The equation of a parallel line will have the same x-term, but a different constant (y-intercept). The required value can be found by putting the given point values in to the equation to see what it needs to be.
... y = 5x + ___
... 8 = 5·4 + ___
... 8 - 20 = -12 = ___
Your equation is ...
... y = 5x -12
Subtract y from the fist equation to get an expression for x.
... x + y - y = 6 - y
... x = 6 - y
Assuming your multiplication symbol (×) in the second equation is supposed to be the variable x, performing the substitution gives
... 3(6-y) - y = 6 . . . . . . equation after substitution
... 18 - 4y = 6 . . . . . . . . eliminate parentheses
... 12 - 4y = 0 . . . . . . . . subtract 6
... 3 - y = 0 . . . . . . . . . . divide by 4
... 3 = y . . . . . . . . . . . . .add y
Now, we can use the above expression for x to find x
... x = 6 - y = 6 - 3
... x = 3
The solution of the system of equations is (x, y) = (3, 3).
