| x+3 | < |x-1

[stat643]

| x+3 | < |x-1|

The solution states that this inequality implies x is closer to -3 therefore the solution is -1... why?

 

[cristo]

Huh?

You need to state the exact question, and show some work before we can help you.

 

[stat643]

i dont know how to solve it, the sol is "because x is closer to -3 than it is to 1 Hence every x for which x < −1 satisfies the inequality." Im just trying to see why this is the case ... and why x is closer to -3?

 

[sutupidmath]

i dont know how to solve it, the sol is "because x is closer to -3 than it is to 1 Hence every x for which x < −1 satisfies the inequality." Im just trying to see why this is the case .

Well as i can see you are trying to solve the following inequality, right?
THen here it goes.

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

we devide this in the following intervals

$$1)x\in(-\infty,-3)...2)x\in[-3,1),...and....3) x\in[1,\infty )$$

Then evaluate separately the expression$$|x+3|<|x-1|$$ in each interval. Pay heed to the sign of |x+3| and |x-1|, separately in each interval as you evaluate it. THen at the end you will be able to find the interval for which that inequality holds.

Regards,

P.S. SHow what are your thoughts on the problem first, forum rules! lol...

 

[arildno]

i dont know how to solve it, the sol is "because x is closer to -3 than it is to 1 Hence every x for which x < −1 satisfies the inequality." Im just trying to see why this is the case ... and why x is closer to -3?

|x+3|=|x-(-3)| is the distance between the number "x" and the number (-3).
|x-1| is the distance between the number "x" and the number 1.

Thus, the inequality |x+3| less than |x-1| has as its solutions all those x's whose distances to (-3) is less than their distances to 1.

 

[stat643]

thanks