Answer:
solution is
[-3,-1]
Step-by-step explanation:
we are given
[tex]|10x+20|\leq 10[/tex]
Firstly, we will find critical values
so, let's assume it is equal
[tex]|10x+20|= 10[/tex]
now, we can break absolute sign
For [tex]|10x+20|= -(10x+20)[/tex]:
[tex]-(10x+20)= 10[/tex]
we can solve for x
[tex]-10x-20= 10[/tex]
Add both sides by 20
[tex]-10x-20+20= 10+20[/tex]
[tex]-10x= 30[/tex]
Divide both sides by -10
and we get
[tex]x=-3[/tex]
For [tex]|10x+20|= (10x+20)[/tex]:
[tex](10x+20)= 10[/tex]
we can solve for x
[tex]10x+20= 10[/tex]
Subtract both sides by 20
[tex]10x+20-20= 10-20[/tex]
[tex]10x= -10[/tex]
Divide both sides by 10
and we get
[tex]x=-1[/tex]
so, critical values are
[tex]x=-3[/tex]
[tex]x=-1[/tex]
now, we can draw a number line and locate these values
and then we can check inequality on each intervals
For [tex](-\infty,-3)[/tex]:
We can select any random value from this interval and plug that in inequality
and we get
we can plug x=-5
[tex]|10\times -5+20|\leq 10[/tex]
[tex]|-50+20|\leq 10[/tex]
[tex]30\leq 10[/tex]
so, this is FALSE
For [tex][-3,-1][/tex]:
We can select any random value from this interval and plug that in inequality
and we get
we can plug x=-2
[tex]|10\times -2+20|\leq 10[/tex]
[tex]|-20+20|\leq 10[/tex]
[tex]0\leq 10[/tex]
so, this is TRUE
For [tex](-1,\infty)[/tex]:
We can select any random value from this interval and plug that in inequality
and we get
we can plug x=0
[tex]|10\times 0+20|\leq 10[/tex]
[tex]|0+20|\leq 10[/tex]
[tex]20\leq 10[/tex]
so, this is FALSE
so, solution is
[-3,-1]
