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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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How to Solve: Absolute Value (Basics)


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Absolute Value Basics.pdf [289.69 KiB]
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After this post please go through Absolute Value Problems and Absolute Value + Inequality post


Theory


    What is Absolute Value / Modulus of a number
    Absolute Value on Number Line
    Properties of Absolute Values
    Absolute Value on Number Line Examples


What is Absolute Value / Modulus of a number

• Absolute Value or modulus (|x|) of a real number x is the non-negative value of the number (x), without any consideration to its sign
Ex
|12| = 12
|-12| = 12 (we just the value after ignoring the sign)

• |x| = x for x >0
= -x for x <0
= 0 for x = 0


Q1. Find the value of |-3| + | 2*3 – 4*2| + |25|

Q2. Find the value of | x+y| where x + z = 20 and y – z = -25

Sol1: 3 + | 6-8 | + 25 = 3 + 2 + 25 = 30

Sol2: x + z = 20 and y – z = -25
Adding both of them we get x + y = -5
=> | x+y | = |-5| = 5

Absolute Value on Number Line

• Absolute value of a number x can also be imagined as the distance of that number x from 0 on a number line

Let's say we have two numbers x and y and x is positive and y is negative. What we are saying is
|x| = x = distance of x from origin
|y| = -y = distance of y from origin
As, shown in the image below:

Image

Properties of Absolute Values

• PROP 1: Absolute value of a number is always Non-negative
|a| ≥ 0 for all values of a
Ex: |3| = 3 ≥ 0
|-7| = 7 ≥ 0

• PROP 2: Minimum value of |a| = 0, when a=0
Ex: If |x| =0 => x=0

• PROP 3: Square root of a number is always positive
\(\sqrt{a^2}\) = |?|
Ex:
If x = \(\sqrt{25}\) => x = +5
But if \(x^2\) = 25 => x = ± \(\sqrt{25}\) => x = ±5

• PROP 4: Absolute value of negative of a number is same as absolute value of the number
|-a| = |a|
A derivative of this is
| a-b | = | b-a | because | b-a | = | -(a-b) |

• PROP 5: Product of absolute value of two numbers is same as product of their absolute values
|ab| = |a|*|b|
Ex:
|7*3| = |7| * |3| = 21

• PROP 6: Division of absolute value of two numbers is same as division of their absolute values
\(|\frac{?}{?}| = \frac{{|?|}}{{|?|} }\)
Ex:
\(|\frac{4}{2}| = \frac{{|4|}}{{|2|} }\) = 2

• PROP 7: Sum of absolute value of two numbers is always ≥ absolute value of their sum
|a| + |b| ≥ |a+b|
Ex:
|7| + |3| ≥ |7+3| => 10 ≥ 10
|5| + |-8| ≥ |5 + (-8) | => 13 ≥ 3

• PROP 8: Difference of absolute value of two numbers is always ≤ absolute value of their difference
|a| - |b| ≤ |a-b|
Ex:
|7| - |3| ≤ |7-3| => 4 ≤ 4
|5| - |-8| ≤ |5 - (-8) | => -3 ≤ 13

• PROP 9: Taking absolute value multiple times or taking it once gives the same result
||a|| = |a|
Ex:
||-4|| = |-4| => |4| = |-4| = 4

• PROP 10: If absolute value of difference of two numbers is zero => both numbers are equal
|a-b|=0 => a=b
Ex:
| x-4 | =0 > x=4

Next two will be used a lot to solve absolute values problem!

• PROP 11: If |a| ≤ b => -b ≤ a ≤ b

• PROP 12: If |a| ≥ b => a ≤ -b or a ≥ b

• PROP 13: |\(a^n\)| = \(|a|^n\)
Ex:
|\({-2}^4\)| = \(|-2|^4\) = 16

• PROP 14: |a-b| ≥ ||a|-|b||
Ex:
|7-4| ≥ ||7|-|4|| => 3 ≥ 3
|8-(-2)| ≥ ||8|-|-2|| => 10 ≥ 6

Q1. If |a-3| ≤ 9 then find the range of values of a.

