+619 Given that the graphs of $y=h(x)$ and $y=j(x)$ intersect at $(2,2),$ $(4,6),$ $(6,12),$ and $(8,12),$ there is one point where the graphs of $y=h(2x)$ and $y=2j(x)$ must intersect. What is the sum of the coordinates of that point?
+9479 The given intersection points lie on y = h(x) and y = j(x) .
For example...since (8, 12) is an intersection point.... 12 = h(8) and 12 = j(8)
When x = 4 , h(2x) = h(2(4)) = h(8) = 12 = y
When x = 4 , 2j(x) = 2j(4) = 2(6) = 12 = y
So... (4, 12) is a point on y = h(2x) and on y = 2j(x) .
4 + 12 = 16
Sorry that this isn't a very good explanation.....If it doesn't make sense, it might help to try different numbers for x , such as 1, 2, or 6...to see why they don't work. 
+9479 The given intersection points lie on y = h(x) and y = j(x) .
For example...since (8, 12) is an intersection point.... 12 = h(8) and 12 = j(8)
When x = 4 , h(2x) = h(2(4)) = h(8) = 12 = y
When x = 4 , 2j(x) = 2j(4) = 2(6) = 12 = y
So... (4, 12) is a point on y = h(2x) and on y = 2j(x) .
4 + 12 = 16
Sorry that this isn't a very good explanation.....If it doesn't make sense, it might help to try different numbers for x , such as 1, 2, or 6...to see why they don't work.
