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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
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
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
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
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
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)
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
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
Re: How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
[#permalink]
08 Dec 2014, 03:12
Expert Reply
nktdotgupta wrote:
How to Solve: Inequalities
Hi All,
I have learned a lot from gmatclub and am done with my gmat too. So, i have decided to contribute back. As part of this i have decided to share the knowledge i have regarding various topics related to gmat quant. Hope it will be useful. This post is about how to solve "Inequalities"
Theory
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 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
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 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
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 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
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 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
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
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)
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
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
Re: How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
[#permalink]
08 May 2015, 07:27
In number 4 - why did you multiply 7 by (x-1)? I multiplied the whole left hand side of the expression by (x-1) to get rid of (x-1) in the denominator but it seems as if you kept the (x-1) in the denominator when you did so.
Re: How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
[#permalink]
08 May 2015, 09:52
Expert Reply
1
Bookmarks
Hi All,
GMAT Quant questions are typically designed in a way that allows the Test Taker more than one way to solve. In this article, the solutions are essentially all based on "doing algebra", but each of these questions can ALSO be solved by TESTing VALUES.
In many cases, TESTing VALUES is an easier and faster way to get to the correct answer, so beyond learning how to "do the math", you should also put in the proper time to learn (and practice) tactics. You'll find that it's easier to score at a higher level when you have more approaches to choose from.
Re: How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
[#permalink]
07 Jan 2020, 07:14
Top Contributor
Expert Reply
Apologies for replying after 3+years! Yes we can multiply with x-1 on both the sides but since x-1 is negative so we need to make sure that we change the sign of the inequality once we multiply both sides with a negative number (i.e (x-1)). (I did not want to make the solution complex so i did not multiply both the sides with x-1.) Hope it helps!
healthjunkie wrote:
In number 4 - why did you multiply 7 by (x-1)? I multiplied the whole left hand side of the expression by (x-1) to get rid of (x-1) in the denominator but it seems as if you kept the (x-1) in the denominator when you did so.
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