#### Answer

Both methods are easier, so, it depends on the question.

#### Work Step by Step

Let us consider the following example of the substitution method to see which method is easier:
$y=3-3x$ ( $1$ )
$3x+4y=6$ ( $2$ )
Put the value of y from the first equation in the second equation as follows:
$\begin{align}
& 3x+4y=6 \\
& 3x+4\cdot \left( 3-3x \right)=6 \\
& 3x+12-12x=6
\end{align}$
And subtract like terms as follows:
$12-9x=6$
Add $-12$ to both sides as follows:
$\begin{align}
& 12-9x-12=6-12 \\
& -9x=-6 \\
& x=\frac{2}{3}
\end{align}$
Then, put the value of x in the first equation as follows:
$\begin{align}
& y=3-3x \\
& y=3-3\cdot \frac{2}{3} \\
& y=1 \\
\end{align}$
Now, consider the following example of the addition method to compare between the addition method and the substitution method:
$3x+5y=-2$ ( $1$ )
$2x+3y=0$ ( $2$ )
And multiply by $-2$ on both sides of the first equation as follows:
$\begin{align}
& 3x+5y=-2 \\
& -2\left( 3x+5y \right)=-2\cdot -2
\end{align}$
$-6x-10y=4$ (a)
Also, multiply by $3$ on both sides of the second equation as follows:
$\begin{align}
& 2x+3y=0 \\
& 3\left( 2x+3y \right)=0\cdot 3
\end{align}$
$6x+9y=0$ (b)
Then by the addition method just add (a) and (b) as follows:
$\begin{align}
& -6x-10y+6x+9y=4+0 \\
& -y=4 \\
& y=-4
\end{align}$
Put the value of $y$ in (b) as follows:
$\begin{align}
& 6x+9\cdot -4=0 \\
& 6x-36=0
\end{align}$
Add to both sides $36$ as follows:
$\begin{align}
& 6x-36+36=0+36 \\
& 6x=36
\end{align}$
Multiply both sides by $\frac{1}{6}$ as follows:
$\begin{align}
& \frac{1}{6}\cdot 6x=\frac{1}{6}\cdot 36 \\
& x=6 \\
\end{align}$
We see that it depends on the question which method is easier to follow; the steps required in both methods are the same.