You need to sketch the graph of the first linear equation `y=(2/3)x-4` , hence, you need to find two points to draw the line. You may find the intersections to x and y axis such that:

`y = 0 => (2/3)x-4 = 0 => (2/3)x = 4 => x = 6 => A(6,0)` (x intercept)

`x = 0 => y = -4 => B(0,-4)` (y intercept)

The line that passes through the points `A(6,0)` and `B(0,-4)` represents the graph of the function ` y` `=(2/3)x-4` .

You need to sketch the graph of the second linear equation `y=-2x+4` , using the same coordinates system, hence, you need to find two points to draw the line. You may find the intersections to x and y axis such that:

`y = 0 => -2x + 4 = 0 => -2x = -4 => x = 2 => C(2,0)`

`x = 0 => y = 4 => D(0,4)`

The line that passes through the pointsÂ `C(2,0)` and `D(0,4)` represents the graph of the function `y=-2x+4` .

You need to notice that the red line intercepts the black line at `x = 3` and `y = -2` , hence, you need to test these values substituting them in equations of the system, such that:

`(2/3)*3 - 4 = -2 => 2 - 4 = -2 => -2 = -2` valid

`-2=-2*3 + 4 => -2 = -6 + 4 => -2 = -2` valid

Thus, the coordinates` x = 3` and` y=-2` check both equations of the system.

**Hence, evaluating the solution to the system of linear equations, under the given conditions, yields `x = 3` and `y = -2` .**