Q2. If |b+5| ≥ 10 then find the range of values of b.

Sol1: -6 ≤ a ≤ 12
Check Video For solution

Sol2: b ≤ -15 or b ≥ 5
Check Video For solution

Absolute Value on Number Line Example

If a and b are two variables given then:
|a-b| always means the distance between points a and b
|a+b| = |a| + |b| when a and b have the same sign and
|a+b| = |b| - |a| when a and b have different sign and |b| > |a|


Case 1: a and b are positive and a > b

Image

|a-b | = |a| - |b|
|a+b| = |a| + |b|

Case 2: a is positive and b is negative

Image

Given: |a| > |b|
|a-b| = |a| + |b|
|a+b| = |a| - |b|

Case 3: Both a and b are negative

Image

Given: |a| > |b|
|a-b| = |a| - |b|
|a+b| = |a| + |b|

Q1. Given the information (below), Simplify |b-a| + |c-b|

Image

Sol:
Method 1
|b-a| = Distance between a and b = AB
|c-b| = Distance between c and b = BC
=> |b-a| + |c-b| = AB = BC = AC = |c-a|

Method 2
|b-a| = |b| - |a|
|c-b| = |c| - |b|
=> |b-a| + |c-b| = |b| - |a| + |c| - |b| = |c| - |a| = |c-a|
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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How to Solve: Absolute Value + Inequality Problems


Attachment:
Absolute Values + Inequality.pdf [270.16 KiB]
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Before reading this post read Absolute Value Basics and Absolute Value Problems



Theory


    How to open |x|
    Basic Property of Absolute value + Inequality
    Problems with one Absolute Value (+ Inequality)
    * Substitution
    * Algebra
    Problems with two Absolute Values (+ Inequality)
    * Substitution
    * Algebra


How to open |x|

|x| = x for x >0
= -x for x <0
= 0 for x = 0

Basic Property of Absolute value + Inequality

• If |x| ≤ a => -a ≤ x ≤ a
• If |x| ≥ a => x ≤ -a or x ≥ a

Problems with one Absolute Value (+Inequality) : Substitution

• In this method we are going to take smart numbers, numbers which will help us eliminate one or more option choices together, to solve the problem.

Q1. Find all possible range of values of x which satisfy |x-3| < 3x - 5?

A. x ≥ 3 and x ≤ 3
B. x ≥ 3 and x < 2
C. x > 3 and x < 2
D. x ≥ 3
E. x > 2


Sol1:
We see that Option B and Option C have x < 2 and A has x ≤ 3 So, if we are able to prove that for x=0 the equation is not satisfied then we can eliminate all three options together.
|x-3| < 3x - 5 => |0-3| < 3*0 - 5 => 3 < -5
Which is not true, so A, B, C are eliminated
For D and E let's pick an option choice between 2 and 3 let's say 2.5 and if it satisfies
|x-3| < 3x - 5 => |2.5-3| < 3*2.5 - 5 => 0.5 < 2.5
which is true, so D is the answer

Q2. Find all possible range of values of x which satisfy |3x-1| ≥ 2x + 4?

A. x ≥ 1/3 and x ≤ 1/3
B. x ≥ 1/3 and x < -3/5
C. x > 5 and x ≤ 1/3
D. x ≥ 5 and x ≤ -3/5
E. x > 5 and x < -3/5

Sol2:

In this case we can see that A and B have x ≥ 1/3 and C,D,E have x > 5, so let's try a value of x between 1/3 and 5 (Say x=1) and see if it satisfies
|3x-1| ≥ 2x + 4 => |3*1-1| ≥ 2*1 + 4 => 2 ≥ 6 which is false
So, A and B are eliminated
For C, D and E we see that D has x ≥ 5 and C and E have x>5. So, let's try x=5 and see
|3x-1| ≥ 2x + 4 => |3*5-1| ≥ 2*5 + 4 => 14 ≥ 14 which is true
So, D is the answer

Problems with one Absolute Value (+Inequality) : Algebra

• In this method we are going to consider two cases, in one case we will assume that the value inside the absolute value is ≥ 0 and in the second case we will assume that the value inside the absolute value is < 0 and solve the two cases.

