(1) 7y = x + 2
(2) 4x - 8y = 12

To solve this system, let's choose a variable from on of the equations to solve for (it doesn't matter which equation or which variable, so let's make it easy). I'll choose to solve for x in eqn 1.

x = 7y - 2

Now substitute for x in eqn 2

4x - 8y = 12

4(7y - 2) - 8y = 12

Notice that now we have an equation that's only in terms of one variable, y. Solve for y.

28y - 8 - 8y = 12
20y = 20
y = 1

Now that we have a value for y, substitute into any equation to find x (I'll use x = 7y - 2 since it was already solved for x)

x = 7y - 2
x = 7(1) - 2 = 5

So the point of intersection of these two lines is (5 , 1)

The equations are, 7y=x+2.........(1) and 4x-8y=12......... (2)

If we multiply each sides of (1) with 4, we get 28 y= 4x+8,

Rearranging, we get 4x-28y  = -8   ......(3)

Subtracting (3) from  (2) we have  -8y-(-28y)= 12-(-8)

Thus, -8y+28y = 12+8

So 20 y = 20 and thus y= 1

Plugging the value of y=1 in (1),

we have 7= x +2

therefore, x=7-2=5