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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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TooLong150 wrote:
Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

Responding to pm.

Absolute value properties:

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.

STEP BY STEP SOLUTION:

We have three transition points for $$|x+3| - |4-x| = |8+x|$$: -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. $$x<-8$$;
2. $$-8\leq{x}\leq{-3}$$;
3. $$-3<x<4$$
4. $$x\geq{4}$$

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When $$x<-8$$, then $$x+3$$ is negative, $$4-x$$ is positive and $$8+x$$ is negative. Thus $$|x+3|=-(x+3)$$, $$|4-x|=4-x$$ and $$|8+x|=-(8+x)$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$-(x+3) - (4-x) =-(8+x)$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x<-8$$).

2. When $$-8\leq{x}\leq{-3}$$, then $$x+3$$ is negative, $$4-x$$ is positive and $$8+x$$ is positive. Thus $$|x+3|=-(x+3)$$, $$|4-x|=4-x$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$-(x+3) - (4-x) =8+x$$: --> $$x=-15$$. This solution is NOT OK, since $$x=-15$$ is NOT in the range we consider ($$-8\leq{x}\leq{-3}$$).

3. When $$-3<x<4$$, then $$x+3$$ is positive, $$4-x$$ is positive and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=4-x$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (4-x) =8+x$$: --> $$x=9$$. This solution is NOT OK, since $$x=9$$ is NOT in the range we consider ($$-3<x<4$$).

4. When $$x\geq{4}$$, then $$x+3$$ is positive, $$4-x$$ is negative and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=-(4-x)=x-4$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (x-4) =8+x$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x\geq{4}$$).

Thus no value of x satisfies $$|x+3| - |4-x| = |8+x|$$.

Hope it's clear.

Is this the fastest way to do this problem? It took me 4 minutes to do this problem.

I did the question by a graphical method. With absolute values, it is sometimes easier to draw graphs and evaluate the questions graphically.

The given question will have 'n' solutions if the 3 lines given by the equations:

y=|x+8|
y=|x+3|
y=|4-x|

once you do that, it becomes apparent that there are no points that are points of intersections of 3 lines (for us to get a solution, we need to have 3 of the lines intersecting at some common points!). Attached is the graph for the same (Sets of parallel lines are: {A||B||C} and {D||E||F}). This method will be an overkill for simpler problems though.

Hope this helps
Attachments Alternate method.png [ 14.38 KiB | Viewed 3096 times ]

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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I was going to reply and say that the graphical/visual way may work best for many cases but I saw that the poster above me has already covered it.

For many absolute values that are simple addition and subtraction, you might get to the answer quicker and more accurately if you just draw it out, especially if its in the form of abs(x-b) + abs(x+a) etc without a 4x-c or kx-c
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|x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.

Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)

d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help
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Posts: 8164
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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riteshpatnaik wrote:
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)

d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help

Hi,

you have to pick a VALUE falling in the range you are taking and see what happens to the value within the MOD..
1) if the solution of the MOD is a negative number, add a -ive sign..
2)if the solution of the MOD is a positive number, add a +ive sign..

let me show with some examples --
C) -3 <= x < 4 ----- (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)
take x as 0 as it falls in the given range -3 <= x < 4
see what happens to each MOD at this value
i) |x+3|.. 0+3=3 so + sign in front of MOD .. (x+3)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

take the value as -5..
|x+3|.. -5+3=-2 so a negative sign.. -(x+3)
|4-x|.. 4-(-5)=9.. so +sign.. (4-x)
|8+x|.. 8-5=3 so +sign... (8+x)

Hope it helps
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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riteshpatnaik wrote:
Hi, I know how to go about the above problem but the problem is with the signs. I saw this solution also at places but I am confused with the signs

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x. why |4-x| is –(4-x) and |8+x| is (8+x)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> here x < 0 hence |x| = -x and x>0, hence |x| = x . why |4-x| is –(4-x) and |8+x| is (8+x)

d) x >=4. (x+3) + (4-x) = (8+x) --> here x>0, hence |x| = x why |4-x| is +(4-x)
The signs are all reverse for – |4-x| & |8+x|of what I started off with. Please help

Let me first tell you that in this question, x < 0 has no significance.

|x| = x when x >= 0
|x| = -x when x < 0

On the same lines,
|x + 4| = x + 4 when (x + 4) >= 0
|x + 4| = -(x + 4) when (x + 4 ) < 0

In the first definition, x is just a placeholder for any expression.

|x^2 - 8| = x^2 - 8 when (x^2 - 8) >= 0
|x^2 - 8| = -(x^2 - 8) when (x^2 - 8) < 0

So, how do you get rid of |x + 3| in the original question? You take two cases: (x + 3) >= 0 or (x + 3) < 0
|x + 3| = x + 3 when (x + 3) >= 0 (i.e. when x >= -3)
|x + 3| = -(x + 3) when (x + 3) < 0 (i.e. when x < -3)
That is how you get -3 as a transition point.

