Soling an inequality using Algebraic method

[Plutonium88]

Homework Statement

Solve
|3x-7|-|x-8|>4

The Attempt at a Solution

so i made columns... and using the columns i made a number line..

7/3 on the left as a point, with a column on its left, and 8 with a column on its right and sharing a coumn in the middle with 7/3

so i have..

|3x-7| => (3x-7), X>OREQUAL 7/3
-(3x-7) x<7/3


|x-8| => (x-8) x>or equal 8
-(x-8) x< 8

* i algebraeicly solved 3 different inequalities after this
1. -(3x-7)-[-(x-8)]>4 (this was not crossed out)-- the answer was crossed x<-2
2. -(3x-7) - (x-8)>4 (this WAS CROSSED OUT)
3. (3x-7) - (x-8) >4 (this was not crossed out) - the answered was crossed x>2/3

i understand i am supposed to compare the three inequalities after to get the answer, but i don't understand the thinking of how to go about the questoin in terms of CREATING the THREE inequalities..

So how exactly do i... evaluate the different possibilites can some one give me something to run with or think about, so i can solve this :(

 

[HallsofIvy]

Homework Statement

Solve
|3x-7|-|x-8|>4

The Attempt at a Solution

so i made columns... and using the columns i made a number line..

7/3 on the left as a point, with a column on its left, and 8 with a column on its right and sharing a coumn in the middle with 7/3

so i have..

|3x-7| => (3x-7), X>OREQUAL 7/3
-(3x-7) x<7/3


|x-8| => (x-8) x>or equal 8
-(x-8) x< 8

* i algebraeicly solved 3 different inequalities after this
1. -(3x-7)-[-(x-8)]>4 (this was not crossed out)-- the answer was crossed x<-2

Unfortunately, you left out the crucial part- HOW you got "-2" as the answer. Distributing the "-"s, -3x+ 7+ x- 8= -2x- 1> 4. Can you solve that?

2. -(3x-7) - (x-8)>4 (this WAS CROSSED OUT)
3. (3x-7) - (x-8) >4 (this was not crossed out) - the answered was crossed x>2/3

Again, you haven't shown how you got that. "x> 2/3" is wrong. Where did that "3" in the denominator come from?

i understand i am supposed to compare the three inequalities after to get the answer, but i don't understand the thinking of how to go about the questoin in terms of CREATING the THREE inequalities..

So how exactly do i... evaluate the different possibilites can some one give me something to run with or think about, so i can solve this :(

 

[Plutonium88]

Unfortunately, you left out the crucial part- HOW you got "-2" as the answer. Distributing the "-"s, -3x+ 7+ x- 8= -2x- 1> 4. Can you solve that?

- -2x -1 >4
x < -5/3

lol so basically i rushed and made smoe stupid simple agebraeic errors... Uhh i feel really embarassed i didn't look more carefully at this before i posted... x.x

 

[Plutonium88]

Again, you haven't shown how you got that. "x> 2/3" is wrong. Where did that "3" in the denominator come from?

QUOTE]

(3x-7) - (x-8) > 4

3x-7 -x + 8 > 4

2x 1> 4

x > 3/2

lol even when i look at the stepso n my tests.. my steps are actually correct up until i put

x > 2/3

I CANt BELEIVE this right now... these mistakes are so pathetic...man... :'(

Okay so now that i have 2 separate intervals...

Do i plug in my numbers 3/2 and -5/2 to determine if LS> RS and if it is true? then determinme what ever my interval is using interval notation.?

Also here's a link to the structure of how i did the question originally... Please give me some tips on if you tihnk this is a good way to do the question or if you have any other ideas of how to structure it?

http://postimage.org/image/9q1ec1rhj/

(i didn't even show the calulation where i plugged in my numbers to determine LS>RS Originally) i think i just did it in my head and just tried to figure out whether the number was greater or less than 4... i was lazy..

Ah and last but not least, i am really sorry (if you happen to see the image) for the ugliness of my writing. :(