I wrote the following code:
Solve[{x == VIN/VOUT, VOUT == 5 VIN}, {x}]
Mathematica does not solve it. How can I get the result x=1/5?
Thank you so much for your willingness.
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asked Jul 26, 2018 at 9:12
2 Answers
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Solve[{x == VIN/VOUT, VOUT == 5 VIN}, {x, VOUT}]
(*{{x -> 1/5, VOUT -> 5 VIN}}*)
answered Jul 26, 2018 at 9:42
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$\begingroup$ Hello @UlrichNeumann, your solution is very useful. Thanks. $\endgroup$ Commented Jul 26, 2018 at 9:44
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1$\begingroup$ In recent versions, you can also just use
Solve[{x == VIN/VOUT, VOUT == 5 VIN}]$\endgroup$– Mark S.Commented Jul 26, 2018 at 12:17 -
$\begingroup$ ... also
Solve[{x == VIN/VOUT, VOUT == 5 VIN}, {x}, MaxExtraConditions -> 1]$\endgroup$ Commented Jul 26, 2018 at 14:24 -
2$\begingroup$ Or
Solve[{x == VIN/VOUT, VOUT == 5 VIN}, x, {VOUT}]$\endgroup$ Commented Jul 26, 2018 at 15:38 -
1$\begingroup$ @rhermans - This syntax was documented in earlier versions but the documentation stopped mentioning it. However, it appears to be kept for backward compatibility. The variables to be eliminated must be in list brackets, e.g.,
{VOUT}. If you were to enter justVOUTthen Mma would first try to interpreted it as a domain resulting in the warning:Solve::bdomv: Warning: VOUT is not a valid domain specification. Assuming it is a variable to eliminate.and then give the correct output. $\endgroup$ Commented Jul 30, 2018 at 20:55
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Reduce[
{
x == VIN/VOUT,
VOUT == 5 VIN
}, {x}
]
(* VIN == VOUT/5 && x == 1/5 && VOUT != 0 *)
Solve@Reduce[
{
x == VIN/VOUT,
VOUT == 5 VIN
}, {x}
]
(* {{VOUT -> 5 VIN, x -> 1/5}} *)
Solve[Eliminate[{x == VIN/VOUT, VOUT == 5 VIN}, {VOUT}], {x}]giving{{x -> 1/5}}$\endgroup$