Do the same for other expressions.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !
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Joined: 16 Oct 2010
Posts: 9783
Location: Pune, India
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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ameyaprabhu wrote:
I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

Note that $$(a + b)^2 = a^2 + b^2 + 2ab$$

So $$(|x+3| - |4-x|)^2 =|x + 3|^2 + |4 - x|^2 - 2*|x + 3|*|4 - x|$$

You missed out the last term. You would need to square it yet again and that will complicate the question further.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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ohh right...thanks.

but if there is a situation where |a-b| = |e+f| (variables or constants), then I can safely square them right? Or do I need to keep certain things in mind before doing that.

VeritasPrepKarishma wrote:
ameyaprabhu wrote:
I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

Note that $$(a + b)^2 = a^2 + b^2 + 2ab$$

So $$(|x+3| - |4-x|)^2 =|x + 3|^2 + |4 - x|^2 - 2*|x + 3|*|4 - x|$$

You missed out the last term. You would need to square it yet again and that will complicate the question further.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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Bunuel wrote:
guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

Responding to pm.

Absolute value properties:

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.

STEP BY STEP SOLUTION:

We have three transition points for $$|x+3| - |4-x| = |8+x|$$: -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. $$x<-8$$;
2. $$-8\leq{x}\leq{-3}$$;
3. $$-3<x<4$$
4. $$x\geq{4}$$

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When $$x<-8$$, then $$x+3$$ is negative, $$4-x$$ is positive and $$8+x$$ is negative. Thus $$|x+3|=-(x+3)$$, $$|4-x|=4-x$$ and $$|8+x|=-(8+x)$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$-(x+3) - (4-x) =-(8+x)$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x<-8$$).

2. When $$-8\leq{x}\leq{-3}$$, then $$x+3$$ is negative, $$4-x$$ is positive and $$8+x$$ is positive. Thus $$|x+3|=-(x+3)$$, $$|4-x|=4-x$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$-(x+3) - (4-x) =8+x$$: --> $$x=-15$$. This solution is NOT OK, since $$x=-15$$ is NOT in the range we consider ($$-8\leq{x}\leq{-3}$$).

3. When $$-3<x<4$$, then $$x+3$$ is positive, $$4-x$$ is positive and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=4-x$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (4-x) =8+x$$: --> $$x=9$$. This solution is NOT OK, since $$x=9$$ is NOT in the range we consider ($$-3<x<4$$).

4. When $$x\geq{4}$$, then $$x+3$$ is positive, $$4-x$$ is negative and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=-(4-x)=x-4$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (x-4) =8+x$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x\geq{4}$$).

Thus no value of x satisfies $$|x+3| - |4-x| = |8+x|$$.

Hope it's clear.

Hi Bunuel,

I am getting confused at the sign changes. Please explain a bit clearly.

We have three transition points for $$|x+3| - |4-x| = |8+x|$$: -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. $$x<-8$$;
2. $$-8\leq{x}\leq{-3}$$;
3. $$-3<x<4$$
4. $$x\geq{4}$$

Now we have four ranges:

As per property...................... |x| < 0 ... | some expression | = - some expression ---> Property 1
|x| > 0 ... | some expression | = some expression ----> Property 2.

Confirm the below:
Now case 1 : x < -8 then |x+3| = - (x+3) and |x-4| = - ( x-4) ( this has to -ve but you mentioned as positive ) and |x+8| = - (x +8 )

and confused for the other three ranges as well on how the expressions got +ve and -ve signs.
What all need to be checked to give positive and negative signs to the expressions, as you mentioned in the solution for the above four ranges.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.

Responding to a pm:
[quote]

Request you to throw some more light on this concept
I have confusion why we have taken x<-8 and why not x>-8

[/quote

The transition points are -8, -3 and 4.

