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

##### 1 Answer

#### Answer:

#### Explanation:

Your goal here is to isolate *compound inequality*.

Start by multiplying all sides by

#-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

#-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 **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 areflipped!

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 * and* smaller than or equal to

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]#