How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method

How to Solve: Inequality Problems

Attached pdf of this Article as SPOILER at the top! Happy learning!

Hi All,

I have posted a video on YouTube to discuss two method of solving Inequality Problems



Following is Covered in the Video

Theory

Combining Inequalities
Recap of 4 types of Inequality Problems
Solving Linear Inequalities : Method 1: Algebra
Solving Linear Inequalities : Method 2: Sine Wave Method / Wave Method / Wavy method

Before going through this post it is recommended to Check out Inequality Basics Postfirst and then start with this

Combining Inequalities

We are discussing this because we will use this in solving problems using the algebra method



If after solving an inequality equation we are getting x ≥ -2 and x ≥ 1 as two solutions then our final solution will be x ≥ 1
As it is the intersection/common part of both the inequalities
(As shown in orange in above figure)

4 types of Inequality Problems solved using Algebra and Sine Wave Method

There are mainly four types of inequality problems which you would need to solve:--

TYPE 1
x*y > 0
When xy > 0 then we know that both x and y can be either positive or both can be negative
i.e. both x and y have the same sign
so, we have
x>0, y>0 or x<0,y<0

Example Problem
(x-1)*(x-2) > 0
Method 1: Algebra
So, we have two cases
Case 1
both (x-1) and (x-2) are positive
so, x-1 > 0 => x > 1
and x-2 > 0 => x > 2
Intersection of the two cases is x >2

Case 2
both (x-1) and (x-2) are negative
so, x-1 < 0 => x < 1
and x-2 < 0 => x < 2
Intersection of the two cases is x < 1

So, Solution to the question is x < 1 or x > 2

Method 2: Sine Wave Method / Wave Method / Wavy method

In this method we are going to use a sine wave method to solve the problem. Just a quick preview, sine wave is a continuous curve which oscillates between a minimum and a maximum value below and above the base line respectively. Sample image below:



Let's attempt to solve (x-1)*(x-2) > 0 using Sine Wave Method

* Remember that in order to solve the problems using the sine wave method we need to have the coefficient of x positive. [ Check out the last part of the video to go through this ]

To solve an inequality using this method we find out the intersection points by equating the inequality to 0
=> (x-1)*(x-2) = 0
=> x = 1 or 2

Now, we plot these two points on the number line as shown below



Then we are going to draw a sine curve

Starting from right top
Going down at the first solution which is 2 in this case and then
Coming up in the second solution which is 1 in this case and
Going down in the third solution if it is there (in this it is not there

Now we will start marking + and - as mentioned below:

Any Area (in-between) above the number line and below the sine curve is marked as "+" and
Any Area (in-between) below the number line and above the sine curve is marked as "-" as shown below

Now, get your answer as below:

If the inequality in the question is > 0 then pick all the ranges which are "+"
If the inequality in the question is < 0 then pick all the ranges which are "-"

In our case the question was (x-1)*(x-2) > 0 so we will pick all "+" areas which are
x > 2 and x < 1

If the question was (x-1)*(x-2) < 0 then we will pick all "-" areas which are
1 < x < 2

Note that if the question has ≥ or ≥ then we need to check for the border conditions too
Ex: if question was (x-1)*(x-2) ≥ 0 then we need to check the border condition of x = 1 and x = 2 manually and see if we want to include it in the answer or not.


TYPE 2
x/y > 0
When x/y > 0 then we know that both x and y can be either positive or both can be negative
i.e. both x and y have the same sign
so, we have
x>0, y>0 or x<0,y<0

Example Problem
\(\frac{(x-3)}{(x-4)}\) > 0
Method 1: Algebra
So, we have two cases
Case 1
Both (x-3) and (x-4) are positive
=> x-3 > 0 => x>3
And x-4 > 0 => x>4
Intersection of the two cases is x > 4

Case 2
Both (x-3) and (x-4) are negative
=> x-3 < 0 => x < 3
and x-4 < 0 => x < 4
Intersection of the two cases is x < 3

So, solution to the question is x < 3 or x > 4

Method 2: Sine Wave Method / Wave Method / Wavy method

Point of intersections:
x - 3 = 0 and x-4 = 0
=> x = 3, 4

Refer below image



Since question is \(\frac{(x-3)}{(x-4)}\) > 0
So, we will pick "+" area regions
So, answer is x < 3 and x > 4

TYPE 3
x*y < 0
When x*y < 0 then we know that that
(x can be positive and y will be negative) or (x can be negative and y will be positive)
i.e. x and y have opposite signs
so, we have
x>0, y<0 or x<0,y>0

Example Problem
(x+1)(x-1) < 0
Method 1: Algebra
So, we will have two cases
Case 1
(x+1) is positive and (x-1) is negative
=> x + 1 > 0 => x > -1
And x - 1 < 0 => x < 1
Intersection of the two cases is
-1 < x < 1

Case 2
(x+1) is negative and (x-1) is positive
=> x+1 < 0 => x < -1
And x-1 > 0 => x > 1
The two cases have no intersection. So, no solution from this case

