anonymous
  • anonymous
|w-6|=|3-2w|
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
r u trying to solve for w ?
anonymous
  • anonymous
ya
apoorvk
  • apoorvk
w is a complex no. as in 'omega' or just any variable?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

apoorvk
  • apoorvk
anyways when you remove mod, add +- to any one side. like........|dw:1332639411896:dw| because x could originally be +ve or -ve, but its modulus would still give a positive 'a' always.. so use this info to remove mod from your equation. you ll get two different equations then. use union of the values of 'w' for getting complete set of solutions.
apoorvk
  • apoorvk
so here, w-6=(3-2w) and also, w-6=-(3-2w) solve these two and proceed to get union of the values for complete set of solutions.
myininaya
  • myininaya
w-6=0 when w=6 3-2w=0 when w=3/2 Draw number line: -------------|------------|-------------- 3/2 6 Test intervals for + and - -------------|------------|--------------- 3/2 6 Check the following numbers: 0 4 8 w-6: 0-6=-6 4-6=-2 8-6=2 3-2w: 3-2(0)=3 3-2(4)=-5 3-2(8)=-13 ------------------------------------------- So we have the following w-6 negative negative positive 3-2w positive negative negative --------------|------------|------------- 3/2 6 ============================== So checking that first interval and by the way we have for the first interval that |w-6|=-(w-6) (since we got it was negative on that interval) and |3-2w|=3-2w (since we got it was positive on that interval) Now solving -(w-6)=3-2w gives you what?
myininaya
  • myininaya
and then you do the same thing for the other two intervals
myininaya
  • myininaya
On the second interval you have |w-6|=-(w-6) and |3-2w|=-(3-2w) Which means you have -(w-6)=-(3-2w) or w-6=3-2w
myininaya
  • myininaya
For the last interval you get same thing as the first since one is negative and one is positive
myininaya
  • myininaya
What I'm saying is for the last interval we have |w-6|=w-6 since w-6 is positive on the last interval |3-2w|=-(3-2w) since 3-2w is negative on the last interval So that means we are looking to solve w-6=-(3-2w) which is the same as -(w-6)=3-2w if we multiply both sides by -1 We already solve this in the very first interval
.Sam.
  • .Sam.
what i so is i square both sides (w-6)^2=(3-2w)^2 w^2-12w+36=9-12w+4w^2 -3w^2=-27 w^2=9 w=+-3
.Sam.
  • .Sam.
works for every absolute equations
myininaya
  • myininaya
you might get extra solutions though should check both
myininaya
  • myininaya
but yes +-3 is the answer here
apoorvk
  • apoorvk
yeah thats what i wanted to say too... squaring may give extra solutions, which we hafta check.
myininaya
  • myininaya
but squaring is a fine and dandy way :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.