To solve this, first, expand the terms in parenthesis by multiplying each term with the parenthesis by the term outside the parenthesis being careful to manage the signs correctly:

#color(red)(5)(x - 3) <= color(blue)(-2)(x + 1)#

#(color(red)(5) xx x) - (color(red)(5) xx 3) <= (color(blue)(-2) xx x) + (color(blue)(-2) xx 1)#

#5x - 15 <= -2x + (-2)#

#5x - 15 <= -2x - 2#

Next, add #color(red)(15)# and #color(blue)(2x)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#5x + color(blue)(2x) - 15 + color(red)(15) <= -2x + color(blue)(2x) - 2 + color(red)(15)#

#(5 + color(blue)(2))x - 0 <= 0 + 13#

#7x <= 13#

Now, divide each side of the inequality by #color(red)(7)# to solve for #x# while keeping the inequality balanced:

#(7x)/color(red)(7) <= 13/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) <= 13/7#

#x <= 13/7#

We can write this in interval notation as:

#(-oo, 13/7]#