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|x| means the absolute value of x, or the distance of x away from 0. Therefore, if x<0, |x| returns -x if x > 0 , |x| returns x
ok but i must compare the two hamburgers by comparing the volume of the beef in each burger. the volume V of a burger is V= Ah, whee A is the area and h is the height. the area of the lite burger is nr^2, where r is the radius (half the distance across the middle) of the burger. for the lite burger, r= 2 inches, so its area is
In other words or at least more words: positive numbers return themselves negative numbers turn positive \[\left| -2 \right| = 2 and \left| 2 \right| = 2\]
Thank you, I understand that. But then how come for a question like this |5x -2| < 4 has two answers, x < 6/5 or x > -2/5 ?
This is easier to see on a number line so let me try something easier |x| > 4 the answer is x<-4 or x>4.... does it make sense that you have to go more than 4 units away from zero in either direction because the || symbols ignore positive/negative?
who says |5x-2|<4 has two answers (solutions)? first of all there are an infinite number of solutions, but they are contained in one interval. the interval is -2/5 < x < 6/5 which is an AND statement: x is greater than -2/5 and less than 6/5
Oh okay, I understand that now. But then how do we define whether x is > or < ? This is my initial question, maybe before I was framing it wrong.
let me give you a simple example. start with | x -1 | < 2. this says x is within 2 units of 1. so if you draw a number line you can see you can go up to three and down to -1, so the solution is -1 < x < 3
wheres if we say | x - 1| > 2 this means x is greater than two units from 1. so we can be below -1 or above 3. the answer is therefore two intervals, all numbers less than -1 or all numbers greater than 3. you have to write two intervals: x < -1 or x > 3
if you draw a number line you can see it easily. | x - a | < b will be one interval (assuming b is greater than or equal to zero, otherwise it makes no sense) and | x - a | > b will be two intervals.
it is really not that complicated if you remember that | x - a | is the distance between x and a. so if i am on a highway that runs east west (like the number line) and I say i am within 5 miles from mile marker 100, that means i must be somewhere between mile marker 95 and mile marker 105. i.e. the solution to | x - 100 | < 5 is 95 < x < 105 but if i say i am further than 5 miles from mile marker 100 i must be either east of marker 105 or west of marker 95 i.e. |x -100| > 5 means x < 95 or x > 105
I understand this too. But when it comes to complex questions, I get confused. Like for example: |1-3x| > 8. So in the solution set, how do we determine that x is < -7/3 and x > 3 when we started with 1-3x >8 and 1-3x < -8 respectively?
ok first of all i hope it is clear that |1-3x| = |3x-1| try it with numbers, or just know that |a-b|=|b-a|
Okay I get that.
so there is no difference between solving |1-3x|>8 and |3x-1|>8
second of all, as for my example above, you know that if you are solving |x-a| b you will have two solutions.
and something they often forget to mention is that there is really know difference between solving |x - a| < b and |x - a | > b after all, if i know one i know the other. for example if the solution to |x - a| < b is 5 < x < 10 then i know the solution to |x - a|> b is x < 5 or x > 10 after all those are the logical choices right? either greater than or less than.
so how would i solve |1-3x|> 8 ? first of all i hate having to remember to switch the inequality when dividing by a negative number so i would rewrite it as |3x-1|>8
then i have a choice. i could solve |3x-1|< 8 and take everything else, like this: -8 < 3x -1 < 8 -7 < 3x < 9 -7/3 < x < 3 and that is exactly what i don't want so i say x < -7/3 or x > 3
or i could do what they teach you and say |3x - 1|> 8 means two intervals (we know that) therefore two inequalities to start: 3x-1< -8 or 3x - 1 > 8 and solve these separately. 3x -1 < -8 3x < -7 x < -7/3 OR 3x-1 > 8 3x > 9 x > 3 either way you need to know when you will have one interval (compound inequality) or two
(just to be sure) ..and that is when |x-a| < b there will be one interval and when |x-a| > b two intervals?
yes exactly. if you can remember that you will be in good shape
thank you so much. this explains much better!
|5x+17|>9 two intervals |3x+7|<12 on interval
and while i have your attention one more thing" Like for example: |1-3x| > 8. So in the solution set, how do we determine that x is < -7/3 and x > 3 when we started with 1-3x >8 and 1-3x < -8 respectively? in this sentence your "ands" should be "OR"s x < 2 OR x > 7 there is not such thing as x < 2 and x > 7
Lol oh yeah, thanks.