Question

How do you solve for $n$ in the equation $4k+mn=n-3$?

Answer 1

Hint: We have to solve for $n$ in the given equation, which means that $n$ is the variable and rest all are the constants. So we have to separate all the constant and the variable terms to get a simplified equation. On solving the equation with the help of the algebraic manipulations, we will get the required solution of the given equation.

Complete step by step answer:
The equation to be solved is given in the question as
$4k+mn=n-3..........(i)$
As stated in the question, we have to solve the above equation for $n$. So the variable in the above equation is $n$. The rest of the terms, that is $k$ and $m$, are both constants. So for solving the equation, we have to separate the variable and the constant terms. For this, we first add $3$ on both sides of the equation (i) to get
$\begin{align}
  & \Rightarrow 4k+mn+3=n-3+3 \\
 & \Rightarrow 4k+mn+3=n \\
\end{align}$
Now, we subtract $n$ from both sides of the above equation to get
$\begin{align}
  & \Rightarrow 4k+mn+3-n=n-n \\
 & \Rightarrow 4k+mn+3-n=0 \\
\end{align}$
Writing the constant terms and the variable terms separately, we get
$\Rightarrow mn-n+4k+3=0$
Subtracting $\left( 4k+3 \right)$ from both sides of the above equation, we get
$\begin{align}
  & \Rightarrow mn-n+4k+3-\left( 4k+3 \right)=0-\left( 4k+3 \right) \\
 & \Rightarrow mn-n=-\left( 4k+3 \right) \\
\end{align}$
Taking $n$ common on the left hand side of the above equation, we get
$\Rightarrow n\left( m-1 \right)=-\left( 4k+3 \right)$
Finally, dividing by $\left( m-1 \right)$ on both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow \dfrac{n\left( m-1 \right)}{\left( m-1 \right)}=-\dfrac{\left( 4k+3 \right)}{\left( m-1 \right)} \\
 & \Rightarrow n=-\dfrac{\left( 4k+3 \right)}{\left( m-1 \right)} \\
\end{align}\]

Hence, the solution of the given equation is \[n=-\dfrac{\left( 4k+3 \right)}{\left( m-1 \right)}\].

Note: From the value of the solution of the given equation obtained, we can impose a condition for the existence of the solution. The solution of the given equation comes out to be \[n=-\dfrac{\left( 4k+3 \right)}{\left( m-1 \right)}\]. As we can see that the denominator of the solution is equal to $\left( m-1 \right)$. Since the denominator must be non zero, we have \[m-1\ne 0\] which in turn means that \[m\ne 1\]. Hence, we can say that the solution of the given equation will exist only if \[m\ne 1\].