Question

Given $\left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)+\left( \dfrac{z}{c} \right)=1$, how do you solve for a?

Answer 1

Hint: We will multiply the term ‘abc’ to both of the sides of the equation in order to get towards the right answer. After this we will cancel common terms and take a as a common variable for finding the value of ‘a’. For the right answer we will take care of the sighs also as the positive sign will become negative when shifted towards the right side of the equal sign.

Complete step-by-step answer:
We will consider the given equation $\left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)+\left( \dfrac{z}{c} \right)=1$ and multiply this whole equation by abc. This will result into the following,
$\begin{align}
  & \left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)+\left( \dfrac{z}{c} \right)=1 \\
 & \Rightarrow \left( abc \right)\left[ \left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)+\left( \dfrac{z}{c} \right) \right]=\left( abc \right)1 \\
 & \Rightarrow \left( \dfrac{abcx}{a} \right)+\left( \dfrac{abcy}{b} \right)+\left( \dfrac{abcz}{c} \right)=abc \\
\end{align}$
After this we are going to cancel common terms from the numerator and denominator both. This will give us the following new equation.
\[\begin{align}
  & \left( \dfrac{abcx}{a} \right)+\left( \dfrac{abcy}{b} \right)+\left( \dfrac{abcz}{c} \right)=abc \\
 & \Rightarrow bcx+acy+abz=abc \\
\end{align}\]
Now, we will take all the terms to the left side of the equal sign. Therefore, we get\[bcx+acy+abz-abc=0\]. After this take the term ‘bcx’ to the right side of the equation. Thus, we get acy + abz – abc = -bcx. At this stage we need to take ‘a’ as a common variable. This results in a(cy + bz – bc = - bcx. It is clear from now on that it is time to divide the whole equation by cy + bz – bc and get \[a=\dfrac{-bcx}{cy+bz-bc}\].
Hence, the desired value of \[a=\dfrac{-bcx}{cy+bz-bc}\].

Note: Tackling such a type of question, it is important to make all the variables to one side of the equation. It is our choice to take them to the left or right side of the equation. After this we will apply the statement of the question and figure out which variable needs to be taken as a common one. In this question, it is ‘a’ which is taken as a common variable because of the fact that we need to find out the value of ‘a’. If instead of ‘a’, the question asked to find the value of ‘b’, then it was ‘b’ which we will take as a common variable. Similarly, the question can ask for the value of ‘c’ as well. We need to focus on signs for such shifting of the terms to either side of the equation. There is also an alternative method to solve this question, which is to take the l.c.m. of left hand side and use cross multiplication like so,
\[\begin{align}
  & \left( \dfrac{x}{a} \right)+\left( \dfrac{y}{b} \right)+\left( \dfrac{z}{c} \right)=1 \\
 & \Rightarrow \dfrac{xbc+yac+zab}{abc}=1 \\
 & \Rightarrow xbc+yac+zab=abc\,\left[ \text{Since, by}\,\text{cross}\,\text{multiplication} \right] \\
\end{align}\]