How many solutions are possible for the inequality |x-1| + |x-6| < 2?

For example,

| x - 1 | = 3

This means all the values of x whose distance from 1 on the number line is 3 units. So, the two points are 4 and -2

Similarly, if | x - 6 | = 3

This means all values of x whose distance from 6 on the number line is 3 units. So, the two points are 9 and 3.

In the question, we need to find a common point which is placed on the number line such that it's distance from 1 and 6 should add up to less than 2.

The difference between 1 and 6 is 5. Any point between the two on a number line would divide the line segment between 1 and 6 into two parts whose sum would be 5, which is not less than 2.

If we take the negative value for | x - 1 |, the distance further increases. Hence, no solution. Option D

msk0657

i actually tested all 4 options:
+ +
+ -
- +
- -

1st one:

x-1 + x-6 < 2
2x-7 < 2
2x < 9
x < 4.5
suppose x=4
3 + 2 = 5
5>2
not working.
E is out.

2nd one:
x-1 -x +6 < 2
x cancels, and we get 5<2 - nonsense.
A is out.

3rd one:
-x+1 +x-6 <2
x cancels, and we get an inequality that is always true -> -5<2. but we don't have an answer for X....
so this one is kind of blurry...

4th one:
-x+1 -x+6 < 2
-2x+7 < 2
5<2x
2.5<x
so x>2.5
suppose x=3
3-1 = 2.
3-6 = -3, absolute value is 3.

2+3 = 5
5 is not < 2.
so out.

in this case, it's either D or C.
since the 3rd case doesn't actually provide an exact solution...I would go with D...

hi

I have also learnt this technique from Bunuel. its simply great, no doubt out there ...

you can, however, apply to this problem the following technique...I hope you will also like it...

-2 < x - 1 + x - 6 < 2

solve this inequality within a moment to get 5/2 <x <7/5

here you can see that virtually there is no solution for x..

hope this helps ...

Why 7/5?

For me is 5/2 < x < 9/2 where my mistake?

hdwnkr wrote gmatcracker2017 wrote:
guilherme28 , if you are asking about 7/5, see my question to gmatcracker2017 below. I do not think you made a mistake on the arithmetic. I got 9/2, too, see below.

I am not sure, but maybe also you forgot to test your answers?

I think you removed absolute value bars and did the math for the four cases.

FT = first term and ST = second term

1st: +FT + ST < 2
2nd: +FT - ST < 2
3rd: -FT + ST < 2
4th: -FT - ST < 2

1st:
x - 1 + x - 6 < 2
2x < 9
x < \(\frac{9}{2}\)

2nd:
x - 1 - x + 6 < 2
NO SOLUTION, LHS --> x - x = 0

3rd:
-x + 1 + x - 6 < 2
NO SOLUTION, LHS --> -x + x = 0

4th:
-x + 1 - x + 6 < 2
-2x < -5
x > \(\frac{5}{2}\), or \(\frac{5}{2}\) < x

Put 1st and 4th cases' solutions together

\(\frac{5}{2}\) < x < \(\frac{9}{2}\)

Just one problem: with absolute value inequalities, you should check every single solution.

Test \(\frac{5}{2}\) < x . Okay, suppose x is 3. Plug 3 into original equation

|3 - 1| + |3 - 6| < 2 ???

2 + 3 = 5, and 5 is NOT less than 2. That means the solution is not valid. Eliminate it.

Now try x < \(\frac{9}{2}\). So x could be 4. Plug in:

|4 - 1| + |4 - 6| < 2 ???

3 + 2 = 5, and 5 is NOT less than 2. That solution is not valid either. Eliminate it.

There are ZERO solutions.

gmatcracker2017 , you wrote:

I'm not sure how you got the value in bold from the inequality you posited.

I think RHS should be my 1st case, above, x < \(\frac{9}{2}\)

You are correct; there is no solution. But how would we know that from the inequality you wrote?

Am I missing something?

