How do you write the equation $y = - \dfrac{3}{2}x + \dfrac{4}{3}$in standard form?

Hint: We need to convert into the form of standard. The standard form is $Ax + By = C$. We find the coefficient of $x$ and $y$ in the given equation.
Use L CM, clear the fraction if any.
Simplify both sides of the equation
Isolate the variable and constant.
Verify your answer.

Complete step by step answer:
The given equation is, $y = - \dfrac{3}{2}x + \dfrac{4}{3}$
The standard form of a linear equation is:
$Ax + By = C$
Where, if at all possible, $A,B$ and $C$ are integers, and $A$ is non-negative, and, $A,B$ and $C$ have no common factors other than $1$.
First, we will multiply each side of the equation by$6$to eliminate the fractions because by the definition above all the coefficients and the constant must be integers:
\[ \Rightarrow 6 \times y = 6\left( { - \dfrac{3}{2}x + \dfrac{4}{3}} \right)\]
Now, RHS (Right Hand Side) multiply by$6$each term
$ \Rightarrow 6y = \left( {6 \times - \dfrac{3}{2}x} \right) + \left( {6 \times \dfrac{4}{3}} \right)$
Now divide first term $6$ by $2$ and second term$6$ by $3$, hence we get
\[ \Rightarrow 6y = \left( {3 \times ( - 3x} \right) + \left( {2 \times (4)} \right)\]
Now multiply first term $3$ by$ - 3x$ and second term $2$ by $4$
$ \Rightarrow 6y = - 9x + 8$
Now, we will add $9x$ to each side of the equation to put this equation into the standard form:
$ \Rightarrow 9x + 6y = 9x - 9x + 8$
Subtract in RHS (Right Hand Side) the first term
$ \Rightarrow 9x + 6y = 0 + 8$
The zero is vanish
$ \Rightarrow 9x + 6y = 8$

Hence the standard form is $9x + 6y = 8$

Note: The standard form for a linear equation in two variables, $x$ and $y$, is usually given as $Ax + By = C$. Where, if at all possible, $A,B$ and $C$ are integers, and $A$ is non-negative, and $A,B$ and $C$ have no common factor other than $1$. If we have a linear equation in slope-intercept form $y = mx + c$. We can change that equation into standard form. To do this we need to express the slope and the ordinate of the $y$-intercept in rational number form.