Draw them on the number line:

------------------- (-8) ----------- (-3) --------------------------- (4) -------------

In different sections of this number line, the terms are going to behave differently.
When x < -8,
|x + 8| = - (x + 8)
For the other terms too, when we remove the absolute value sign, we need a negative sign.

At x = -8, the sign for |x +8| turns.
When x >= -8 but less than -3, then
|x+8| = x + 8
For the other terms, when we remove the absolute value sign, we need a negative sign.

Hence each of the four sections of the number line are considered separately.

For more on this, check: http://www.veritasprep.com/blog/2014/06 ... -the-gmat/
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

A faster way to solve this is to notice that:
|x+3|-|4-x|=|8+x|>=0

this give us 2 equations:
1. |x+3|-|4-x|>=0 - > solving 4 cases here (which 2 cases are not possible) give us x>=1/2 and x<=1/2 -> 1/2 is the only value for all the 4 options
2. |8+x|>=0 -> give us x>=-8 or x<=-8 -> which give us 1 value x=1/2

Those 2 equations should have a shared area on the number line, and since they do no have it, there is no solution for this euqation -> 0 values.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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johnwesley wrote:
You do it in the proper way. This method is called the "critical values" method. And once you have the critical values (by doing each absolute term equal to zero), you have to place them in the Real numbers line to make all the possible intervals. Then you just do the intervals as follows: x<lowest number in your real line, and then you take the intervals from each critical value to before the next one: i.e. x<-8, -8<=x<-3, -3<=x<4, x<=4. Therefore, you are getting all the possible intervals in the real line, and splitting the intervals from one critical value (including it) to before the next critical value (not including it).

Then, as you have done, you just set the predominant sign for each term under each condition, you solve the equation, and finally you check if the result saqtisfies the condition.

--------------------------------------------------------------------------------

I find the above technique easier for these problems.

Can someone please confirm if my below approach is correct :

The roots/criticial values for the given equations are : -8,-3,4.

Now placing these values on the no. line we have four ranges & checking for each ranges :

1. For x greater than 4

Randomly selecting any value greater than 4 , lets say x = 10

|x+3| - |4-x| = |8+x|

|10+3| - |4-10| is not equal to |8+10|

So any value greater than 4 doesnot satisfy the equation.

2. For x between -3 & 4

Let x = 2
|x+3| - |4-x| = |8+x|

|2+3| - |4-2| is not equal to |8+2|

So any value between -3 & 4 doesnot satisfy the equation.

3. For x between -8 & -3,

Let x = -2

|x+3| - |4-x| = |8+x|

|-2+3| - |4-(-2)| is not equal to |8+(-2)|

So any value between -8 & -3 doesnot satisfy the equation.

4. For x lesser than -8,

Let x = -9

|x+3| - |4-x| = |8+x|

|-9+3| - |4-(-9)| is not equal to |8+(-9)|

So any value less than -8 doesnot satisfy the equation.

So the total solution is 0.

Please confirm if my above approach is correct .

Thanks
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|x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.

hi mam

in "a" when assumed that x is less than -8, the RHS modulus (8+x) is correctly negated. I am okay with this operation. Side by side, however, the LHS modulus (x + 3) is also negated...

what is the main logic underlying this negation ..? is this such that to make the LHS negative, some number should be subtracted from some bigger negatives..? if so, how can we conclude that (x+3) is bigger in value than is (4-x) ..?

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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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ssislam wrote:
VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .

Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?

Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 4)

thanks !

|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.

hi mam

in "a" when assumed that x is less than -8, the RHS modulus (8+x) is correctly negated. I am okay with this operation. Side by side, however, the LHS modulus (x + 3) is also negated...

what is the main logic underlying this negation ..? is this such that to make the LHS negative, some number should be subtracted from some bigger negatives..? if so, how can we conclude that (x+3) is bigger in value than is (4-x) ..?

wow..i think i have got it...

when x is less than -8, (x+3) got negative and hence this is negative ... hooray..
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|x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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Hello Bunuel, Can you explain me the following.

When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?

Posted from my mobile device
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Joined: 02 Sep 2009
Posts: 59039
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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parthabar wrote:
Hello Bunuel, Can you explain me the following.

When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?