Q1. Find all possible range of values of x which satisfy |x-3| < 3x - 5?

A. x ≥ 3 and x ≤ 3
B. x ≥ 3 and x < 2
C. x > 3 and x < 2
D. x ≥ 3
E. x > 2


Sol1:
Case 1
x - 3 ≥ 0 => x ≥ 3
=> | x-3 | = x-3
=> x-3 < 3x - 5
=> x > 1
But one condition was x ≥ 3. So, out final answer will be the intersection of x > 1 and x ≥ 3 which is x ≥ 3
[Check out the link for Inequalities in my signature to understand this part in bit more detail]

Case 2
x - 3 < 0 => x < 3
=> | x-3 | = -(x-3)
=> -(x-3) < 3x - 5
=> 4x > 8
=> x > 2
But our condition was x < 3
So, Answer is x ≥ 3

Q2. Find all possible range of values of x which satisfy |3x-1| ≥ 2x + 4?

A. x ≥ 1/3 and x ≤ 1/3
B. x ≥ 1/3 and x < -3/5
C. x > 5 and x ≤ 1/3
D. x ≥ 5 and x ≤ -3/5
E. x > 5 and x < -3/5

Sol2:

Case 1
3x-1 ≥ 0 => x ≥ 1/3
=> |3x-1|= 3x-1
=> 3x-1 ≥ 2x + 4
=> x ≥ 5
Which is inside the range so x ≥ 5 is a solution

Case 2
3x-1 < 0 => x < 1/3
=> |3x-1|= -(3x-1)
=> -(3x-1) ≥ 2x + 4
=> 5x ≤ -3
=> x ≤ -3/5
Which is inside the range so x ≤ -3/5 is a solution
So, D is the answer

Problems with two Absolute Value (+Inequality) : Substitution

• In this method we are going to take smart numbers, numbers which will help us eliminate one or more option choices together, to solve the problem.

Q1. Find all possible range of values of x which satisfy |x+1| + |x+2| > 3x+1 ?

A. x ≥ 2 and x ≤ -2
B. x ≥ -1 and x < -2
C. -2 ≤ x < 2
D. x < 2
E. -2 ≤ x < -1

Sol1:

For A and B we will pick a value of x > 2 (let's say x=3) to eliminate both of these option choices.
|x+1| + |x+2| > 3x+1 => |3+1| + |3+2| > 3*3+1 => 9 > 10
which is not true => A and B are eliminated
For C,D and E Let's pick a value of x which is < -2 (let's say x=-10)
=> |x+1| + |x+2| > 3x+1 => |-10+1| + |-10+2| > 3*-10+1
=> 17 => -29
Which is true => D is the answer

Q2. Find all possible range of values of x which satisfy |2x+3| + |3x+4| < 6x+5 ?

A. x > 2 and x ≤ -3/2
B. -3/2 ≤ x ≤ -4/3
C. x >2 and -3/2 ≤ x ≤ -4/3
D. x ≥ -4/3
E. x > 2


Sol2:
For A, B, C we can substitute x = -3/2 and check if it satisfies
|2x+3| + |3x+4| < 6x+5 => |2*-3/2 + 3| + |3*-3/2 + 4| < 6*-3/2 + 5
which will be false as left hand side will be non-negative and right hand side is negative
So, A,B and C are eliminated
For D and E we can pick a value of x > 2 (Say x=3)and check if it satisfies
|2x+3| + |3x+4| < 6x+5 => |2*3+3| + |3*3+4| < 6*3+5 => 9 + 13 < 23
which is true => E is the answer

Problems with two Absolute Value (+Inequality) : Algebra

• In this method we are going to assume that the value inside the two absolute value is zero and take down the points. We will plot the points on the number line and divide the number line into three parts and then solve after opening the absolute value in these three cases.

Q1. Find all possible range of values of x which satisfy |x+1| + |x+2| > 3x+1 ?

