rubiojc1985
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i need help with this question and can somebody explain
x<2/x-1
x<2/x-1
inequality
- Thread starter rubiojc1985
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Solving Inequalities is only BARELY different from solving equalities. In fact, the very first thing I would do is ignore that it is an INequality.rubiojc1985 said:
x = 2/(x-1)
Domain Considerations: x != 1
Multiply by x-1, which we now know is not zero.
x*(x-1) = 2
x^2 - x = 2
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
Check your answers
2/(2-1) = 2/1 = 2 -- Good.
-1/(2-1) = -1/1 = -1 -- Good.
OK, that's where we quit with an equality. Since this is an INequality, there is one more thing to do.
Draw a Number Line, marking x = 2 and x = -1 AND, since we have a problem with x = 1, mark that, too.
<--------(-1)---------(1)----(2)---------->
See how this splits up the number line into four chunks? Well, each of these chunks either works in the inequality or it doesn't. Just pick a number from each region and check it out.
x < 2/(x-1)
Pick the left-most region, (-big,-1). I'll use x = -2
-2 < 2/(-2-1) = 2/-3 = -2/3 -- Is -2 < -2/3? Yes!
Pick the next region, (-1,1). I'll use x = 0
0 < 2/(0-1) = 2/-1 = -2 -- Is 0 < -2? No!
Pick the next region, (1,2). I'll use x = 3/2
3/2 < 2/((3/2)-1) = 2/(1/2) = 4 -- Is 3/2 < 4? Yes!
Pick the right-most region, (2,big). I'll use x = 3
3 < 2/(3-1) = 2/2 = 1 -- Is 3 < 1? No!
Write down the "Yes!" regions and you are done.
(-big,-1)U(1,2)
Note: Generally, you can assume that the regions alternate. For now, though, I'd just test them all.
Note: See how important the Domain considerations were? They did more for us than may have been imagined originally.
arthur ohlsten
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x<2/[x-1]
if you multiply or divide by a minus you reverse <or> sign
1) let x-1>0 or x>1
multiply by x-1
x[x-1]<2
x^2-x-2<0 let us examine x^2-x-2=0
[x-2][x+1]=0
x-2<0 or x+1<0 but x must be greater than 1
x<2 or x<-1
1<x<2 partial answer
2) let x-1<0 or x<1
multiply by x-1
x^2-x> 2 changed < to >
x^2-x-2>0
[x-2][x+1]>0
x>2 or x>-1 but x must be less than 0
-1>x>0 partial answer
x must be between -1<x<0 or 1<x<2 answer
Arthur
if you multiply or divide by a minus you reverse <or> sign
1) let x-1>0 or x>1
multiply by x-1
x[x-1]<2
x^2-x-2<0 let us examine x^2-x-2=0
[x-2][x+1]=0
x-2<0 or x+1<0 but x must be greater than 1
x<2 or x<-1
1<x<2 partial answer
2) let x-1<0 or x<1
multiply by x-1
x^2-x> 2 changed < to >
x^2-x-2>0
[x-2][x+1]>0
x>2 or x>-1 but x must be less than 0
-1>x>0 partial answer
x must be between -1<x<0 or 1<x<2 answer
Arthur