Hi All,

I have posted a video on YouTube to discuss two method of How to Solve 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 Post first 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)

**$ 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

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 "-"

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

(x-3)/(x-4)> 0

Method 1: Algebra

*Example Problem*(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 (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

(x+1)(x-1) < 0

Method 1: Algebra

*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

(x-2)/(x+3)< 0

Method 1: Algebra

*Example Problem*(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 (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

Source: Unknown

**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)

Source: Economist GMAT Tutor

**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

Source: OG

**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

Source: Economist GMAT Tutor

**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

Source: Unknown

Hope it helps!

Good Luck!