Solve for X. -8x + 3 ≥ 27 AND −13x + 5 ≥ 57
Choose ONE and ONLY the CORRECT answer.
A: x ≤ −4
B: x ≤ −3
C: −4 ≤ x ≤ −3
D: There are NO solutions.
E: All values of X are solutions.

A: x ≤ −4

Step-by-step explanation:

Hi there!

We're given the two inequalities -8x + 3 ≥ 27 and  −13x + 5 ≥ 57

we need to find the set of solutions that make the two inequalities true (the intersection)

First, let's solve the two inequalities, starting with -8x + 3 ≥ 27

-8x + 3 ≥ 27

subtract 3 from both sides

-8x≥24

divide both sides by -8 and remember to FLIP the inequality sign, as we're dividing with a negative

x ≤ -3

now for the other inequality:

−13x + 5 ≥ 57

subtract 5 from both sides

-13x ≥ 52

divide both sides by -13 and remember to FLIP the sign

x ≤ -4

please see below for the graph to find the intersection, as well as the final answer

Hope this helps! :)

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'.

what’s the question?