SOLUTION: How can I isolate R1? 1/1.33 = 1/(R1-6) + 1/R1 This is what I've done: 1/1.33 = R1 + (R1-6) -----------

SOLUTION: How can I isolate R1? 1/1.33 = 1/(R1-6) + 1/R1 This is what I've done: 1/1.33 = R1 + (R1-6) ----------- - (R1-6)(R1) 1/1.33 = 2R1-6 ---

Question 47642This question is from textbook
: How can I isolate R1?
1/1.33 = 1/(R1-6) + 1/R1
This is what I've done:
1/1.33 = R1 + (R1-6)
-------------
(R1-6)(R1)
1/1.33 = 2R1-6
------
(R1-6)(R1)
I did not know what to do after this. I also tried the following
1/1.33 = R1/ (R1-6)(R1) + (R1-6)/(R1-6)(R1)
1/1.33 = 1/(R1-6) + 1 / R1
This as far as I've gone
ISBN 0471-15183-1 CHAPTER 20 PROBLEM 54 This question is from textbook

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1/1.33 = 1/(R1-6) + 1/R1
I'm going to replace R1 by "x" as it is easier to type.
1/1.33 = 1/(x-6) + 1/x
1/1.33 = [x+x-6]/[x(x-6)]
1/1.33 = [2x-6)]/[x^2-6x]
Cross-multiply to get;
x^2-6x=2.66x-7.98
x^2-8.66x+7.98=0
x=[8.66+sqrt(8.66^2-4(7.98)]/2 or x=[8.66-sqrt(8.66^2-4(7.98)]/2
x=[8.66+sqrt43.0756]/2 or x=[8.66-sqrt(43.0756)]/2
Reverting now to x=R1 you get:
R1=7.61160022... or R1=2.09679956...

Only R1=7.61160022... is a valid answer
The other is an extraneous result of the way the
quadratic equation was formed. That means the
2nd answer is valid for the quadratic equation but
not for your original equation.
Cheers,
Stan H.
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