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# Equations

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1) C = K(Rr/(R-r)) Solve for K

2) C = K(Rr/(R-r) Solve for R

3) F = fg/(f+g-d) Solve for f

waffles  Feb 14, 2018
edited by waffles  Feb 14, 2018
edited by waffles  Feb 14, 2018

#1
+1720
+1

To solve for a variable, in this case, simpl means to isolate it. That is the goal of solving for a variable.

1)

 $$C=K\left(\frac{Rr}{R-r}\right)$$ Let's first multiply both sides by (R-r). This will eliminate the denominator, which will ease the process of solving for "K." $$C(R-r)=KRr$$ Divide by Rr to isolate K. $$K=\frac{C(R-r)}{Rr}\\$$ You could distribute, if you so desired, but it is not necessary. This just creates more work for yourself, anyway. This is the final answer.

2) This one will be harder since two "R's" exist in the original equation. This generally signals for the use of grouping.

 $$C=K\left(\frac{Rr}{R-r}\right)$$ Let's do the exact same thing as before; multiply by R-r. $$C(R-r)=KRr$$ Let's distribute the C into the binomial. This will allow us to have two terms with "R's" $$CR-Cr=KRr$$ Let's subtract KRr from both sides. Let's add Cr to both means meanwhile. $$CR-KRr=Cr$$ Using grouping, we can change this into one "R." $$R(C-Kr)=Cr$$ Now, divide by C-Kr to isolate "R." $$R=\frac{Cr}{C-Kr}$$ This is finished because we have isolated the R.

3) We will have to utilize the same technique in number 2 in order to isolate "f."

 $$F=\frac{fg}{f+g-d}$$ Let's multiply the denominator on both sides. This will get rid of any and all pesky denominators in this equation, $$F(f+g-d)=fg$$ There is a "g" in the trinomial, so we will have to expand that part. $$Ff+Fg-Fd=fg$$ Move all terms with a factor of f on the left hand side. Any other terms are moved to the right hand side. $$Ff-fg=-Fg+Fd$$ Factor out an "f" from both terms from the left hand side. $$f(F-g)=-Fg+Fd$$ Finally, divide by F-g to isolate the wanted variable completey. $$f=\frac{-Fg+Fd}{F-g}$$ No more simplification is possible here. Leave the fraction as is.
TheXSquaredFactor  Feb 14, 2018
Sort:

#1
+1720
+1

To solve for a variable, in this case, simpl means to isolate it. That is the goal of solving for a variable.

1)

 $$C=K\left(\frac{Rr}{R-r}\right)$$ Let's first multiply both sides by (R-r). This will eliminate the denominator, which will ease the process of solving for "K." $$C(R-r)=KRr$$ Divide by Rr to isolate K. $$K=\frac{C(R-r)}{Rr}\\$$ You could distribute, if you so desired, but it is not necessary. This just creates more work for yourself, anyway. This is the final answer.

2) This one will be harder since two "R's" exist in the original equation. This generally signals for the use of grouping.

 $$C=K\left(\frac{Rr}{R-r}\right)$$ Let's do the exact same thing as before; multiply by R-r. $$C(R-r)=KRr$$ Let's distribute the C into the binomial. This will allow us to have two terms with "R's" $$CR-Cr=KRr$$ Let's subtract KRr from both sides. Let's add Cr to both means meanwhile. $$CR-KRr=Cr$$ Using grouping, we can change this into one "R." $$R(C-Kr)=Cr$$ Now, divide by C-Kr to isolate "R." $$R=\frac{Cr}{C-Kr}$$ This is finished because we have isolated the R.

3) We will have to utilize the same technique in number 2 in order to isolate "f."

 $$F=\frac{fg}{f+g-d}$$ Let's multiply the denominator on both sides. This will get rid of any and all pesky denominators in this equation, $$F(f+g-d)=fg$$ There is a "g" in the trinomial, so we will have to expand that part. $$Ff+Fg-Fd=fg$$ Move all terms with a factor of f on the left hand side. Any other terms are moved to the right hand side. $$Ff-fg=-Fg+Fd$$ Factor out an "f" from both terms from the left hand side. $$f(F-g)=-Fg+Fd$$ Finally, divide by F-g to isolate the wanted variable completey. $$f=\frac{-Fg+Fd}{F-g}$$ No more simplification is possible here. Leave the fraction as is.
TheXSquaredFactor  Feb 14, 2018

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