CookieMonster18
  • CookieMonster18
1.Solve | x-1/5|=3 A.x=-14/5 or x=16/5 B.x=14/5 or x=-16/5 C.x=-14/5 or x=-16/5 D.x=14/5 or x=16/5 ***** 2.The table shows the relationship between two variables. Which selection describes the relationship? X=1, 2, 3, 4 Y=2, -3, -8, -13 A.increasing; linear B.increasing; nonlinear C.decreasing; linear ***** D.decreasing; nonlinear please correct me if im wrong
Mathematics
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SOLVED
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katieb
  • katieb
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PhantomCrow
  • PhantomCrow
Is that \[\frac{ x-1 }{ 5 }=3\] ?
CookieMonster18
  • CookieMonster18
no @PhantomCrow I corrected it. (I edited it) please look up at #1 for the correct problem I had to fix it since I done it wrong :)
PhantomCrow
  • PhantomCrow
So is it \[|x-\frac{ 1 }{ 5 }|=3\] ?

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PhantomCrow
  • PhantomCrow
If so, you can begin by negating the absolute value and adding a positive/negative sign to three.
CookieMonster18
  • CookieMonster18
yes :)
PhantomCrow
  • PhantomCrow
We do this because the absolute value of anything has to be positive, regardless of the value within the bars. So we should account for both negative and positive values.
PhantomCrow
  • PhantomCrow
After this, we split the equation up into two separate solutions and solve for x in each one.
CookieMonster18
  • CookieMonster18
how do I do that @PhantomCrow
PhantomCrow
  • PhantomCrow
\[|x-\frac{ 1 }{ 5 }|=3\] \[x-\frac{ 1 }{ 5 }=\pm3\] \[x-\frac{ 1 }{ 5 }=3\] and \[x-\frac{ 1 }{ 5 }=-3\]
CookieMonster18
  • CookieMonster18
??? @PhantomCrow
PhantomCrow
  • PhantomCrow
What don't you understand?
PhantomCrow
  • PhantomCrow
Remember that the absolute value of anything must be positive. So no matter what we put into x with the absolute value bars around it, it ill output a positive number. There can only be two possible values for x-1/5 --3 and negative 3 because we are taking the absolute value of it and regardless of its sign, we will get a positive number so we should account for the negative values in the absolute value bars as well as the positive ones. So if the absolute value bars were not there, the answer could remain as 3 or potentially change to -3.
CookieMonster18
  • CookieMonster18
oh okay but for #1 I thought it was D and for #2 I thought it was C and im not sure if I was correct or not @PhantomCrow
PhantomCrow
  • PhantomCrow
Did you solve the two equations we derived for the first one?
Crystal_Bliss
  • Crystal_Bliss
Break down the problem into these 2 equations \[x-1/5=3 -(x-1/5)=3\]
Crystal_Bliss
  • Crystal_Bliss
the first one is not D @CookieMonster18
PhantomCrow
  • PhantomCrow
\[x=3+\frac{ 1 }{ 5 }\] \[x=-3+\frac{ 1 }{ 5 }\]
mathmale
  • mathmale
| x-1/5|=3 can be broken down into two separate equations: \[x-\frac{ 1 }{ 5 }=3, and\]
mathmale
  • mathmale
\[-(x-\frac{ 1 }{ 5 })=3\]
CookieMonster18
  • CookieMonster18
A?? because is x=16/5 -1/5 it'll equal 3
PhantomCrow
  • PhantomCrow
We have two solutions for x.
mathmale
  • mathmale
Eliminating the fraction 1/5 will make the solution easier. Multiply both equations by 5 to do this (multiply every term). Sove the resulting equations. You'll get two separate solutions.
Crystal_Bliss
  • Crystal_Bliss
yes it would be A you just forgot the negative
mathmale
  • mathmale
Again, expect TWO solutions, and be sure to check both.
CookieMonster18
  • CookieMonster18
@Crystal_Bliss what do you mean I forgot the negative?? for #1 A?? you mean x=-14/5??
CookieMonster18
  • CookieMonster18
@pooja195 am I correct? #1 is A and #2 is C??

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