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Question: how to solve system of equation to get answer without having to "guide" it with intermediate steps?

Question:how to solve system of equation to get answer without having to "guide" it with intermediate steps?

jrive 195 Maple 2023
solve radical rootof simplification
April 16 2024
0

I'm trying derive the equation for a matching resistor pad.  In order for me to get a solution (eg for R2), I had to "manually" solve the one equation and then substitute the resulting value in the other.  Why doesn't solve do the same thing?  

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restart
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NULLeq1 := (R2+Rhi)*R1/(R1+R2+Rhi) = Rlo

(R2+Rhi)*R1/(R1+R2+Rhi) = Rlo

 (1)

NULL

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eq2 := R1*Rlo/(R1+Rlo)+R2 = Rhi

R1*Rlo/(R1+Rlo)+R2 = Rhi

 (2)
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``
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R1sol := solve(eq2, R1)

-Rlo*(R2-Rhi)/(R2-Rhi+Rlo)

 (3)
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R2sol := solve(subs(R1 = R1sol, eq1), R2)

(Rhi^2-Rhi*Rlo)^(1/2), -(Rhi^2-Rhi*Rlo)^(1/2)

 (4)
> 

restart
> 

NULLeq1 := (R2+Rhi)*R1/(R1+R2+Rhi) = RloNULL

(R2+Rhi)*R1/(R1+R2+Rhi) = Rlo

 (5)

 

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NULLeq2 := R1*Rlo/(R1+Rlo)+R2 = RhiNULL

R1*Rlo/(R1+Rlo)+R2 = Rhi

 (6)
> 

s := `assuming`([solve({eq1, eq2}, {R1, R2})], [`and`(Rhi > 0, Rlo > 0)])

{R1 = RootOf((Rhi-Rlo)*_Z^2-Rhi)*Rlo, R2 = -(RootOf((Rhi-Rlo)*_Z^2-Rhi)*Rhi-RootOf((Rhi-Rlo)*_Z^2-Rhi)*Rlo-Rhi)/(RootOf((Rhi-Rlo)*_Z^2-Rhi)-1)}

 (7)
 

``

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