Evaluating Absolute Value

[Marciamay]

Problem: If a = -3, evaluate: /1-a/-2/a+2/+1
I think it is: /1-(-3)/-2/-3+2/+1
=/4/-2/-1/+1
=/4/2/1/1
= 8
I'm using a study guide for a test i have to take next Friday and i know the answer is 3. I just don't know what I'm doing wrong and I can't find any examples quite like mine. Thank you for your help! Marciamay

 

[Denis]

Problem: If a = -3, evaluate: /1-a/-2/a+2/+1

You're using a very POOR way to show absolutes...

ABS(1 - a) - 2*ABS(a + 2) + 1
4 - 2*1 + 1
3 - 1 + 1
3

 

[lookagain]

Problem: If a = -3, evaluate: /1-a/-2/a+2/+1
I think it is: /1-(-3)/-2/-3+2/+1
=/4/-2/-1/+1
=/4/2/1/1
= 8

Marciamay,

there is a vertical symbol used for the absolute value. If you hit "Shift" while
holding the "| \" key on the keyboard, you will get the "|" character. And you
should use vertical spaces in between certain characters for more readability:

If a = -3, evaluate: |1 - a| - 2|a + 2| +1

|1 - (-3)| - 2|(-3) + 2| + 1 =

|1 + 3| - 2|-3 + 2| + 1 =

|4| - 2|-1| + 1 =

4 - 2(1) + 1 =

4 - 2 + 1 =

3

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The effect can be improved by using some Latex, and you can
look at my characters used by clicking "quote." For instance:


\(\displaystyle If \ a \ = \ -3, \ evaluate: \ \ \big|1 - a \big| \ - \ 2\big|a + 2\big| \ + \ 1\)


\(\displaystyle \big|1 - (-3)\big| \ - \ 2\big|(-3) + 2\big| \ + \ 1 \ =\)

\(\displaystyle \big|1 + 3\big| \ - \ 2\big|-3 + 2\big| \ + \ 1 \ =\)

\(\displaystyle \big|4\big| \ - \ 2 \big|-1\big| \ + \ 1 \ =\)

\(\displaystyle 4 \ - \ 2(1) \ + \ 1\ =\)

\(\displaystyle 4 \ - \ 2 \ + \ 1 \ =\)

\(\displaystyle 3\)

 

[NikkiJ24]

\(\displaystyle > \ > \ > \ The \ following \ commentary \ is \ by \ lookagain: < \ < \ <\)

"In my homework I came across the equation 2|3-2x| -6>or equal to 18
this is the work i did to try to find the solution set
2|3-2x|-6>18
+6 +6
2|3-2x|>24
/2 /2
|3-2x|>12 or |3-2x|<-12
-3 -3 -3 -3
-2x >9 or -2x <-15
/-2 /-2 /-2 /-2
x> -9/2 x<15/2


The book says those are the correct answers, but it shows the graph as

\(\displaystyle <-|--|--|--|--|--|--|--|--|--|--|-]-|--|--|--|--|--|--|--|--|--|--|--|--|-[-|--|--|--|-->\)
\(\displaystyle -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8\)


and the solution set written as -infinity, -9/2] and [15/2, infinity. i don't understand how that can be when the answer is
x is greater than or equal to -9/2. from the answer I've gotten the brackets should be turned the other way.


\(\displaystyle > \ > \ >\)have i done a step wrong or missed something? \(\displaystyle < \ < \ <\)I'm so confused

\(\displaystyle > \ > \ > \ Please \ \ put \ a \ new \ problem \ in \ a \ different \ thread \ by \ clicking \ on \ "NEW TOPIC*."\)


\(\displaystyle \big|3 - 2x \big| \ > \ 12 \ \ becomes\)


\(\displaystyle 3 - 2x < -12 \ \ or \ \ 3 - 2x > 12 \ \ \ \ There \ are \ no \ absolute \ value \ signs \ at \ this \ step.\)


\(\displaystyle -2x < -15 \ \ or \ \ -2x > 9\)


\(\displaystyle \frac{-2x}{-2} > \frac{-15}{-2} \ \ or \ \ \frac{-2x}{-2} < \frac{9}{-2} \ \ \ \ The \ direction \ of \ the \ inequality \ symbols \ change \ at \ \ *this* \ \ step.\)


\(\displaystyle x > \frac{15}{2} \ \ or \ \ x < \ \ \frac{-9}{2}\)


In interval notation, then, the solution is:

\(\displaystyle \bigg(-\infty, \frac{-9}{2} \bigg) \ \ or \ \ \bigg(\frac{15}{2}, \infty \bigg)\)