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enyata [817]
3 years ago
8

Find the solution set of this inequality|10x+20| ≤10

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
7 0

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

|10x+20|\leq 10

Firstly, we will find critical values

so, let's assume it is equal

|10x+20|= 10

now, we can break absolute sign

For |10x+20|= -(10x+20):

-(10x+20)= 10

we can solve for x

-10x-20= 10

Add both sides by 20

-10x-20+20= 10+20

-10x= 30

Divide both sides by -10

and we get

x=-3

For |10x+20|= (10x+20):

(10x+20)= 10

we can solve for x

10x+20= 10

Subtract both sides by 20

10x+20-20= 10-20

10x= -10

Divide both sides by 10

and we get

x=-1

so, critical values are

x=-3

x=-1

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For (-\infty,-3):

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

|10\times -5+20|\leq 10

|-50+20|\leq 10

30\leq 10

so, this is FALSE

For [-3,-1]:

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

|10\times -2+20|\leq 10

|-20+20|\leq 10

0\leq 10

so, this is TRUE

For (-1,\infty):

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

|10\times 0+20|\leq 10

|0+20|\leq 10

20\leq 10

so, this is FALSE

so, solution is

[-3,-1]

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