How to Solve for r and c in Two Equations with Known Quantity w

[zekester]

i have two equations r^2/wc
divided by r^2 + (1/(wc)^2)
equals 13026

and r/wc
divided by r^2 + (1/(wc)^2)
equals 951

where w is a known quantity. how do I solve for r and c.

 

[HallsofIvy]

You want to solve
\frac{\frac{r^2}{wc}}{r^2+ \frac{1}{w^2c^2}}= 13026
and
\frac{\frac{r}{wc}}{r^2+ \frac{1}{w^2c^2}}= 13026
for r and c given that w is a known constant.

The first thing I would do is notice that the left hand side of the first equation is just the left hand side of the second equation times r:
\frac{\frac{r^2}{wc}}{r^2+ \frac{1}{w^2c^2}}= r\frac{\frac{r}{wc}}{r^2+ \frac{1}{w^2c^2}}= r(951)= 13026
so r= 13026/951= 13.7 (approximately). That was easy!

If we let x= wc then the second equation can be written
\frac{\frac{r}{x}}{r^2+\frac{1}{x^2}}= 951.
Multiply both numerator and denominator of the fraction by x² to get
\frac{rx}{r^2x^2+ 1}= 951
Now, I would be inclined to let y= rx so that
\frac{y}{y^2+ 1}= 951
Multiplying on both sides by y²+1 gives the quadratic equation
951y²- y- 951= 0. That can be solved by the quadratic formula:
the two solutions are approximately 0.0329 and -0.0319. Since I don't know where this problem came from I don't know if both of these are valid or not.
Now we have rx= 0.0329 and rx= -0.0319 and we know that r= 13.7 so
x= 0.0329/13.7= 0.0024 and x= -0.0319/13.7= -0.0023.
If x= wc= 0.0024, c= 0.0024/w.
If x= wc= -0.0023, c= -0.0023/w.

 

[zekester]

thanks for the help it is appreciated