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X + y = 5           and                  y = 3x - 1
 x - y = -3                          2x + y = 14

Solve using substitution.

First, isolate one variable.

x + y = 5
      x = 5 - y

Then, plug its expression into the other equation and solve.

     x - y = -3
5 - y - y = -3
   5 - 2y = -3
       -2y = -8
          y = 4

Next, plug your new value into one of the original equations to solve for the second variable.

x + y = 5
x + 4 = 5
      x = 1

Finally, check work with substitution.

x + y = 5   =>   1 + 4 = 5   <--True
x - y = -3   =>   1 - 4 = -3   <--True

Next.

        y = 3x - 1
2x + y = 14

         2x + y = 14
2x + (3x - 1) = 14
   2x + 3x - 1 = 14
           5x - 1 = 14
                5x = 15
                  x = 3

y = 3x - 1
y = 3(3) - 1
y = 9 -1
y = 8

y = 3x - 1   =>   8 = 3(3) - 1   =>   8 = 9 - 1   <--True
2x + y = 14   =>   2(3) + 8 = 14   =>   6 + 8 = 14   <--True

4.
x = 1
y = 4

5.
x = 3
y = 8