Question

How do you solve $\dfrac{1}{3}\left( n+1 \right)=\dfrac{1}{6}\left( 3n-5 \right)$?

Answer 1

Hint: In this problem we need to solve the given equation i.e., we need to calculate the value of the variable for which the given equation is satisfied. We can observe that the numbers $3$, $6$ are in the denominators. So, we will first multiply with the $6$ on both sides of the given equation. After multiplication we will simplify the equation by cancelling the possible terms. Now we will apply the distribution law of multiplication over addition and over subtraction and simplify the equation. Now we will shift the terms in the equation so that all the variables are at one side and all the constants are one side. Now we will apply arithmetic operations to solve the given equation.

Complete step by step answer:
Given equation, $\dfrac{1}{3}\left( n+1 \right)=\dfrac{1}{6}\left( 3n-5 \right)$
We can observe that the numbers $3$, $6$ are in the denominators. So, multiplying $6$ on both sides of the above equation, then we will get
$\Rightarrow 6\times \dfrac{1}{3}\left( n+1 \right)=6\times \dfrac{1}{6}\left( 3n-5 \right)$
Simplifying the above equation by cancelling the all the possible terms, then we will get
$\Rightarrow 2\left( n+1 \right)=\left( 3n-5 \right)$
The distribution law of multiplication over addition says that $a\left( b+c \right)=ab+ac$. Applying this formula in the above equation, then we will have
$\begin{align}
  & \Rightarrow 2n+2\times 1=3n-5 \\
 & \Rightarrow 2n+2=3n-5 \\
\end{align}$
Rearranging the term in the above equation such that all the variables are one side and all the constants are at one side.
$\begin{align}
  & \Rightarrow 3n-2n=5+2 \\
 & \Rightarrow n=7 \\
\end{align}$
Hence the solution for the given equation $\dfrac{1}{3}\left( n+1 \right)=\dfrac{1}{6}\left( 3n-5 \right)$ is $n=7$.

Note:
We can also check whether the obtained solution is correct or wrong by substituting the calculated $n$ value in the given equation, then we will have
$\begin{align}
  & \dfrac{1}{3}\left( n+1 \right)=\dfrac{1}{6}\left( 3n-5 \right) \\
 & \Rightarrow \dfrac{1}{3}\left( 7+1 \right)=\dfrac{1}{6}\left( 3\times 7-5 \right) \\
 & \Rightarrow \dfrac{8}{3}=\dfrac{16}{6} \\
 & \Rightarrow \dfrac{8}{3}=\dfrac{8\times 2}{3\times 2} \\
 & \Rightarrow \dfrac{8}{3}=\dfrac{8}{3} \\
\end{align}$
Hence our solution is correct.