Answer
\[x=4,\ y=1\]
Work Step by Step
Add the given equations from both RHS and LHS as follows:
\[\begin{align}
& \underline{\begin{align}
& 3x+2y=14 \\
& 3x-2y=10
\end{align}} \\
& 6x\ \ \ \ \ \ \ \ =24 \\
& \ \ \ \ \ \ \ \ \ x=4 \\
\end{align}\]
Put \[x=4\]in\[3x+2y=14\], to get:
\[\begin{align}
& 3\left( 4 \right)+2y=14 \\
& 12+2y=14 \\
& 2y=2 \\
& y=1
\end{align}\]
Put\[x=4\]and \[y=1\]in any of the given equations to check the solution:
\[\begin{align}
& 3\left( 4 \right)-2\left( 1 \right)=10 \\
& 12-2=10 \\
& 10=10
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
Now, check with 3x + 2y = 14 too.
3(4) + 2(1) = 14
12 + 2 = 14
14 = 14
Here LHS = RHS.So, this is a solution for this system of equations.
