3x+y=1 6x+6y=18

Solve the following system:
{3 x+y = 1 |     (equation 1)
6 x+6 y = 18 |     (equation 2)
Swap equation 1 with equation 2:
{6 x+6 y = 18 |     (equation 1)
3 x+y = 1 |     (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{6 x+6 y = 18 |     (equation 1)
0 x-2 y = -8 |     (equation 2)
Divide equation 1 by 6:
{x+y = 3 |     (equation 1)
0 x-2 y = -8 |     (equation 2)
Divide equation 2 by -2:
{x+y = 3 |     (equation 1)
0 x+y = 4 |     (equation 2)
Subtract equation 2 from equation 1:
{x+0 y = -1 |     (equation 1)
0 x+y = 4 |     (equation 2)
Collect results:
Answer: | x = -1     and       y=4
 

3x+y=1     rearrange as    y = 1 - 3x     (1)

6x+6y=18     divide through by 6 →   x + y = 3    (2)

Put (1)  into (2)

x + [ 1 - 3x ]  = 3     simplify

-2x + 1 =  3     subtract 1 from both sides

-2x  = 2       divide by -1

x = -1

And using (1)    y = 1 - 3(-1)   =  1 + 3    = 4

So  (x , y)  = ( -1, 4)