2 < I x+2 I < 6 How do you solve this?

Question

Wouldn't the absolute vaule make this into two?: 2 < x+2 < 6
2 < x -2 < 6
But then how would you graph this on a numberline?
Can someone explain how you solve this inequality?

Answer 1

Here is a very basic approach:

If (x+2) is positive (meaning that x > -2), then the inequality is:
2 < x+2 < 6, which means (after subtracting 2 from each term):
0 < x < 4
We made the assumption that x > -2, which is true over the entire range from 0 to 4, so all values in that range satisfy the original inequality.

If (x+2) is negative (meaning that x < -2), then the inequality is:
2 < -(x+2) < 6, or 2 < -x -2 < 6
which means (after adding 2 to each term):
4 < -x < 8
We can multiply by -1, but that means we need to reverse the "sense" of the inequalities:
-4 > x > -8, so x lies between -8 and -4.
We made the assumption that x < -2, which is true over the entire range from -8 to -4, so all values in that range satisfy the original inequality.

Final answer:
The inequality is solved by any x such that
0 < x < 4 OR -4 > x > -8

Graphical way to approach it. The graph of x+2 is a straight line with a slope of 1 and a y-intercept of 2. It goes below the x axis (that is, y becomes negative) for values of x below -2. So the graph of y = I x+2 I has a slope of -1 below x = -2.

Having pictured this graph, ask the question: For what values of x will the graph be between 2 and 6? (i.e., 2 < I x+2 I < 6)

Well, since the slope is 1 (and -1), this will occur where x is between 2 and 6 units from x = -2 (the point where the graph is 0). So the answer is "from -8 to -4, and from 0 to 4," as we found before.

Answer 2

You have to divide the equation into two parts because it's a compund inequality. It would look like this:
2< I x+2 I and I x+2 I < 6
Then you solve them separately.
2 < I x+2 I
2-2 < x +2-2
0 So the absolute value of x has to be greater than 0.
And the other: I x+2 I < 6
x +2-2 < 6-2
x < 4
And then you put them back together when you finish:
0
And to graph it on a number line, you make a number line with numbers from -1 to 6 (just for space), and put an open circle on 0 and on 4. Then connect the dots. By the way, an open circle is just a circle. This means that the number that will replace x has to be greater than 0, but less than 4.

Hope this helps!!

Answer 3

You have two problems into one:

(1) solve 2 < x+ 2 < 6, for x+2 >= 0

and

(2) solve 2 < -(x+ 2) < 6, for x+ 2 < 0
--- --- ---
Solving (1),

2 < x+ 2 < 6, for x >= -2
0 < x < 4, for x >= -2

Solving (2),

2 < -x- 2 < 6, for x < -2
4 < -x < 8, for x < -2
-8 < x < -4, for x < -2
--- --- ---
So the answers are: x between 0 and 4 except both 0 and 4, or x between -8 and -4 except both -8 and -4.

Answer 4

First, enable's merely seem at those solutions. endure in ideas, {x I x.. } ability {x, such that x is... }. no longer something frightening there. so as that they seem intimidating, yet, merely concentration on the final area. Now, as on your equation. 7x + 6 < 3(x - 2) you are attempting to get x on my own. Now individually, i can not resolve for x and bypass away it on the applicable, when you consider that that messes me up while staring on the extra effective than/under sign. So i will resolve for the left (additionally sturdy when you consider that each and every of the solutions have x to the left) 7x + 6 < 3(x - 2) First, distribute to do away with parenthesis: 7x + 6 < 3x - 6 Now, get x on my own on one area by way of subtracting x from the applicable and subtracting 6 from the left: 4x < -12 Now, get x on my own by way of branch: x < -3 {x I x < -3} it is your answer! On a area be conscious (and that i do no longer advise to confuse you, yet it is significant to be responsive to), fortunate you we did it this way- if we had solved for x on the applicable, we'd have had a topic properly worth addressing. There could have been a detrimental x on the applicable- this implies that to do away with it, you should divide by way of a detrimental variety. Why? think of of the variety line- by way of dividing by way of a detrimental variety and flipping the sign, you're genuinely flipping the sign to point a diverse way. that's when you consider which you replaced the sign or the numbers, and subsequently the trail they attitude. hence the 1st answer is there- to confuse you. consistently you should alter the sign while dividing (or additionally multiplying) by way of a detrimental.

Answer 5

Okay, first for 2
Now, we need to look at 2<-x-2<6.
First add +2, giving you 4<-x<8.
Then divide by -1, giving you -4>x>-8.

So, the answer is 0x>-8.