solve l x+1 l < l 3x-1 l?

Question

please show the workings... thank you very much!!! ^v^

Answer 1

|x+1| < |3x-1|

Because both sides are positive, we can remove the modulus signs straightaway, and square both sides.

(x+1)^2 < (3x-1)^2
(x+1)^2 - (3x-1)^2 < 0

Using the formula (a-b)(a+b) = a^2 - b^2, I take a = x+1 and b = 3x-1. I will get:

(x+1+3x+1)(x+1-3x+1) < 0
4x(-2x+2) < 0
-8x^2 + 8x < 0
8x^2 - 8x > 0 Notice the change in signs
8x (x-1) > 0

Now, find out which values of x will the expression equal to 0. You will get x = 0 and x = 1. Now, using your calculator, try values greater than 1, values between 0 and 1, and values less than 0 and substitute them into the question.

You will find that the expression above is ONLY valid for values of x greater than 1 and less than 0. Therefore, this is your answer.

x > 1 OR x < 0 (Ans)

Answer 2

If x>=1/3 both are positive and
x+1<3x-1 so 2x >2 and x>1 solution
if -1 x+1<1-3x so 4x<0 and -1 finally if
x<-1
-x-1<1-3x and 2x<2 and x<1 so x<-1 is also solution
s0 x>1 and x<0 are the solution

Answer 3

Absolute value definition : (a million) |x| = x <=> x >= 0 (2) |x| = -x <=> x < 0 (a million) |7x + 11| = |3x + a million| 7x + 11 = 3x + a million 4x = -10 x = -5/2 (2) -(7x + 11) = -(3x + a million) -7x - 11 = -3x -a million -4x = 10 x = -5/2 verification : x = -5/2 => -13/2 = -13/2 answer ; x = -5/2

Answer 4

either
x+1>0 and 3x-1>0 and x+1<3x-1 which gives x>1
or
x+1<0 and 3x-1<0 and x+1>3x-1 which gives x<-1

Answer 5

|x+1| < |3x+1|
x+1 < 3x+1
-x -x
1 < 2X+1
-1 -1
0 < 2x
/2 /2
0 < x or x > 0

So, take any number greater than 0, plug it into the equation on both sides, and if it comes out correct, you have your solution.

Answer 6

i think the 2 one i greater cuz the absolute value of x+1 is x+1 and 3x-1 would be 3x+1.cuz the absolute value always changes negatives to positives but never positives to negatives.
hope i helped

Answer 7

x+1<3x-1 --> 2< 2x --> 1 x+1 <1-3x --> 4x < 0 --> x<0
(-infinity, 0) U (1, infinity)

Answer 8

The easiest way to calculate the answer would be to graph.
You get every value except for [0,1].
Answer would be: (-infinity,0)U(1,infinity)
or
all reals except for numbers 0, 1, and numbers between 0 and 1

Answer 9

the first one is greater lol

Answer 10

let me fire up the time machine and ask al gebra