+2446 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. |
+2446 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. |
