Solve the following inequality using both the graphical and algebraic approach: StartFraction x Over 2 EndFraction + 1 less-than-or-equal-to 3 x + 2 Graph A On a coordinate plane, a line goes through (0, 0) and (1, 3). Another line goes through (0, 0) and (2, 1). The lines intersect at (0, 0). Graph B On a coordinate plane, a line goes through (negative 2, 0) and (0, 1). Another line goes through (negative 1, negative 1) and (0, 2). The lines intersect around (negative 0.5, 0.9). a. x greater-than-or-equal-to negative 0.4 Graph A b. x less-than-or-equal-to negative 0.4 Graph A c. x greater-than-or-equal-to negative 0.4 Graph B d. x less-than-or-equal-to negative 0.4 Graph B

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'.

answer: x=60

is there more to this question?

the answer on ed. is:

c. "the graph of f(x) has exactly two x-intercepts."

hope i could someone! : )