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.
