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Mathematics 7 Online
OpenStudy (anonymous):

please help!; 2[x-5]+4=34 its an absolute value problem

OpenStudy (anonymous):

is x-5 the abs value part? ie: 2|x-5|+4=34

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

We are able to write this as two separate equations if we limit the domain, but first we need to know where to separate it. at what x value does the abs value portion = exactly 0?

OpenStudy (anonymous):

not sure. i dont really know how to isolate this varible- what to do with the 2 divide or not..

OpenStudy (anonymous):

don't worry about that yet, look at just |x-5|. at what x value will this equation = 0?

OpenStudy (anonymous):

5, right

OpenStudy (anonymous):

right, so now we know when the absolute value part of the equation flips. so now we need to test each side. when x=4 and when x = 6 the equation equals 1. we can see here that when x=6, the x is not changed: |6-5|=1 && 6-5=1 however when x=4: |4-5|=1 but 4-5=-1 so from this we can tell that for all x>5, x=x for all x<5, x=-x does that make sense so far?

OpenStudy (anonymous):

im sorry, x does not equal -x in this one

OpenStudy (anonymous):

sort of but i dont get how this has to do with the rest of the problem..

OpenStudy (anonymous):

the idea is to turn this one equation into a system of inequalities, ie: when x>5 then (equation 1) when x<5 (equation 2)

OpenStudy (anonymous):

right, i get that there has to be two outcomes..

OpenStudy (anonymous):

actually when x<5, x=((5-x)*2) and when x > 5, x=x with this we can make a piece-wise function

OpenStudy (anonymous):

but there is an easier way to do it than changing x, i remember now, im sorry

OpenStudy (anonymous):

no problem

OpenStudy (anonymous):

we can treat |x-5| as parentheses, and just put a negative sign in front, so that it distributes: x>5: |x-5| = (x-5) x<5: |x-5| = -(x-5) does this part make sense?

OpenStudy (anonymous):

oo i think so

OpenStudy (anonymous):

now we have: x>5 then 2(x-5)+4=34 x<5 then -2(x-5)+4=34 (the negative goes in front of the 2)

OpenStudy (anonymous):

right

OpenStudy (anonymous):

so now solve for x for both, and make sure it is in the domain, ie: for the first one, if you get 4 as your x, it is not >5

OpenStudy (anonymous):

right so both equations (outcomes)= 30 right?

OpenStudy (anonymous):

no, when i solve for x, the first equation x=20, the second x=-10

OpenStudy (anonymous):

my book says the answers are 12, and -18..

OpenStudy (anonymous):

It looks like i made a mistake somewhere, let me find it.

OpenStudy (anonymous):

Check the question that you gave me, because it should be x=-10,20

OpenStudy (anonymous):

ya im confused, do you distribute the 2 into the perentices?

OpenStudy (anonymous):

you can do it a few ways: first move over the 4 2(x-5)+4 (-4) = 34 (-4) 2(x-5)=30 you can then either distribute, or divide out 2

OpenStudy (anonymous):

ya my bad you got it right -i was looking at a diff answer.

OpenStudy (anonymous):

sry about that and thank you sooo much for your help!!

OpenStudy (anonymous):

no problem, do you understand it better now?

OpenStudy (anonymous):

yup!

OpenStudy (anonymous):

Glad I could help, remember to use parentheses for abs value, and then through a negative sign on one of the inequalities to make it work =)

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