First, multiply each side of the inequality by #color(blue)(-2)# to eliminate the fraction while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operator:
#color(blue)(-2)(6x - 5) color(red)(>) color(blue)(-2) xx (10 - x)/-2#
#(color(blue)(-2) xx 6x) - (color(blue)(-2) xx 5) color(red)(>) cancel(color(blue)(-2)) xx (10 - x)/color(blue)(cancel(color(black)(-2)))#
#-12x - (-10) color(red)(>) 10 - x#
#-12x + 10 color(red)(>) 10 - x#
Next, add #color(red)(12x)# and subtract #color(blue)(10)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:
#-12x + 10 + color(red)(12x) - color(blue)(10) > 10 - x + color(red)(12x) - color(blue)(10)#
#-12x + color(red)(12x) + 10 - color(blue)(10) > 10 - color(blue)(10) - x + color(red)(12x)#
#0 + 0 > 0 - 1x + color(red)(12x)#
#0 > (-1 + color(red)(12))x#
#0 > 11x#
Now, divide each side of the inequality by #color(red)(11)# to solve for #x# while keeping the inequality balanced:
#0/color(red)(11) > (11x)/color(red)(11)#
#0 > (color(red)(cancel(color(black)(11)))x)/cancel(color(red)(11))#
#0 > x#
To state the solution in terms of #x# we can reverse or "flip" the entire inequality:
#x < 0#
