Question

How do you solve for \[r\] in \[S=L(1-r)\]?

Answer 1

Hint: To solve the given equation we will need the expansion of the expression \[a(b-c)\]. The expression \[a(b-c)\] is simplified by expanding the bracket, the term outside the bracket is multiplied separately to the terms inside the bracket. After this, the product of the second term inside the bracket and term present outside is subtracted from the product of the first term inside the bracket and the term present outside. It is done as follows,
\[\Rightarrow a(b-c)=ab-ac\]

Complete step by step answer:
We are given the equation \[S=L(1-r)\], and we have to solve the equation for \[r\], which means that we have to find the \[r\] in terms of other terms present in the equation. To do this, we have to take all the terms in the equation except \[r\] to one side of the equation, leaving only \[r\] to the other side. We will do this in the given equation as follows,
\[S=L(1-r)\]
The RHS is of the form \[a(b-c)\], we know that it can be simplified as \[ab-ac\]. Using this in the RHS of the above equation we get,
\[\Rightarrow S=L-Lr\]
Adding Lr to both sides of the above equation, we get
\[\Rightarrow S+Lr=L-Lr+Lr\]
Subtracting S from both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow S+Lr-S=L-S \\
 & \Rightarrow Lr=L-S \\
\end{align}\]
Dividing both sides of the above equation by L, we get
\[\Rightarrow \left( Lr \right)\dfrac{1}{L}=\left( L-S \right)\dfrac{1}{L}\]
Using the expansion of \[a(b-c)\] in the RHS of the above equation we get,
\[\begin{align}
  & \Rightarrow r=\left( L \right)\dfrac{1}{L}-\left( S \right)\dfrac{1}{L} \\
 & \Rightarrow r=1-\dfrac{S}{L} \\
\end{align}\]
Hence, by solving the equation for \[r\], we get \[r=1-\dfrac{S}{L}\].

Note:
These types of problems are not too difficult and can be solved by remembering the expansions of the simple expressions like \[a(b-c),a(b+c)\] etc. One should keep attention to sign conventions and should avoid making mistakes.
The above question can also be solved as follows,
\[S=L\left( 1-r \right)\]
Dividing both sides of above equation by \[L\], we get
\[\begin{align}
  & \Rightarrow \dfrac{S}{L}=\dfrac{L\left( 1-r \right)}{L} \\
 & \Rightarrow \dfrac{S}{L}=1-r \\
\end{align}\]
Adding \[r\] to both sides of the above equation we get,
\[\Rightarrow \dfrac{S}{L}+r=1-r+r\]
Subtracting \[\dfrac{S}{L}\] from both sides of the above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{S}{L}+r-\dfrac{S}{L}=1-\dfrac{S}{L} \\
 & \therefore r=1-\dfrac{S}{L} \\
\end{align}\]
We get the same answer from both methods.