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How to Solve: Absolute Value + Inequality Problems
Hi All,
I have posted a video on YouTube to discuss two method of solving Absolute Value + Inequality Problems
Before reading this post read
Absolute Value Basics and
Absolute Value ProblemsAttached pdf of this Article as SPOILER at the top! Happy learning! Absolute Values Theory and Solved Problems Playlist Link hereFollowing is Covered in the Video
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 > 2Sol1: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 E 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 > 2Sol1:Case 1x - 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 2x - 3 < 0 => x < 3
=> | x-3 | = -(x-3)
=> -(x-3) < 3x - 5
=> 4x > 8
=> x > 2
But our condition was x < 3
So, Answer in this case will be 2 < x < 3
So, final answer is 2 < x < 3 and x ≥ 3
So, Answer is x > 2
So, E is correct 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:Case 13x-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 23x-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 > 2Sol2: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
Case 1x > -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 3x < -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 > 2Sol2: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
Case 1x > -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 3x < -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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