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Could use some math help... a lot of help.

post #1 of 15
Thread Starter 
It's been a while since I needed some math help but I have a test tomorrow and missed yesterday's class because I had a migraine. I emailed the instructor about getting an office visit with him to go over some things tomorrow but I probably won't hear back until morning.

The following problems involve Equations and Inequalities Involving Absolute Value:

3|y+6|= 10 (the answer I come up with doesn't jive with the book)

|6y-2|+4= 32

7|5x|+2= 16

|2x+6/3|< 2 (I hope you can read this one 2x+6 over 3)

I have others like these but if I can get some instruction on how to do these, I should be able to finish the rest of the problems. Thanks.
post #2 of 15
Don't believe the back of text books you wouldn't believe how often they are wrong!!

I tried to do the problems but it's been so long I can't remember anything past BEDMAS (ok it's been like 3 years, that's really really sad...)
post #3 of 15
Thread Starter 
Quote:
Originally Posted by Nes View Post
Don't believe the back of text books you wouldn't believe how often they are wrong!!

I tried to do the problems but it's been so long I can't remember anything past BEDMAS (ok it's been like 3 years, that's really really sad...)
I've been wondering if there are some typos in the answers section.
post #4 of 15
you have to assume y is either positive or negative so you will have 2 possible answers.
For example the first one, you have to solve 3|y+6|= 10 AND 3|y-6|= 10
That will give you y= -8/3 or -2.66 and y= 28/3 or 9.33
Does that make sense? I don't think i'm wrong but it's been 2 years since i've taken any math.
post #5 of 15
the more you post these problems, the gladder I am I skipped an algebra refresher and went straight to Calculus I probably would have flunked Algebra
post #6 of 15
Oh ut0pia you made my day!! That's why I got! I was sure I was doing something backwards (I'm really bad at math!)

Actually I used to have a teacher who co-wrote a math text book for the province . So I have it on good authority frequently there are typos, sometimes they are just plain wrong!!
post #7 of 15
Quote:
Originally Posted by katiemae1277 View Post
the more you post these problems, the gladder I am I skipped an algebra refresher and went straight to Calculus I probably would have flunked Algebra

I hate algebra. It's so simple but the way they teach it is confusing. I was so bad at algebra in high school. In college I went straight to calculus too and I liked calculus much better than algebra or any other math I had taken in high school. and it isn't until calculus that I realized that all those algebra and geometry teachers made my life hell for NO REASON at all because you never use those ridiculous things anymore.
post #8 of 15
Quote:
Originally Posted by essayons89 View Post
it's been a while since i needed some math help but i have a test tomorrow and missed yesterday's class because i had a migraine. I emailed the instructor about getting an office visit with him to go over some things tomorrow but i probably won't hear back until morning.

The following problems involve equations and inequalities involving absolute value:

3|y+6|= 10 (the answer i come up with doesn't jive with the book)

|y+6|=10/3 ==> what's in the absolute value brackets must equal +/- 10/3 ==> y = -8/3 or y = -28/3

|6y-2|+4= 32

|6y-2|=28 ==>y=5 or y = -26/6

7|5x|+2= 16

7|5x|=14 ==> |5x|=2 ==> x=+/- 2/5

|2x+6/3|< 2 (i hope you can read this one 2x+6 over 3)

are you saying |(2x+6)/3| < 2?

Assuming you are:

|2x+6| < 6 {multiplication/division isn't affected by the abs. Value brackets, so multiply both sides by 3}

==> |x+3| < 3 {and divide both sides by 2 just to make it a little clearer}

solving for |x+3| = 3, x = 0 or x = -6. These are your critical values. You have to figure out if |x+3| is greater or less than 3 between these values. If you put in, say, x = -3, you get |-3+3| = 0, which is less than 3, so your values for x are -6 < x < 0


i have others like these but if i can get some instruction on how to do these, i should be able to finish the rest of the problems. Thanks.
123456789012
post #9 of 15
Quote:
Originally Posted by Nes View Post
Oh ut0pia you made my day!! That's why I got! I was sure I was doing something backwards (I'm really bad at math!)

Actually I used to have a teacher who co-wrote a math text book for the province . So I have it on good authority frequently there are typos, sometimes they are just plain wrong!!
lol I hope we're not wrong
I would think typos are common too..
Essayons is this stuff for a grade or just practice for tests? If so I wouldn't worry too much if one or two of them are wrong b/c it might be a typo.
post #10 of 15
LOL I'm so confused. This is what happens when I use my own logic to solve problems rather than the rules of math- I end up wrong Grogs is right but I'm still trying to find a flaw in my method and i can't lol
post #11 of 15
Good grief!

*goes off to take some Tylenol!!*
post #12 of 15
I tried to answer them as I remember, and well - we're not gonna post my logic.
post #13 of 15
Quote:
Originally Posted by Essayons89 View Post

3|y+6|= 10 (the answer I come up with doesn't jive with the book)

|6y-2|+4= 32

7|5x|+2= 16

|2x+6/3|< 2 (I hope you can read this one 2x+6 over 3)
Divide both sides by 3 to isolate the absolute value.
3 |y+6| = 10/3
3

|y+6| = 10/3

You can think of this as two equations: y+6 = 10/3 and y+6 = -10/3
Solve both.
If you get a common denominator, 6 = 18/3 so that your equations become

y + 18/3 = 10/3 and y + 18/3 = -10/3

For the first one: y = 10/3 - 18/3 = -8/3

For the second one: y = -10/3 - 18/3 = -28/3

The trick is to isolate whatever is in absolute value brackets. Then, solve two equations: one where the other side is negative and one where the other side is positive. This works because 10/3 and -10/3 have the same absolute value. Just solve the other problems in a similar fashion.
post #14 of 15
Quote:
Originally Posted by Going Nova View Post
Divide both sides by 3 to isolate the absolute value.
3 |y+6| = 10/3
3

|y+6| = 10/3

You can think of this as two equations: y+6 = 10/3 and y+6 = -10/3
Solve both.
If you get a common denominator, 6 = 18/3 so that your equations become

y + 18/3 = 10/3 and y + 18/3 = -10/3

For the first one: y = 10/3 - 18/3 = -8/3

For the second one: y = -10/3 - 18/3 = -28/3

The trick is to isolate whatever is in absolute value brackets. Then, solve two equations: one where the other side is negative and one where the other side is positive. This works because 10/3 and -10/3 have the same absolute value. Just solve the other problems in a similar fashion.
You're right.

I just spent the last 26 WEEKS in math I had to take a developmental class, and a college algebra class for the nursing school I started 3 days ago. HATED IT. Thankfully the only math I have to do now is dosage calculations. And after having to deal with radicals and the quadratic formula- dosage calculation is a breeze.

Which leads to this question. Why on earth do we need to know it anyhow? From what I've seen, the hospital that I'm going to be doing clinical at (it's a hospital based program) has a computer program that does all the math for you on the floor. With all of the "stuff" I've got cram in my head over the next 16 weeks, I'm really not looking forward to having to do math stuff too. And the bad thing? If you fail the math test 3x- they can kick you out of the program, even if you're passing theory and clinical!

Cheryl
post #15 of 15
Thread Starter 
Thanks.

I had the teacher go over some of them in class. I understand them better now.

I bombed my test. A migraine and not sleeping well didn't help my preparedness. We get to drop one of our four tests, this will be the one I drop. Our final is next Thursday, I need to catch up on learning quadratics.
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