#(x+7)/(x-4)-{x-7}/(x+4)<= 0#
#hArr((x+7)(x+4)-(x-7)(x-4))/((x-4)(x+4))<= 0#
or #((x^2+7x+4x+28)-(x^2-7x-4x+28))/((x-4)(x+4))<= 0#
or #(22x)/((x-4)(x+4))<= 0#
or #x/((x-4)(x+4))<= 0#
Let us now use sign chart.
For simplicity let us avoid equality sign temporarily. It is apparent that sign of #x/((x-4)(x+4))# is negative and that sign of #x+4#, #x# and #x-4# will change around the values #-4#, #0# and #4# respectively. In sign chart we divide the real number line below #-4#, between #-4# and #0#, between #0# and #4# and above #4# and see how the sign of #x/((x-4)(x+4))# changes.
Sign Chart
#color(white)(XXXXXXXX)-4color(white)(XXXXX)0color(white)(XXXXX)4#
#xcolor(white)(XXXX)-ive color(white)(XXX)-ive color(white)(XX)+ive color(white)(XXX)+ive#
#(x-4)color(white)(X)-ive color(white)(XXX)-ive color(white)(XX)-ive color(white)(XXX)+ive#
#(x+4)color(white)(X)-ive color(white)(XXX)+ive color(white)(XX)+ive color(white)(XXX)+ive#
#x/((x-4)(x+4))#
#color(white)(XXXXX)-ive color(white)(XXX)+ive color(white)(XX)-ive color(white)(XXX)+ive#
It is observed that #x/((x-4)(x+4)) < 0# when either #x< -4# or #0 < x < 4# , which is the solution for the inequality.
For adding equality observe tat function is not defined at #x=-4# and #x=4#, but #0# is permissible.
Hence, Solution is #x < -4# or #0 <= x < 4#
