Two boundary lines of a feasible region are 2x + y = 14 and x + 3y = 17…

Step 1. Eliminate $y$. Multiply the first equation by $3$: $6x + 3y = 42$. Subtract $x + 3y = 17$ to get $5x = 25$, so $x = 5$.

Step 2. Find $y$. $2(5) + y = 14$, so $y = 4$.

Step 3. Check. $2(5) + 4 = 14$ and $5 + 3(4) = 17$. Both are true.