Graph the solution set for the absolute value of 2x+5 less than 15 Solution We need to find the solution set for |2x+5| < 15. We divide the real line into two parts... Part1 x >= -5/2 In this interval 2x+5 >=0, therefore|2x+5|=2x+5 So for2x+5<15 x<5 So the solution to the inequality in this interval is [-5/2,infinity) intersection (-infinity,5) So [-5/2,5) Part2 x<-5/2 In this interval2x+5 < 0, therefore|2x+5|=-(2x+5) So for|2x+5| < 15 or, -2x-5 < 15 or, x > -10 So the solution to the inequality is the interval (-10,infinity) intersection (-infinity,-5/2) So the requited interval is (-10,-5/2). So from part 1 and part 2 we get x belongs to(-10,-5/2) U[-5/2,5) or x belongs to (-10,5). .

1. Graph the solution set for the absolute value of 2x+5 less than 15
Solution
We need to find the solution set for |2x+5| < 15.
We divide the real line into two parts...
Part1
x >= -5/2
In this interval 2x+5 >=0, therefore|2x+5|=2x+5
So for2x+5<15
x<5
So the solution to the inequality in this interval is [-5/2,infinity) intersection (-infinity,5)
So [-5/2,5)
Part2
x<-5/2
In this interval2x+5 < 0, therefore|2x+5|=-(2x+5)
So for|2x+5| < 15
or, -2x-5 < 15
or, x > -10

2. So the solution to the inequality is the interval (-10,infinity) intersection (-infinity,-5/2)
So the requited interval is (-10,-5/2).
So from part 1 and part 2
we get x belongs to(-10,-5/2) U[-5/2,5)
or x belongs to (-10,5).