So, solution of the problem is -1 < x < 1

Method 2: Sine Wave Method / Wave Method / Wavy method

Point of intersections:
x + 1 = 0 and x - 1 = 0
=> x = -1, 1

Refer below image



Since question is (x+1)(x-1) < 0
So, we will pick "-" area regions
So, answer is -1 < x < 1

TYPE 4
x/y < 0
When x/y < 0 then we know that that
(x can be positive and y will be negative) or (x can be negative and y will be positive)
i.e. x and y have opposite signs
so, we have
x>0, y<0 or x<0,y>0

Example Problem
\(\frac{(x-2)}{(x+3)}\) < 0
Method 1: Algebra
So, we will have two cases
Case 1
(x-2) is positive and (x+3) is negative
=> x-2 > 0 => x > 2
And x+3 < 0 or x < -3
There is no intersection of the two cases. So, no solution from this case

Case 2
(x-2) is negative and (x+3) is positive
=> x-2 < 0 => x < 2
And x+3 > 0 => x > -3
Intersection of the two cases is -3 < x < 2

So, Solution of the question is -3 < x < 2

Method 2: Sine Wave Method / Wave Method / Wavy method

Point of intersections:
x - 2 = 0 and x + 3 = 0
=> x = 2, -3

Refer below image



Since question is \(\frac{(x-2)}{(x+3)}\) < 0
So, we will pick "-" area regions
So, answer is -3 < x < 2

SUGGESTION: Try solving inequalities, they are not tough after all!

Problems:

1. x(x-1) > 0. Then value of x will be?

A. x > 0 and x > 1
B. x < 0 and x > 1
C. x < 0 and x < 1
D. x > 0 and x < 1

Solution:
x*(x-1) > 0
this is of the form xy>0 i.e. x and y have the same sign
so,
(1) either, x > 0 and x-1 >0
i.e. x >0 or x>1
taking intersection of the two possibilities we have x >1

(2)or x <0, and x-1 < 0
i.e. x <0 or x<1
taking intersection of the two possibilities we have x < 0

So, Answer will be B

link to the problem:
https://gmatclub.com/forum/x-x-189656.html


2. Which of the following describes all the values of y for which y < y^2 ?

A. 1 < y
B. −1 < y < 0
C. y < −1
D. 1/y < 1
E. 0 < y < 1

Solution:
The question can be written as
y^2 - y > 0
s=> y*(y-1) > 0
It is of the form xy > 0
So, we will have two cases
Case 1
Both y and y-1 are positive
=> y > 0
And y-1 > 0 => y > 1
Intersection of the two cases is y > 1

Case 2
Both y and y-1 are negative
=>y < 0
And y -1 < 0 => y < 1
Intersection of the two cases is y <0

So, solution to the problem is y < 0 or y > 1

So, Answer will be D
(As option D can be written as
1/y - 1 < 0
or, (1-y)/y < 0
or (y-1)/y > 0
And solution to this will be same as that of y*(y-1) > 0)

Link to the problem:
https://gmatclub.com/forum/which-of-the- ... 61602.html


3. Which of the following describes all values of x for which 1–x^2 >= 0?

(A) x >= 1
(B) x <= –1
(C) 0 <= x <= 1
(D) x <= –1 or x >= 1
(E) –1 <= x <= 1

Solution:
Question can be written as
x^2 - 1 <=0
=> (x+1)*(x-1) <=0
Case 1
x+1 is positive or 0 and x-1 is negative or 0
=> x+1 >= 0 => x >= -1
And x-1 <= 0 => x <= 1
Intersection is -1 <= x <= 1

Case 2
x+1 is negative or 0 and x-1 is positive or 0
x+1 <=0 => x <= -1
And x-1 >= 0 => x >= 1
No intersection in this case

So, solution to the problem is -1 <= x <= 1
So, Answer will be E

Link to the problem
https://gmatclub.com/forum/which-of-the- ... 44461.html


4. If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Solution:
(3+5y)/(x−1) < −7
=> (3+5y)/(x−1) + 7 < 0
=> ((3+5y) + 7*(x-1) )/ (x-1) < 0
=> ( 7x + 5y -4 )/ (x-1) < 0
Now, we know that x < 0 so, x- 1 < 0
in (7x + 5y -4 )/ (x-1) < 0
we know that x - 1 < 0
=> (7x + 5y -4 ) > 0
So, Answer will be B

Link to the problem
https://gmatclub.com/forum/if-y-0-x-and- ... 55220.html


5. Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2

Solution:
k^2 + k - 2 > 0
=> (k+2)*(k-1) > 0
So, we will have two cases
Case 1
Both k+2 and k -1 positive
k+2 > 0 and k-1 > 0
=> k > -2 and k > 1
Intersection is k > 1

Case 2
Both k+2 and k-1 negative
k+2 < 0 and k -1 < 0
=> k < - 2 and k < 1
intersection is k < -2
So, Solution to the problem is k> 1 or k < -2

So, STAT1 is not SUFFICIENT
STAT2 is SUFFICIENT

So, Answer will be B

Link to the problem
https://gmatclub.com/forum/is-k-2-k-147133.html

Hope it helps!
Good Luck!