2x+y=6 x-y=2

To solve the system of equations:

1. \(2x + y = 6\) (Equation 1)
2. \(x - y = 2\) (Equation 2)

We can use the substitution or elimination method. Here, I'll use the elimination method.

First, let's solve Equation 2 for \(y\):

\[ x - y = 2 \implies y = x - 2 \]

Now, substitute \(y\) in Equation 1:

\[ 2x + (x - 2) = 6 \]

Combine like terms:

\[ 2x + x - 2 = 6 \]

\[ 3x - 2 = 6 \]

Now, add 2 to both sides:

\[ 3x = 8 \]

Divide by 3:

\[ x = \frac{8}{3} \]

Now that we have \(x\), we can find \(y\) using the expression we derived for \(y\):

\[ y = x - 2 = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \]

So, the solution to the system of equations is:

\[ x = \frac{8}{3}, \quad y = \frac{2}{3} \]

In ordered pair notation, the solution is \(\left(\frac{8}{3}, \frac{2}{3}\right)\).