Question

How do you solve the following equation for \[s?\;p = \dfrac{1}{3}r(q + s)\]

Answer 1

Hint:To solve the given equation for value of $s$, when an equation is being solved for the value of a particular variable then the expression is simplified such that the particular variable will shift on the left hand side of the equation and rest of the terms on the right hand side. This shifting can be done with algebraic operations taking BODMAS rule in consideration.

Complete step by step solution:
 In order to solve the given equation \[\;p = \dfrac{1}{3}r(q + s)\] for $s$, we have to simplify the equation with help of BODMAS rule and distributive property of multiplication as follows
\[p = \dfrac{1}{3}r(q + s) \\
\Rightarrow p = \dfrac{1}{3}rq + \dfrac{1}{3}rs \\ \]
Now, as asked in the question, we have to solve the given equation for the value of $s$ so we have to shift the term consist of $s$ to the left hand side and rest all the terms to the right hand side, we can do this with help of algebraic operations such as addition and subtraction, as follows
\[p = \dfrac{1}{3}rq + \dfrac{1}{3}rs\]
Subtracting $p$ from both sides of the equation in order to shift it on the right hand side, we will get
\[p - p = \dfrac{1}{3}rq + \dfrac{1}{3}rs - p \\
\Rightarrow 0 = \dfrac{1}{3}rq + \dfrac{1}{3}rs - p \\ \]
Now, subtracting \[\dfrac{1}{3}rs\] from both sides to shift term consist of $s$ to left hand side,
\[0 - \dfrac{1}{3}rs = \dfrac{1}{3}rq + \dfrac{1}{3}rs - p - \dfrac{1}{3}rs \\
\Rightarrow - \dfrac{1}{3}rs = \dfrac{1}{3}rq - p \\ \]
To find the solution for $s$ dividing both sides with coefficient of $s$
\[\dfrac{{ - \dfrac{1}{3}rs}}{{ - \dfrac{1}{3}r}} = \dfrac{{\dfrac{1}{3}rq - p}}{{ - \dfrac{1}{3}r}} \\
\Rightarrow s = - q + \dfrac{{3p}}{r} \\
\therefore s = \dfrac{{3p}}{r} - q \\ \]
Therefore \[\dfrac{{3p}}{r} - q\] is the required solution.

Note:When shifting the terms to either of the sides according to the need of the hour, take care of their signs (positive or negative) and also when performing algebraic operation to shift terms, always perform the operation to both sides of the equation so that the balance of the equation is maintained.