Conclusion: if you remove absolute value bars, you should test every solution.

guilherme28 , does that help? :-)

of course you will have to test the viability of every solution.....
and yes, x is less than 9/2 ....

thanks

Ans is D

case1 x>6 then (x-1)+(x-6)<2 gives x<9/2 hence no soln
case2 x~(1,6) then (x-1)+(6-x)<2 gives no soln
case3 x<1 then (1-x) + (6-x) <2 gives x>5/2 hence no soln

therefore no solutions are possible

gmatcracker2017 - Hmm, hope I didn't offend. I certainly did not intend offense. I did not suggest that you failed to test viability of solutions.

I simply could not figure out how you got 7/5. I also thought: if 7/5 were correct (which if true = zero solutions b/c no number is greater than 2.5 and simultaneously less than 1.4), no need to test four cases as I had and as guilherme28 might have.

That's all.

I rarely use the critical points method. There was a decent chance I was missing something.

And just in case there is seeming offense elsewhere: Neither did I say that guilherme28 failed to test solutions. I wondered whether or not that might be the case.

If it were, I decided it was better to cover the bases than to leave the impression that the inequality in question led to valid solutions.

IMO, better safe than sorry for those like I who aren't 100 percent fluent in absolute value inequalities.

Regards.

hey man

Its okay...
I am okay with the matter. You know, its possible to solve a PS in many ways. Like you, I also work with critical point method or squaring to help me solve inequalities. Truth be told, I have learnt these methods from gmatclub, from you people. Side by side, however, I was also finding some other way available. So far I know inequality is widely tested on the GMAT. So, I am planing a dexterity on the topic, if not such by now ...

Finally, hope to discuss such issues from time to time in upcoming days. Again I aver I am okay with your kudos ...

thanks

Notice that x - 1 = 0 means x = 1 and x - 6 = 0 means x = 6, so we have 3 cases: (1) x ≥ 6, (2) 1 ≤ x < 6, and (3) x < 1. We need to see what happen in each of these cases. Let’s do that now:

Case 1:

If x ≥ 6, then both x - 1 and x - 6 are positive, so |x - 1| = x - 1 and |x - 6| = x - 6, so |x - 1| + |x - 6| < 2 becomes:

x - 1 + x - 6 < 2

2x - 7 < 2

2x < 9

x < 9/2

However, x < 9/2 contradicts our assumption that x ≥ 6, so there are no solutions when x ≥ 6.

Case 2:

If 1 ≤ x < 6, then x - 1 is positive but x - 6 is negative, so |x - 1| = x - 1 and |x - 6| = -(x - 6) = -x + 6, so |x - 1| + |x - 6| < 2 becomes:

x - 1 + (-x + 6) < 2

5 < 2

We see that 5 < 2 is a false statement. Therefore, there are no solutions when 1 ≤ x < 6, either.

Case 3:

If x < 1, then both x - 1 and x - 6 are negative, so |x - 1| = -(x - 1) = -x + 1 and |x - 6| = -(x - 6) = -x + 6, so |x - 1| + |x - 6| < 2 becomes:

-x + 1 + (-x + 6) < 2

-2x + 7 < 2

-2x < -5

x > 5/2

However, x > 5/2 contradicts our assumption that x < 1, so there are no solutions when x < 1.

We see that there no solutions that satisfy the given inequality.

Answer: D

Hi! In case 3 : doesnt the last statement becomes 2x - 7 < 2. Hence x < 4.5? How did you get x = 7?

|x - 1| + |x - 6| < 2

You can read the absolute value expression as follows:

|x - 1| = distance on the number line between X and +1

|x - 6| = distance on the number line between X and + 6

And the entire inequality tells us that if we add up both distances it should be less than 2 units on the number line.


Now, if you set up a number line and plug in the points:

[—————-(+1)—————————-(+6)—]

If we place x to the left of (+1), the distance from X to +6 is much greater than 2 units

The same if we place X to the right of (+6) on the number line

In fact, the minimum value we can obtain for the expression is if we place X anywhere between 1 and 6, inclusive.

The sum of the two absolute value expressions will equal 5, which is still greater than 2

There is no way we can find a real value of X, such that the sum of the distances on the number line is less than < 2 units

0 solutions

Posted from my mobile device

Hello, can anyone link some sources for understanding the line approach with abs values?

Thanks