how do you solve |x-1| < 2|x+2| ?

Answer 1

Two ways.

you can graph the two functions and look at the graph to see where |x-1| is less than 2 * |x+2|

algebraically....

look at the critical points, x = 1, x = - 2

for the range -infinity to x = -2 the expression will look like this:

-x + 1 < 2 * ( -x - 2)
-x +1 < -2x - 4
x + 1 < -4
x < -5

sure enough |x-1| < 2|x+2| is true for x < -5

for the range between the two critical points, -2 < x < 1 the equation will look like this:

-x + 1 < 2 * (x + 2)
-x + 1 < 2x + 4
1 < 3x + 4
-3 < 3x
x > -1

the expression |x-1| < 2|x+2| is true for -1 < x < 1 as we've seen in this range.

the final range to look at is x > 1 the equation is going to look like this:

x - 1 < 2 * (x + 2)
x - 1 < 2x + 4
-1x < 5
x > -5

so when combining all that info you find that the expression |x-1| < 2|x+2| is true for x < -5 and x > -1

Answer 2

well first of all you need to find out what X is,, and to do that you have to get the variable X on one side of the equation. So for the time being treat the Less than sign as an Equal sign, and treat the Absolute Value Brackets as Parenthasis. So it will look like this (x+1)= 2(x+2) What you need to do is use the distributive property on the right side of the equation to make 2x + 4, and then you subract 2x from the right side and also from the left side you and you have... -1x -1= 2. And then to get the variable X all by itself you must add 1 to both sides, and you get -1x = 3, and then divide both sides by -1 and you get x= -3. TA DA!
Now you plug in the value of x into the original problem and it will look something like this,and I'm still going to use parenthasis for the absolute value brackets because i don't know where those are on my computer.
((-3)-1) < 2((-3) +2)
and then just work it out
(-4) < -6 + 12
-4 < 6

and it's true,, so that's the answer

Answer 3

treat it like its an equan sign

l x - 1 l = 2 l x + 2l

+/- l x - 1 l = 2 * +/- (x + 2)

take postive for both sides
x - 1 = 2(x + 2)
x - 1 = 2x + 4
-1 = x + 4
x = -5


take negative for one side and positive for the other side
-(x - 1) = 2 (x + 2)
-x + 1 = 2x + 4
1 = 3x + 4
-3 = 3x
x = -1

so you have two roots: -5 and -1

plug in some number to test:
try -6
l -6 - 1 l l -7 l 7 7 < 8

the statement is true, thus, x < -5

try 0
l 0 - 1 l l -1 l 1 < 4

the statement is true, thus, x > -1

answer: x < -5 or x > -1

Answer 4

-2|x+2| < x-1 < 2|x+2|

Case 1. x ≤ -2
2x+4 < x-1 < -2x-4
x < -5, or x < -1
Solution: x < -5

Case 2. x > -2
-2x-4 < x-1 < 2x+4
x > -1, or x > -5
Solution: x > -1

Answer 5

square both sides

x^2-2x+1 < 4x^2+16x+16

3x^2+18x+15 > 0

x^2+6x+5 > 0

(x+5)(x+1) > 0

x<-5, x>-1

Answer 6

x-1<2x+4
-5 < x, and
x-1< -2x-4
3x< -3
x<-1
So x >-5 and x < -1, or
-5