - but is there another way around this?
or can you run multiple versions of solver e.g. to determine 100+ values of z
for 100+ different equations? even 20+ would do!


Sheeloo[_3_]

Solving multiple equations
 
It should be
=(B1/A1-C1)/(1-B1/A1)

sorry for the typo...
--
Always provide your feedback so that others know whether the solution worked
or problem still persists ...


"Sheeloo" wrote:

if you just need to calculate z for each set of x,y and a then use a formula...
enter x values in Col A, y in B and a in C then enter this in D1 and copy down
=(B1/A1-1)/(1-B1/A1)
[check my solution and correct if wrong, but you get the idea, right?]
You can enter as many sets as you want, one set per row... and copy down the
formula in D1



--
Always provide your feedback so that others know whether the solution worked
or problem still persists ...


"Simon Jowitt" wrote:

Hi, I have an equation in the following form: x = (y*(z+1))/(z+a) which I
have the values for x, y and a, but need to find a value for z. I can do this
one at a time using the SOLVER add-in - but is there another way around this?
or can you run multiple versions of solver e.g. to determine 100+ values of z
for 100+ different equations? even 20+ would do!


ShaneDevenshire

Solving multiple equations
 
Hi,

you need to rearrange the formula, if I did it correctly it is

Z=(Y/X-A)*X/(X-Y)


--
Thanks,
Shane Devenshire


"Simon Jowitt" wrote:

Hi, I have an equation in the following form: x = (y*(z+1))/(z+a) which I
have the values for x, y and a, but need to find a value for z. I can do this
one at a time using the SOLVER add-in - but is there another way around this?
or can you run multiple versions of solver e.g. to determine 100+ values of z
for 100+ different equations? even 20+ would do!


Dana DeLouis

Solving multiple equations
 
Same solution as the others, but with just 1 division

z = (y - a*x)/(x - y)

--
Dana DeLouis


"Simon Jowitt" wrote in message ...

Hi, I have an equation in the following form: x = (y*(z+1))/(z+a) which I
have the values for x, y and a, but need to find a value for z. I can do this
one at a time using the SOLVER add-in - but is there another way around this?
or can you run multiple versions of solver e.g. to determine 100+ values of z
for 100+ different equations? even 20+ would do!

muddan madhu

Solving multiple equations
 
suppose u have values x,y & a in cell A1,B1,C1

in then use this formula Z=(B1-A1*C1)/A1-B1


On Oct 21, 9:03*am, ShaneDevenshire
wrote:
Hi,

you need to rearrange the formula, if I did it correctly it is

Z=(Y/X-A)*X/(X-Y)

--
Thanks,
Shane Devenshire

"Simon Jowitt" wrote:
Hi, I have an equation in the following form: x = (y*(z+1))/(z+a) which I
have the values for x, y and a, but need to find a value for z. I can do this
one at a time using the SOLVER add-in - but is there another way around this?
or can you run multiple versions of solver e.g. to determine 100+ values of z
for 100+ different equations? even 20+ would do!




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