How do you solve and write the following in interval notation: # -4<=( -2x+5)/3<=3#?
1 Answer
Explanation:
Your goal here is to isolate
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
#(-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
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]#