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]#
