Question

How do you solve for \[p\] in \[\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{f}\]?

Answer 1

Hint: Here, we will solve for the variable. We will arrange the terms in an expression by equating the variables which have to be solved on one side of the equation and the rest variables on the other side of the equation. Then by taking LCM and solving the terms of the equation, we will find the variable.

Complete step by step solution:
We are given an equation\[\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{f}\].
Now, we will isolate the term of the variable which has to be solved on one side and the other terms of the different variable on the other side, we get
\[ \Rightarrow \dfrac{1}{p} = \dfrac{1}{f} - \dfrac{1}{q}\]
The L.C.M. of the denominators of \[\left( {f,q} \right)\] on the Left hand side of the equation is \[fq\].
So, by cross multiplying with the denominators to equalize the denominator, we get
\[ \Rightarrow \dfrac{1}{p} = \dfrac{1}{f} \cdot \dfrac{q}{q} - \dfrac{1}{q} \cdot \dfrac{f}{f}\]
So, by multiplying the numerators and the denominators, we get
\[ \Rightarrow \dfrac{1}{p} = \dfrac{q}{{fq}} - \dfrac{f}{{qf}}\]
Since the denominators are equal, thus the terms on the numerator have to be subtracted, then we get
\[ \Rightarrow \dfrac{1}{p} = \dfrac{{q - f}}{{fq}}\]
Now, we will solve for the variable\[p\], so the term on the Right-hand side of the equation gets reciprocal.
Thus, we get
\[ \Rightarrow p = \dfrac{{fq}}{{q - f}}\]

Therefore, the solution for the variable \[p\] is \[\dfrac{{fq}}{{q - f}}\].

Note:
We know that if all the terms in the given expression are variables, then we will isolate the terms which have to be solved on one side and all the other variables on the other side and then solving, we will find the solution of the expression, which is similar to the steps in solving the Linear equation. We should know that Multiplication is always commutative i.e., \[ab = ba\]. We should always remember that whenever Reciprocal is done on one side, then it has to be done on the other side simultaneously.