Answer
\[x=2,\ y=-1\]
Work Step by Step
Add the given equations from both RHS and LHS as follows:
\[\begin{align}
& \underline{\begin{align}
& x+y=1 \\
& x-y=3 \\
\end{align}} \\
& 2x\ \ \ \ =4 \\
& \ \ \ \ \ x=2 \\
\end{align}\]
Put \[x=2\]in\[x+y=1\], to get:
\[\begin{align}
& 2+y=1 \\
& y=-1
\end{align}\]
Put\[x=2\]and \[y=-1\]in any of the given equations to check the solution:
\[\begin{align}
& 2-\left( -1 \right)=3 \\
& 2+1=3 \\
& 3=3
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
For solution of system of equations, we shall check it by putting values of x and y in both the equations. So put x = 2 and y = -1 in x + y = 1 too:
2 -1 = 1
1 = 1
Here LHS = RHS ,
So x = 2.y = -1 is the correct solution of this system of equations.
