First, divide each segment of the system of inequalities by #color(blue)(-2)# to eliminate the need for parenthesis while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:
#10/color(blue)(-2) color(red)(>=) (-2(x - 2))/color(blue)(-2) color(red)(>=) 20/color(blue)(-2)#
#-5 color(red)(>=) (color(blue)(cancel(color(black)(-2)))(x - 2))/cancel(color(blue)(-2)) color(red)(>=) -10#
#-5 color(red)(>=) x - 2 color(red)(>=) -10#
Now, add #color(red)(2)# to each segment to solve for #x# while keeping the system balanced:
#-5 + color(red)(2) >= x - 2 + color(red)(2) >= -10 + color(red)(2)#
#-3 >= x - 0 >= -8#
#-3 >= x >= -8#
Or
#x <= -3# and #x >= -8#
Or
#x >= -8# and #x <= -3#
Or, in interval notation:
#[-8, -3]#
