1) 6(d-2) - 2(d+2) = 2) (e-3) - 4 (2e +1) =
3) (f -1) - (f-1) + 1 =
4) 2(7-9g) - 3(4-2g) =
5) 5a(6 + 2b) + 6ab - 7 =
6) 2x(3y - 1) + 2x(3y + 1) =
7) 6p(q+9) - 3p(2p-1) =

hope it helps

plz mark as brainliest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.