A. x ≥ 2 and x ≤ -2
B. x ≥ -1 and x < -2
C. -2 ≤ x < 2
D. x < 2
E. -2 ≤ x < -1

Sol1:

We will assume x+1=0 and x+2=0 and get x=-1 and x=-2 as two points on the number line. We will plot the points on the number line and split the number line into three parts as shown in the image below

Image

Case 1
x > -1
If x > -1 then take any value of x, let's say x=0 and check if the value inside the two absolute values is positive or negative
Both x+1 and x+2 are positive
=> |x+1| = x+1 and | x+2| = x+2
=> x+1 + x+2 > 3x+1
=> x < 2
But our condition was x > -1
So, our solution will be the intersection of these two which is nothing but -1 < x < 2

Case 2
-2 ≤ x ≤ -1
If -2 ≤ x ≤ -1 then take any value of x, let's say x=-1.5 and check if the value inside the two absolute values is positive or negative
x+1 will be negative and x+2 will be positive
=> |x+1| = -(x+1) and | x+2| = x+2
=> -x-1 + x+2 > 3x+1
=> x < 0
But our condition was -2 ≤ x ≤ -1
So, our solution will be the intersection of these two which is nothing but -2 ≤ x ≤ -1

Case 3
x < -2
If x < -2 then take any value of x, let's say x=-3 and check if the value inside the two absolute values is positive or negative
Both x+1 and x+2 are negative
=> |x+1| = -(x+1) and | x+2| = -(x+2)
=> -x-1 - x-2 > 3x+1
=> x < -4/5
But our condition was x < -2
So, our solution will be x < -2
So, our solution is -1 < x < 2 , -2 ≤ x ≤ -1 and x < -2
So, combined solution is x < 2

Q2. Find all possible range of values of x which satisfy |2x+3| + |3x+4| < 6x+5 ?

A. x > 2 and x ≤ -3/2
B. -3/2 ≤ x ≤ -4/3
C. x >2 and -3/2 ≤ x ≤ -4/3
D. x ≥ -4/3
E. x > 2


Sol2:
We will assume 2x+3=0 and 3x+4=0 and get x=-3/2 and x=-4/3 as two points on the number line. We will plot the points on the number line and split the number line into three parts as shown in the image below

Image

Case 1
x > -4/3
If x > -4/3 then take any value of x, let's say x=0 and check if the value inside the two absolute values is positive or negative
Both 2x+3 and 3x+4 are positive
=> 2x+3 + 3x+4 < 6x+5
=> x > 2
Which is in the range, so x > 2 is one solution

Case 2
-3/2 ≤ x ≤ -4/3
If -3/2 ≤ x ≤ -4/3 then take any value of x, let's say x=-1.4 and check if the value inside the two absolute values is positive or negative
2x+3 will be positive and 3x+4 will be negative
=> 2x+3 - (3x+4) < 6x+5
=> 7x > -6
=> x > -6/7
But our condition was -3/2 ≤ x ≤ -4/3
So, No solution here

Case 3
x < -3/2
If x < -3/2 then take any value of x, let's say x=-10 and check if the value inside the two absolute values is positive or negative
Both 2x+3 and 3x+4 will be negative
=> -2x-3 - (3x+4) < 6x+5
=> 11x > -12
=> x > -12/11
But our condition was x < -3/2
So, No solution here
So, Solution is Option E x > 2
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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2
Bookmarks
How to Solve: Absolute Value Problems


Attachment:
Absolute Values Advanced.pdf [246.32 KiB]
Downloaded 220 times



Before reading this post read Absolute Value Basics Post and after reading this post please read Absolute Value + Inequalities Post

Theory


    Absolute Value of x

    Problems with one Absolute Value
    • Substitution
    • Squaring Both Sides
    • Algebra

    Problems with two Absolute Values
    • Substitution
    • Algebra


Absolute Value of x

|x| = x for x >0
= -x for x <0
= 0 for x = 0


Problems with one Absolute Value : Substitution

• In this method we are going to substitute the answer choices and check which answer choice satisfies the answer.

Q1. Which value of x satisfies |x-3| = 3x?

