Question

Solve the linear equation $5x-\dfrac{1}{3}\left( x+1 \right)=6\left( x+\dfrac{1}{30} \right)$?

Answer 1

Hint: For this problem we need to solve the given equation that means we need to calculate the value of $x$ for which the given equation is satisfied. For this we will expand each term which are in the brackets by applying the distribution law of multiplication. After that we will rearrange the terms in the obtained equation so that all the variables are at one side and all the constants are at one side. Now we will perform the arithmetic operations based on the sign that the variables have and simplify the equation to get the required result.

Complete step-by-step solution:

Given that, $5x-\dfrac{1}{3}\left( x+1 \right)=6\left( x+\dfrac{1}{30} \right)$.
Distribution law of multiplication gives the value of $a\left( b+c \right)$ as $ab+ac$. Applying this law in the above equation and expanding the terms in the brackets, then we will get
$\Rightarrow 5x-\dfrac{1}{3}\times x-\dfrac{1}{3}\times 1=6\times x+6\times \dfrac{1}{30}$
Simplifying the above equation, then we will have
$\Rightarrow 5x-\dfrac{x}{3}-\dfrac{1}{3}=6x+\dfrac{1}{5}$
Rearranging the terms in the above equation such that all the variables are at one side and all the constants are at one side, then we will get
$\Rightarrow 6x-5x+\dfrac{x}{3}=-\dfrac{1}{3}-\dfrac{1}{5}$
Simplifying the above equation by taking LCM from denominators, then we will get
$\begin{align}
  & \Rightarrow x+\dfrac{x}{3}=-\dfrac{1}{3}-\dfrac{1}{5} \\
 & \Rightarrow \dfrac{3x+x}{3}=\dfrac{-5-3}{3\times 5} \\
 & \Rightarrow \dfrac{4x}{3}=\dfrac{-8}{3\times 5} \\
\end{align}$
Multiplying the above equation with $\dfrac{3}{4}$ on both sides, then we will get
$\Rightarrow \dfrac{4x}{3}\times \dfrac{3}{4}=\dfrac{-8}{3\times 5}\times \dfrac{3}{4}$
Cancelling the all the possible terms in the above equation, then we will have
$\Rightarrow x=-\dfrac{2}{5}$
Hence the solution of the given equation $5x-\dfrac{1}{3}\left( x+1 \right)=6\left( x+\dfrac{1}{30} \right)$ is $x=-\dfrac{2}{5}$.

Note: In this problem they have only mentioned to solve the given equation, so we have calculated the value of $x$ only. In some cases, they may also ask to justify your solution, then we need to substitute the calculated value of $x$ and check whether the calculated solution is satisfying the given equation or not.