Question

How do you solve this $\dfrac{{2x}}{3} - 9 = 0$ ?

Answer 1

Hint: This problem deals with solving the linear equation with one variable. A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form of $ax + b = 0$, and are usually solved for the variable $x$ using basic algebraic operations.

Complete step-by-step answer:
Given a linear equation one variable which is considered as given below:
$ \Rightarrow \dfrac{{2x}}{3} - 9 = 0$
Now rearrange the terms such that all the constants are on one side of an equation and all the $x$terms are on the other side of the equation.
$ \Rightarrow \dfrac{{2x}}{3} = 9$
Now multiply the above equation with 3, on both sides of the equation, as shown below:
$ \Rightarrow 2x = 9\left( 3 \right)$
Now simplifying the above equation, that is simplifying the product of two constants which are on the right side of the above equation, as shown below:
$ \Rightarrow 2x = 27$
Now divide the above equation by 2, so as to remove the coefficient of the $x$ term on the left hand side of the equation, as shown below:
$ \Rightarrow x = \dfrac{{27}}{2}$
So the solution of $x$ of the given linear equation is $\dfrac{{2x}}{3} - 9 = 0$ is:
$\therefore x = \dfrac{{27}}{2}$

Final answer: The solution of the given expression $\dfrac{{2x}}{3} - 9 = 0$ is equal to $\dfrac{{27}}{2}$.

Note:
Please note that the linear equations in one variable which are expressed in the form of $ax + b = 0$, have only one solution. Where a and b are two integers, and x is a variable. This means that there will be no terms involving higher powers of x, not even the power of 2, which is ${x^2}$.