How do you solve by substitution $2x = 3y\;{\text{and}}\;x = 3y - 3?$

Hint:First solve any one of the given equations for one of the variables present in both the equations. After solving this, you will get an expression, take that expression as the value of the variable and put this value or expression in the second expression then solve it for the second variable and get the value for it. Now re-substitute the value of the second equation in any of the equations and get value for the first one.

Complete step by step solution:
In order to solve the given equations $2x = 3y\;{\text{and}}\;x = 3y -
3$, we will first solve one of the equation for one variable in terms of other as follows
$ \Rightarrow x = 3y - 3 - - - - - (i)$
Here we can see in the second equation, it is already been solved in the terms of $x$
So, now substituting this expressional value of $x$ in the first equation, we will get
$
\Rightarrow 2x = 3y \\
\Rightarrow 2(3y - 3) = 3y \\
$
Opening the parentheses with help of commutative property of multiplication,
\[
\Rightarrow 2 \times 3y - 2 \times 3 = 3y \\
\Rightarrow 6y - 6 = 3y \\
\]
Solving it further, we will get
\[
\Rightarrow 6y - 3y = 6 \\
\Rightarrow 3y = 6 \\
\Rightarrow y = 2 \\
\]
Now substituting this value of $y$ in equation (i) in order to get the value of another variable, that is $x$
$
\Rightarrow x = 3 \times 2 - 3 \\
\Rightarrow x = 6 - 3 \\
\Rightarrow x = 3 \\
$
Therefore $x = 3\;{\text{and}}\;y = 2$ is the required set of solutions for the given equation.


Note: When finding the substitution parameter (that is which variable to select for substitution process) in both equations, select the variable which seems to have less complexity in its expressional value, or the one which is expressed already in terms of other variables in the equation. This tip will make your calculation process easier and help you to solve the question more quickly.