Solve the following linear equation:
\[\dfrac{2(x+1)}{3}=\dfrac{3(x-2)}{5}\]

Hint: The linear equations in one variable is expressed in the form of \[ax+b\], where \[a\] and \[b\] are two integers, and \[x\] is a variable and has only one solution. The linear Equation in one variable is solved by separating variable and constant through equals to sign. Here we will first cross multiplication to simplify the expression.

Complete step by step answer:
The linear Equation in one variable is solved by separating variable and constant through equals to sign. Here we will first cross multiplication to simplify the expression.
\[\begin{align}
  & \Rightarrow \dfrac{2(x+1)}{3}=\dfrac{3(x-2)}{5} \\
 & \Rightarrow 10(x+1)=9(x-2) \\
 & \Rightarrow 10x+10=9x-18 \\
 & \Rightarrow 10x-9x=-18-10 \\
\end{align}\]
After separating constants and variables, we have to divide the coefficient of x by the constant term which is at the right hand side of equals sign.
\[\begin{align}
  & \Rightarrow 10x-9x=-18-10 \\
 & \Rightarrow x=\dfrac{-28}{1} \\
 & \therefore x=-28 \\
\end{align}\]
By solving a given linear equation, we got \[x=-28\] as a solution.

Note:
In these types of questions when we are asked to find the value of a variable and any equation is given when the equation is first sequenced, we take all the variable terms on one side of the equation and the fixed terms on the other side of the equation number. We then divide the fixed word by the coefficient \[x\] to get the required value of \[x\]. But if the given equation is not in line, we will have to use a formula to find a different value of \[x\]. As the equation given is quadratic and is \[a{{x}^{2}}+bx+c\] depending on the formula the different value of \[x\] will be \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] .
Therefore, this will be an effective way to obtain the value of \[x\] from a given equation.