Answer
\[x=3,\ y=0\]
Work Step by Step
Add the given equations from both RHS and LHS as follows:
\[\begin{align}
& \underline{\begin{align}
& 2x+3y=6 \\
& 2x-3y=6
\end{align}} \\
& 4x\ \ \ \ \ \ \ \ =12 \\
& \ \ \ \ \ \ \ \ \ x=3 \\
\end{align}\]
Put \[x=3\]in\[2x+3y=6\], to get:
\[\begin{align}
& 2\left( 3 \right)+3y=6 \\
& 6+y=6 \\
& y=0
\end{align}\]
Put\[x=3\]and \[y=0\]in any of the given equations to check the solution:
\[\begin{align}
& 2\left( 3 \right)-3\left( 0 \right)=6 \\
& 6-0=6 \\
& 6=6
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
Similarly, we shall check it with 2x + 3y = 6.
2(3) +3(0) = 6
6 = 6
LHS = RHS. So, x = 3 and y = 0 is a solution of this system of equations.