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


Sol1:
A. |0-3| = 3*0 => 3 = 0 FALSE
B. |-3/2-3| = 3*-3/2 => 9/2 = -9/2 FALSE
C. |3/4-3| = 3*3/4 => 9/4 = 9/4 TRUE, So C is the answer. WE don't need to check further but still showing the work for completing the solution
D. |1-3| = 3*1 => 2 = 3 FALSE
E. |2-3| = 3*2 => 1 = 6 FALSE

Q2. Which value of x satisfies |x-4| = 5 + 2x?

A. 0 and -1/3
B. -1/3
C. 1 and -1/3
D. 2 and -1/3
E. -9 and -1/3


Sol2: Since -1/3 is in all the option choices so we will not check this value
A. 0 and -1/3 => |0-4| = 5 + 2*0 => 4 = 5 FALSE
B. -1/3 => Leave for now
C. 1 and -1/3 => |1-4| = 5 + 2*1 => 3 = 7 FALSE
D. 2 and -1/3 => |2-4| = 5 + 2*2 => 2 = 9 FALSE
E. -9 and -1/3[/b] => |-9-4| = 5 + 2*-9 => 13 = -13 FALSE

So, Answer is B

Problems with one Absolute Value : Squaring Both Sides

• In this case we will be taking absolute value term to one side and everything else to the other side. Then, we will square both the sides to eliminate absolute value and solve the quadratic. We will then substitute the answer choices back in the question and check if the answer satisfies the question or not.

Q1. Which value of x satisfies |x-3| = 3x?

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


Sol1:
|x-3| = 3x
Squaring both the sides we have
\((x-3)^2\) =\( (3x)^2\)
\(x^2\) - 6x + 9 = 9\(x^2\)
=> 8\(x^2\) + 6x - 9 = 0
=> 8\(x^2\) + 12x - 6x - 9 = 0
=> 4x (2x+3) -3 (2x+3) = 0
=> (4x-3) (2x+3) = 0
=> x = 3/4 or -3/2
Now, substitute both the values in the question and check which one satisfies. As seen in the case of substitution only 3/4 will satisfy.

Q2. Which value of x satisfies |x-4| = 5 + 2x?

A. 0 and -1/3
B. -1/3
C. 1 and -1/3
D. 2 and -1/3
E. -9 and -1/3


Sol2:
|x-4| = 5 + 2x
Squaring both the sides we have
\((x-4)^2\) = \((5 + 2x)^2\)
=> \(x^2\) - 8x + 16 = 25 + 20x + 4\(x^2\)
=> 3\(x^2\) + 28x + 9 =0
=> 3\(x^2\) + x + 27x + 9 =0
=> x(3x+1) + 9(3x+1) = 0
=> (x+9) * (3x+1) = 0
=> x = -9, -1/3
Now, substitute both the values in the question and check which one satisfies. As seen in the case of substitution only -1/3 will satisfy.

Problems with one Absolute Value : Algebra

• In this method we are going to use the property of |x|

|x| = x for x ≥ 0
= -x for x < 0


and we will create two cases and solve both of them individually.

Q1. Which value of x satisfies |x-3| = 3x?

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


Sol1:
Case 1: Value inside the absolute value is positive
x-3 ≥ 0 => x ≥ 3
=> |x-3| = x-3
So, |x-3| = 3x => x-3 = 3x
=> x= -3/2
But our condition was x ≥ 3, so this is NOT a solution

Case 2: Value inside the absolute value is negative
x-3 < 0 => x < 3
=> |x-3| = -(x-3)
So, |x-3| = 3x => -(x-3) = 3x
=> x= 3/4
This is true as our condition was x < 3
So, answer is x = 3/4

Q2. Which value of x satisfies |x-4| = 5 + 2x?