Posted from my mobile device

When $$x\geq{4}$$, then $$x+3$$ is positive, $$4-x$$ is negative and $$8+x$$ is positive. Thus $$|x+3|=x+3$$, $$|4-x|=-(4-x)=x-4$$ and $$|8+x|=8+x$$.

Therefore for this range $$|x+3| - |4-x| = |8+x|$$: transforms to $$x+3 - (x-4) =8+x$$: --> $$x=-1$$. This solution is NOT OK, since $$x=-1$$ is NOT in the range we consider ($$x\geq{4}$$).

Thus no value of x satisfies $$|x+3| - |4-x| = |8+x|$$.

Check step-by-step solution here: https://gmatclub.com/forum/x-3-4-x-8-x- ... l#p1241355

Hope it helps.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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parthabar wrote:

Hello Bunuel, Can you explain me the following.

When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?

Posted from my mobile device

When x≥4
x≥4
, then x+3
x+3
is positive, 4−x
4−x
is negative and 8+x
8+x
is positive. Thus |x+3|=x+3
|x+3|=x+3
, |4−x|=−(4−x)=x−4
|4−x|=−(4−x)=x−4
and |8+x|=8+x
|8+x|=8+x
.

Therefore for this range |x+3|−|4−x|=|8+x|
|x+3|−|4−x|=|8+x|
: transforms to x+3−(x−4)=8+x
x+3−(x−4)=8+x
: --> x=−1
x=−1
. This solution is NOT OK, since x=−1
x=−1
is NOT in the range we consider (x≥4
x≥4
).

Thus no value of x satisfies |x+3|−|4−x|=|8+x|
|x+3|−|4−x|=|8+x|
.

I can understand the solution for x>4 , which turns out to be x=-1 and not in the range we considered.
But I am not able to understand the part when we take boundary value say x=4 , then 4-x=4-4=0 , and we know |x|=x ,x>=0, so why are we not considering this point in solution. So why are we considering |4-x| =-(4-x) only for x>4 ?
I think I am missing something here .Please explain.
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Posts: 59039
Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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parthabar wrote:
I can understand the solution for x>4 , which turns out to be x=-1 and not in the range we considered.
But I am not able to understand the part when we take boundary value say x=4 , then 4-x=4-4=0 , and we know |x|=x ,x>=0, so why are we not considering this point in solution. So why are we considering |4-x| =-(4-x) only for x>4 ?
I think I am missing something here .Please explain.

The solution considers the range when x >= 4, so it must be true for x = 4 too bit let's still consider x = 4 case separately.

If x = 4, then |x+3|−|4−x|=|8+x| will be |4+3|−|4−4|=|8+4| --> |7|−|0|=|12| --> 7 = 12, which is NOT true, so x = 4 is not a solution for given equation. This is exactly what we got in the solution, when we considered x > = 4 range: NO solution in this range.
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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Are there any official questions which have appeared on GMAT Tests or has anyone ever faced such a question in the actual GMAT?
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Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation  [#permalink]

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|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

Bunuel

Kindly check my approach for questions like these

By this i believe we can solve these questions within 1 min and save time for other hard questions.

My approach is more conceptual rather than typical.

Using the Distance approach.

we can plot the critical points as mentioned below

Case -2 Case-1

(X) (X)

-------(-8)---------(-3)-----------(4)--------------

LHS=Dist of X from (-3) - Dist of X from (4)
RHS=Dist of X from (-8)

Here it is interesting to note that since RHS = dist/mod so it has to be positive, so LHS has to be positive as well.(since LHS=RHS)

So for LHS to be Positive-->Dist of X from (-3) > Dist of X from (4)
So X has to be on the RHS of (-3) on the number line.

We can take 2 cases here

Case 1 ;

X is on RHS of (4)

So the above Equality cannot hold ,since Dist of X from (8) is always greater the Diff {of Dist (-3) and (4) from X}

Case 2;

X is between (-3) and (4)

Since we know that Dist of (-3) > Dist of (4) from X
So X has to the RHS of (0.5) [mid point of (-3) and (4)]

Even in this case the above Equality cannot hold ,since Dist of X from (8) is always greater the Diff {of Dist (-3) and (4) from X}

So there is no solution for the equation above.

Bunuel ,do see and correct me , if there is any gap in my understanding.

Regards,

Mahi

Posted from my mobile device Re: |x+3| - |4-x| = |8+x|. How many solutions does the equation   [#permalink] 19 Feb 2019, 05:47

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