<p>If |-2x + 1| < 1, what is one possible value of x?</p>
<p>How do you go about solving this?? Ahhh these absolute values are throwing me off!!!</p>
<p>think about it this way... if |x| < 1, then to satisfy the equation -1<x<1</p>
<p>In this case, x is actually (-2x + 1), so -1<-2x+1<1.</p>
<p>Do the normal algebra to find x.</p>
<p>-2<-2x<0</p>
<p>1>x>0.</p>
<p>Hope that helps.</p>
<p>Oh so any time there is an inequality and absolute value signs such that |x| < y then you can just rewrite it as... -y < x < y ?</p>
<p>If so, that helps a lot. :D</p>
<p>Yup, that's right.</p>
<p>Also, if it's |x|>y, it goes to x>y, x<-y.</p>
<p>But instead of memorizing these two things, it's easier to just think about it like an equality (not an inequality). If |x| = y, then x=-y or x=y. Similarly, if |x|<y, then="" x<y="" or="" x="">-y (the inequality is flipped because of the negative), which can be put together as -y<x<y.</y,></p>
<p>Let me know if I just confused you.</p>
<p>No no that makes perfect sense. I dont think I should have any more problems with this cleared up. :D Your awesome thanks!</p>
<p>wait thats crazy awesome, if its |x| < y.. .then its x < y and x > -y
and if its |x| > y, then its x>y and x<-y ? ? ? ?</p>
<p>
You don't have to "go about solving". :)
You are looking just for one possible value of x out of (possibly) many.
Why not try and check?
x=0 does not work.
x=1 does not work.
Maybe between 0 and 1?</p>
<h1>x=1/2 does work!</h1>
<p>iin77's "algebraically correct" approach can be used when a question is more convoluted.
For example,
1. How many integer values of x satisfy |2x - 1| < 21?
-21 < 2x-1 < 21
-10< x < 11
...
2. If |2x - 1| > 21, what is the largest possible negative integer x?
You can do a quick guess'n'check, or go the iin77's way.
|2x - 1| > 21
2x-1<-21 OR 2x-1>21.
sushant269 - OR is what separates |x|>y scenario from |x|<y.
x<-10
...</p>
<p>So you say when it comes to those type of problems, ALWAYS start with plugging in 0 and 1 and then try 1/2 and not any other number?</p>
<p>Celita, sorry to disappoint you, but it all depends on a problem. For some problems you should avoid using 0 and 1. For most problems you don't use fractions at all. Knowing which numbers to plug comes with experience.
Even if you take December 1 SAT you have more than enough time to go through the whole Blue Book and see for which questions you can use "plug-in"" method and which numbers work.</p>
<p>another way to do it:</p>
<p>square both sides of the inequality to get rid of the absolute value condition (since square of a real number is always positive):
(-2x+1)^2 < 1 => 4x^2 - 4x + 1 < 1 => 4x^2 - 4x < 0 => (4x)(x-1) < 0</p>
<p>Now the zeros are at x = 0, and x = 1, so we test x>1, x<0, and 0<x<1 (number="" line="" helps="" to="" visualize,="" we="" have="" all="" the="" possibilities="" so="" now="" just="" check).="" if="" x=""> 1, we have two positive products, so this is not a solution. If x < 0, we have two negative products that multiply to be positive, so this is not a solution. In the third case, we have a positive product multiplied by a negative product, so this is our solution set. </x<1></p>
<p>The main idea is that squaring helps get rid of absolute values. Dealing with quadratic inequalities isn't that bad once you've had some practice.</p>
<p>Squaring does work, but there is even faster algebraic method of solving | | eaqualities (it also works for | | equations with a slight variation).
1. Find zeroes of all the expressions inside of | | (even if there are more than one).
2. Mark them on the number line. Now you have the number line broken into intervals.
3. Pick a number from each interval and plug it into your inequality. Intervals which "work" are your answer.</p>
<p>Here:
|-2x+1| + |x|< 1
Zeroes are 1/2 and 0 give us three intervals:
x<1/2
1/2=<x<0
0=<x.
...</p>
<p>I doubt you'll ever need either "squaring" or "intervaling" on the SAT.</p>
<p>I don't think I've even seen many absolute values on the two testing dates I took and the numerous practice tests I took back in junior year. Now that I look back at what the question is asking, I would have have simply guessed the right number (one possible value). The math2c would actually ask for the solution set, which is a tad bit more complicated.</p>
<p>Pfft, yours all work, but mine's quite obviously the best way. And apparently, I'm awesome.</p>