A. 0 and -1/3
B. -1/3
C. 1 and -1/3
D. 2 and -1/3
E. -9 and -1/3


Sol2:
Case 1: Value inside the absolute value is positive
x-4 ≥ 0 => x ≥ 4
=> |x-4| = x-4
So, |x-4| = 5 + 2x=> x-4 = 5 + 2x
=> x = -9
But our condition was x ≥ 4, so this is NOT a solution

Case 2: Value inside the absolute value is negative
x-4 < 0 => x < 4
=> |x-4| = -(x-4)
So, |x-4| = 5 + 2x=> -(x-4) = 5 + 2x
=> x= -1/3
This is true as our condition was x < 4
So, answer is x = -1/3

Problems with two Absolute Values : Substitution

• In this problem we are going to substitute the answer choices and check which value satisfies.

Q1. Which value of x satisfies |x+1| + |x+2| = 3x?

A. 3
B. 3 and -1/3
C. 3 and -3/5
D. 3 and 1/3
E. 3, 1/3, -1/3 and -3/5


Sol1: A
Check video for explanation

Q2. Which value of x satisfies |2x-4| + |3x+6| = 6x?

A. 2
B. 2 and -2/11
C. 2 and -10/7
D. 2 , -2/11 and -10/7


Sol2: A
Check video for explanation

Problems with two Absolute Values : Algebra

• In this method we are going to assume that the value inside the two absolute value is zero and take down the points. We will plot the points on the number line and divide the number line into three parts and then solve after opening the absolute value in these three cases.

Q1. Which value of x satisfies |x+1| + |x+2| = 3x?

A. 3
B. 3 and -1/3
C. 3 and -3/5
D. 3 and 1/3
E. 3, 1/3, -1/3 and -3/5


Sol1:
Assuming x+1 = 0 and x+2 =0 we get x = -1 and x = -2
We will plot these two points on the number line and take three cases as shown in the image below

Image

Case 1
x > -1
If x > -1 then take any value of x, let's say x=0 and check if the value inside the two absolute values is positive or negative
Both x+1 and x+2 are positive
=> |x+1| = x+1 and | x+2| = x+2
=> x+1 + x+2 = 3x
=> x = 3
Which is true as our condition was x > -1

Case 2
-2 ≤ x ≤ -1
If -2 ≤ x ≤ -1 then take any value of x, let's say x=-1.5 and check if the value inside the two absolute values is positive or negative
x+1 will be negative and x+2 will be positive
=> |x+1| = -(x+1) and | x+2| = x+2
=> -x-1 + x+2 = 3x
=> x = 1/3
But our condition was -2 ≤ x ≤ -1 so it is NOT a solution

Case 3
x < -2
If x < -2 then take any value of x, let's say x=-3 and check if the value inside the two absolute values is positive or negative
Both x+1 and x+2 are negative
=> |x+1| = -(x+1) and | x+2| = -(x+2)
=> -(x+1) + -(x+2) = 3x
=> x = -3/5
But our condition was x < -2 so it is NOT a solution
So, solution is x = 3

Q2. Which value of x satisfies |2x-4| + |3x+6| = 6x?

A. 2
B. 2 and -2/11
C. 2 and -10/7
D. 2 , -2/11 and -10/7


Sol2:
Assuming 2x-4 = 0 and 3x+6 =0 we get x = 2 and x = -2
We will plot these two points on the number line and take three cases as shown in the image below

Image

Case 1
x > 2
If x > 2 then take any value of x, let's say x=3 and check if the value inside the two absolute values is positive or negative
Both 2x-4 and 3x+6 are positive
=> |2x-4| = 2x-4 and | 3x+6| = 3x+6
=> 2x-4 + 3x+6 = 6x
=> x = 2
Now 2 is in the boundary and if you will check then 2 actually satisfies the answer. (If we would have taken x ≥ 2 then 2 would be in the answer choice, we will get 2 answer in the second case now)

Case 2
-2 ≤ x ≤ 2
If -2 ≤ x ≤ 2 then take any value of x, let's say x=0 and check if the value inside the two absolute values is positive or negative
2x-4 will be negative and 3x+6 will be positive
=> |2x-4| = -(2x-4) and | 3x+6| = 3x+6
=> -2x+4 + 3x+6 = 6x
=> x = 2
Which is true as our range was -2 ≤ x ≤ 2

