This Blog exists for the collective benefit of all algebra students. While the posts are specific to Mr. Chamberlain's class, any and all "algebra-ticians" are welcome. The more specific your question (including your own attempts to answer it) the better.
Here's a HINT: When solving for absolute value equations, "SOLVE for the BLOB"... the BLOB is the Abs Val brackets and everything inside. Once you SOLVE for the BLOB, then you can split the equation into two separate equations and continue...
For example, pg 211 #22 is
-3|2w| = -12 divide both sides by -3 to get: |2w| = 4 and now split the equation into two
okay, for number 33, you would start on the first side, since it's an "OR" problem, which would be f + 14 <9. then subtract 14 from both sdies and you would get f<-5. For the second side: -9f or= -5. So then your answer would be like (-∞, -5) OR you could do bracket not parenthesis: [∞,5)
For #24, the bracket on the left and right simply imply that the borders i.e. -4 & 5 are INCLUDED in the solution set... a ( or ) would imply that the border is EXCLUDED from the solution set.
Graphing t=3 on a number line yields a solid dot (certainly not an open dot, right?)
#7) Solve |3x|+8=5;
We can look at #7 two ways:
First, we can SOLVE for the BLOB by subtracting 8 from both sides, yielding:
|3x|=-3
Think about it, what is the smallest value that can ever "pop" out of AbsVal brackets... zero, right? So this can never work, no matter what value you choose for x, |3x| will ALWAYS be positive, right?
Second, even if you didn't SOLVE for the BLOB, since the smallest value |3x| can ever be is zero, if you add 8 to that number it will always be >= 8. So, there is no value for x such that |3x|+8 can ever = 5... ca-peessh?
Here's a HINT: When solving for absolute value equations, "SOLVE for the BLOB"... the BLOB is the Abs Val brackets and everything inside. Once you SOLVE for the BLOB, then you can split the equation into two separate equations and continue...
ReplyDeleteFor example, pg 211 #22 is
-3|2w| = -12
divide both sides by -3 to get:
|2w| = 4
and now split the equation into two
1) 2w = 4
2) 2w = -4
Questions? You know where to find me!
how would you do number 24???: [-4,5]
ReplyDeleteand how would you do number 33?????
ReplyDeleteokay, for number 33, you would start on the first side, since it's an "OR" problem, which would be f + 14 <9. then subtract 14 from both sdies and you would get f<-5. For the second side: -9f or= -5. So then your answer would be like (-∞, -5) OR you could do bracket not parenthesis: [∞,5)
ReplyDeletethanx
ReplyDeleteFor #24, the bracket on the left and right simply imply that the borders i.e. -4 & 5 are INCLUDED in the solution set... a ( or ) would imply that the border is EXCLUDED from the solution set.
ReplyDeleteCan you draw the graph now?
For #33, I agree with Tyler on the left side that the solution is (-∞, -5)
ReplyDeleteHOWEVER.. the right side where -9f <= -45 simplifies to f>= 5 which would be written in interval notation as: [5, ∞).
In interval notation, the order matters, i.e. you write the intervals from least to greatest, left to right.
Ca-peesh??
(Thanks, Tyler!)
I had a question about #24 but that helped thanks :)
ReplyDeleteI don't understand how to do this one..
ReplyDelete1. |x|=5
How are you supposed to solve that one?
Oh now I understand nevermind. But how would you graph something like t=3? Just a dot on a number line?
ReplyDeleteLastly, I don't understand why #7 does not have a solution?
ReplyDeleteGraphing t=3 on a number line yields a solid dot (certainly not an open dot, right?)
ReplyDelete#7) Solve |3x|+8=5;
We can look at #7 two ways:
First, we can SOLVE for the BLOB by subtracting 8 from both sides, yielding:
|3x|=-3
Think about it, what is the smallest value that can ever "pop" out of AbsVal brackets... zero, right? So this can never work, no matter what value you choose for x, |3x| will ALWAYS be positive, right?
Second, even if you didn't SOLVE for the BLOB, since the smallest value |3x| can ever be is zero, if you add 8 to that number it will always be >= 8. So, there is no value for x such that |3x|+8 can ever = 5... ca-peessh?
Ohhh I see.. Because [as it seems] if x=-1, then it would be 3 x (-1)= -3, but then the absolute value would be 3. And 3 doesn't = -3!!!
ReplyDelete:D
Yup... you got it... hmmm... no one else has ANY questions... I might as well just give the test!!
ReplyDeleteThanx for #7, I looked at it in the wrong way at first , but now i understand.
ReplyDeleteNo No NO TEST YET!!!!!!
ReplyDeletethanx for clearing that up with me MR. C, wont make that mistake again!!! :)
ReplyDeleteWhat test??!! There's no test!! What IDIOT started a rumor about a test??!! (oh, that was me... nevermind)
ReplyDelete