Question

Can someone please help me I'm lost

Solve for x: −5|x + 1| = 10

x = 0

x = −3 and x = 1

x = −1 and x = 3

No solution

Answer 1

Divide both sides by `-5` and you will see that there is no solution to the equation, because you are going to get the absolute value of something equal to a negative, which is not possible.

Answer 2

So it would be no solution

Answer 3

Yup

Answer 4

Thankyou, is it because it was a negative and a positive?

Answer 5

I don't really get what you mean by `it was a negative and a positive`

Answer 6

Um because when I was checking my solution I got, -10=10 and 10=10

Answer 7

Well...

\(\normalsize\color{ purple}{ −5|x + 1| = 10}\) divide both sides by `-5`
and you get, \(\normalsize\color{ purple}{ |x + 1| = -2}\)

Answer 8

And this is where there is no solution, because

\(\normalsize\color{ purple}{ |\rm{~anything~}| ≥ 0}\)

Answer 9

Makes sense, right ?

Answer 10

Ok I get it now, thank you :)

Answer 11

Anytime :)

Answer 12

Are you available for more questions?

Answer 13

Yes, perhaps.

Answer 14

Well standby, because I'm so confused on this absolute value

Answer 15

The minimum and maximum temperature on a cold day in Lollypop Town can be modeled by the equation below:

2|x − 6| + 14 = 38

What are the minimum and maximum temperatures for this day?

x = −9, x = 21

x = −6, x = 18

x = 6, x = −18

No solution

Answer 16

Im lost to

Answer 17

\(\normalsize\color{blue}{ 2|x − 6| + 14 = 38 }\)
\(\normalsize\color{blue}{ 2|x − 6| + 14\color{red}{ -14 } = 38\color{red}{ -14 } }\)
\(\normalsize\color{blue}{ 2|x − 6| = 24 }\)
\(\normalsize\color{blue}{ 2|x − 6|\color{darkgoldenrod}{ \div2 } = 24 \color{darkgoldenrod}{ \div2 } }\)
\(\normalsize\color{blue}{ |x − 6| = 12 }\)

\(\normalsize\color{blue}{ |x − 6| = 12 ~~~~\LARGE\color{black}{ ⇒ }~~~~}\)\(\normalsize\color{blue}{ x − 6 = 12 ~~~~~~~\rm{or}~~~~~~x-6=-12}\)

Answer 18

So, what is the maximum, and what is the minimum ?

Let me know if you need more help.

Answer 19

well first subtract 14 from both sides
then divide by 2 both sides

Answer 20

So No solution?

Answer 21

According to my post there is a solution

Answer 22

Wouldnt it be C?

Answer 23

Close, but the 6 is not positive, and the 18 in not negative.

Answer 24

Ok so B. Thank you so much

Answer 25

Anytime

Answer 26

Solve for x: |x − 2| + 10 = 12

x = 0 and x = 4

x = −4 and x = 0

x = −20 and x = 4

No solution

Answer 27

I got close to the end, then got confused

Answer 28

show me what you got, please

Answer 29

I chose C. Because I substituted

Answer 30

@solomonzelman

Answer 31

I will show the steps, and you will do the work, okay ?

Answer 32

Ok :)

Answer 33

1) Subtract `10` from both sides
2) when you have \(\normalsize\color{blue}{ │\rm{something}│ =number }\) we get,
\(\normalsize\color{blue}{ \rm{something} =number ~~~~~or~~~~\rm{something} =-~number }\)

Appply this here.

Answer 34

questions ?

Answer 35

Ok, so it wouldn't be a solution, because they wouldnt equal