How do you solve and write the following in interval notation: # -4<=( -2x+5)/3<=3#?

1 Answer
Aug 1, 2016

Answer:

#x in [-2, 17/2]#

Explanation:

Your goal here is to isolate #x# in the middle of this compound inequality.

Start by multiplying all sides by #3#

#-4 * 3 <= (-2x + 5)/color(red)(cancel(color(black)(3))) * color(red)(cancel(color(black)(3))) <= 3 * 3#

#color(white)(a)-12 <= color(white)(a)-2x + 5 color(white)(aa)<= 9#

Next, subtract #5# from all sides

#-12 - 5 <= -2x + color(red)(cancel(color(black)(5))) - color(red)(cancel(color(black)(5))) <= 9 - 5#

#color(white)(aaa)-17 <= color(white)(aaa) -2x color(white)(aaaa) <= 4#

Finally, divide all sides by #-2#. Do not forget that when you're multiplying or dividing an inequality by a negative number, the sign of the inequality is flipped!

#(-17)/(-2) color(blue)(>=) color(white)(a)xcolor(white)(a) color(blue)(>=) 4/(-2) -># the signs are flipped !

will get you

#color(white)(-)17/2 >= color(white)(a)xcolor(white)(a) >= -2#

This tells you that in order to be a solution to this compound inequality, a value of #x# must be greater than or equal to #-2# and smaller than or equal to #17/2#.

In interval notation, this is expressed like this

#color(green)(|bar(ul(color(white)(a/a)color(black)(x in [-2, 17/2])color(white)(a/a)|)))#

You can find this solution interval by breaking up the compound inequality into two simple inequalities

#x <= 17/2" "# and #" "x >= -2#

In interval notation, these solution intervals will get you

#x in (-oo, 17/2] nn [-2, + oo) implies x in [-2, 17/2]#