Case 3
x < -2
If x then take any value of x, let's say x=-3 and check if the value inside the two absolute values is positive or negative
Both 2x-4 and 3x+6 are negative
=> |2x-4| = -(2x-4) and | 3x+6| = -(3x+6)
=> -2x+4 - 3x-6 = 6x
=> x = -2/11
But our condition was x < -2 so it is NOT a solution
So, solution is x = 2
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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Absolute Value/Modulus Questions by Tag



Quantitative Comparison Questions



Multiple-choice Questions — Select One Answer Choice




Multiple-choice Questions — Select One or More Answer Choices



Numeric Entry Question



Data Interpretation Sets

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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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1) GRE Lesson: Absolute Value Basics





2) GRE Practice Question: Absolute Values Advanced


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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
thank you for the post
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
Carcass wrote:
[align=center]
3-steps approach for complex problems

Let’s consider following examples,

Example #1
Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?
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) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

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

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

(Optional) The following illustration may help you understand how to open modulus at different conditions.
Image
Answer: 0



Please explain how values for x were solved in a, b, c, d. I've seen this approach (with illustration) a couple of times but could never understand it.
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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computerbot wrote:
Please explain how values for x were solved in a, b, c, d. I've seen this approach (with illustration) a couple of times but could never understand it.


Now the question given to us here is |x+3| - |4-x| = |8+x|

To arrive at the four cases listed above we need to do the following

1. Write down all the expressions inside Absolute value and equate each one of them with 0.
x + 3 = 0
4 - x = 0
8 + x = 0

2. Then find the corresponding values of x in all the cases
=> x = -3
=> x = 4
=> x = -8

3. Draw these points on the number line (as shown below) and take cases with points arrived in step 2. Idea is to cover the entire number line

Attachment:
-8 -3 4.png
-8 -3 4.png [ 5.43 KiB | Viewed 11069 times ]


4. Take the range arrived in step 3 as the conditions for x and proceed as described above

x < -8
-8 ≤ x < -3
-3 ≤ x < 4
4 ≤ x

Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems

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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
BrushMyQuant wrote:
computerbot wrote:
Please explain how values for x were solved in a, b, c, d. I've seen this approach (with illustration) a couple of times but could never understand it.


Now the question given to us here is |x+3| - |4-x| = |8+x|

To arrive at the four cases listed above we need to do the following

1. Write down all the expressions inside Absolute value and equate each one of them with 0.
x + 3 = 0
4 - x = 0
8 + x = 0

2. Then find the corresponding values of x in all the cases
=> x = -3
=> x = 4
=> x = -8

3. Draw these points on the number line (as shown below) and take cases with points arrived in step 2. Idea is to cover the entire number line

Attachment:
-8 -3 4.png


4. Take the range arrived in step 3 as the conditions for x and proceed as described above

x < -8
-8 ≤ x < -3
-3 ≤ x < 4
4 ≤ x


Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems



Thank you for the explanation. However, the thing that I don't get is that
a. How the absolute signs were opened: −(x+3)−(4−x)=−(8+x)
b. How did we get x=-1, x=-15, x=9, and x=-1 in condition a,b,c,d respectively.
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
Q1. Find all possible range of values of x which satisfy |x-3| < 3x - 5?

A. x ≥ 3 and x ≤ 3
B. x ≥ 3 and x < 2
C. x > 3 and x < 2
D. x ≥ 3
E. x > 2

Sol1:
We see that Option B and Option C have x < 2 and A has x ≤ 3 So, if we are able to prove that for x=0 the equation is not satisfied then we can eliminate all three options together.
|x-3| < 3x - 5 => |0-3| < 3*0 - 5 => 3 < -5
Which is not true, so A, B, C are eliminated
For D and E let's pick an option choice between 2 and 3 let's say 2.5 and if it satisfies
|x-3| < 3x - 5 => |2.5-3| < 3*2.5 - 5 => 0.5 < 2.5
which is true, so D is the answer


Answer should be E.
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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Re: GRE Quant - Absolute Values/Modulus Theory [#permalink]
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