Question

absolute value problem please help!

Answer 1

\[3\left| z+7 \right|= 12 \]

Answer 2

@Hero @jim_thompson5910 @myininaya @zzr0ck3r

Answer 3

@zepdrix @dan815

Answer 4

|z+7| = 4

Answer 5

Whenever you come across |A| = B it implies A = B and A = -B.

Solve for z in each case.

Answer 6

In absolute value problems, you can change it into two separate equations. Imagine that there's actually a pair of brackets with a +- in front of them, instead of an absolute value sign. That means you'll have
(z+7) = 4 and -(z+7) = 4
and you just solve these separately to get your answers. Always put the answers back into the original absolute value just to check if they're correct!

Answer 7

case 1
z = -3

case 2
z = -11

Answer 8

so when you solve absolute value yous should have two parts one is negative and one is positive so in your question its should be like this z + 7 = 4 and z+7 = -4

Answer 9

Yes.

Answer 10

yeah right :)

Answer 11

what if you have |x+7|=-4
:p

Answer 12

then what do i do next?

Answer 13

you are done

Answer 14

dont i need to check it or something?

Answer 15

whats extraneous solution?

Answer 16

you can always check your solutions to be sure
i actually suggest it for a beginning algebra student
solutions that do not work qualify as extraneous

Answer 17

so how do i know if my solutions were good or extraneous?

Answer 18

plug them back into you initial equation which in this case was 3|x+7|=12

Answer 19

when i plug in case 1 it works 12=12
but when i plug in case 2 i get -12

Answer 20

so is it extraneous?

Answer 21

in case 2 how did you get -12?

Answer 22

3|-3+7|
3|4|
3(4)
12
---------------------
3|-11+7|
3|-4|
3(4) *when you take the absolute value of some number it will result in 0 or something positive; it will never be negative
12

Answer 23

oh okay i see where i messed up

Answer 24

thank you so much!! :)

Answer 25

so none of the solutions are "extra"

Answer 26

yes got it :)

Answer 27

would you know how to solve |x+7|=-4?
that is a fun one
people always try to solve this one as if it as any solutions when it doesn't

Answer 28

Solving for u:
|u|=a where a is positive will always result in two solutions
|u|=a where a is negative will always result in no solutions
|u|=0 will always result in the solution that is u=0

Answer 29

If the problem had been 3|z + 7| = -12 and you try to solve it you will find all solutions are extraneous because there is NO solution to this problem as the LHS is always positive due to the absolute values and will never equal to a negative number. So when you put the solution back into the original equation to verify you will notice the solutions are not really solutions. So they are extraneous.

Answer 30

case 1 x=-11
case 2 x= 3

@myininaya

Answer 31

and all extraneous means is they are solutions to your "simplified" equations
but not your initial
when people say extraneous solutions they aren't actually solutions
they are solutions you found that do not work

Answer 32

and case 2 would be extraneous

Answer 33

@myininaya

Answer 34

@myininaya "you"
ARE EXTRA SMART

Answer 35

Well are you trying to solve the equation I was talking about?

Answer 36

yes i was

Answer 37

was i correct?

Answer 38

@Nnesha I will try to not take that as an insult :p

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Well aum and I were both trying to say the solutions you would find there won't actually work
they would both be extraneous

Answer 39

you should check them again

Answer 40

GOT IT

Answer 41

lol i see where i messed up again

Answer 42

i really should work slower, i miss things sometimes

Answer 43

thank you again! :)

Answer 44

I know.
I was using what extraneous meant to miss with your meaning of extra there.