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Joined: 02 Jan 2020
Status:GRE Quant Tutor
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GRE Quant - Solving Linear Inequalities Theory
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10 Jun 2020, 12:01
10 Jun 2020, 12:01
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Frequency of the concepts tested: Medium
Theory
⸙ What is Inequality and Types of Inequalities
⸙ Graphing Inequalities
⸙ Properties of Inequalities
⸙ Types of Inequality Problems
● x*y > 0
● x/y > 0
● x*y < 0
● x/y < 0
⸙ Basic Problems on Inequalities
⸙ Combining Inequalities
⸙ Four types of Inequality Problems solved using two methods
● Method 1: Algebra
● Method 2: Sine Wave Method / Wave Method / Wavy method
What is Inequality and Types of Inequalities
Usually we are given discrete values of variables like x=2, y=3 etc. In case of inequalities we are given a range of values. Let's take some examples to understand this:
x > 3 => x can take all real values which are greater than 3, i.e. 3.001, 4, 5, 6, 8, 100, etc...
So, instead of giving a single value in case of inequalities we are given a set of values for the variables.
Let's understand various types of inequalities now:
Greater Than Inequality ( > ) : Ex: x > 3 (We have seen above)
Less Than Inequality ( < ) : Ex: y < 2 => Y can take all read values which are less than 2
Greater Than or Equal to Inequality ( ≥ ): Ex: x ≥ 5 (x can take all real values greater than or equal to 5)
Less Than or Equal to Inequality ( ≤ ): Ex: y ≤ 3 (y can take all real values less than or equal to 3)
In-Between Inequality ( -2 ≤ x < 5 ): Ex: x can take all values which are greater than or equal to -2 and less than 5
Graphing Inequalities
Now let's talk about how to plot and inequality on the number line
Ex 1: Graph a ≤ 2 and b > -3 on a number line
Sol:
To plot a ≤ 2 we need to draw a line starting at 2 and extending till -∞ on the left hand side (Refer Orange line in below image). Note that we need to darken point 2 because it is included (as a ≤ 2, so 2 is included)
To plot b > -3 we need to draw a line starting at -3 and extending till +∞ on the right hand hand. (Refer Green line in below image). Note that we DO NOT darken point -3 as it is excluded (As b > -3 and not ≥ -3 )
Properties of Inequalities
PROP 1: Adding or Subtracting the same number from both the sides of the inequality DOES NOT change the sign of the inequality.
Ex 1: 7 > 3
Add 4 on both the sides we get
7 + 4 > 3 + 4 => 11 > 7 [Which is True and Note that sign of inequality which was > is still > ]
Ex 2: 8 > 4
Subtract 9 from both the sides we get
8 -9 > 4 -9 => -1 > -5 [Which is True and Note that sign of inequality which was < is still < ]
Ex 3: a > b
Add k on both the sides we get
a + k > b + k [ which is true and sign of inequality did not change ]
PROP 2: Multiplying / Dividing an inequality equation with a positive number DOES NOT change the sign of the inequality
Ex 1: 7 > 3
Multiply both the sides by +2 we get
7 * 2 > 3 * 2 => 14 > 6 [ which is true and sign of inequality did not change ]
Ex 2: 8 > 4
Divide both the sides by +2 we get
\(\frac{8}{2}\) > \(\frac{4}{2}\) => 4 > 2 [ which is true and sign of inequality did not change ]
Ex 3: a > b
Multiply both the sides with a positive variable k we get
ak > bk
PROP 3: Multiplying / Dividing an inequality equation with a negative number REVERSES the sign of the inequality
Ex 1: 7 > 3
Multiply both the sides by -2 we get
7 * -2 < 3 * -2 => -14 < -6 [ note the sign of inequality has changed from > to < ]
Ex 2: 8 > 4
Divide both the sides by -2 we get
\(\frac{8}{-2}\) < \(\frac{4}{-2}\) => -4 < -2 [ note the sign of inequality has changed from > to < ]
Ex 3: a > b
Multiply both the sides with a negative variable t we get
at < bt [ note the sign of inequality has changed from > to < ]
PROP 4: We can add two inequalities which have the same inequality sign
Ex 1:
7 > 3 and
8 > 2, Since the two inequalities have same sign of (>) so we can add both of them to get
7 + 8 > 3 + 2 => 15 > 5
Ex 2:
a > b and
c > d
Since the two inequalities have same sign of (>) so we can add both of them to get
a + c > b + d [Note that this is true irrespective of the signs of a, b, c and d]
Ex 3:
If two inequalities have different signs then we can multiply one of them to make the signs same and then add them
a > b
c < d
we can multiple c < d with -1 to get
-c > -d and now we can add a > b and -c > -d to get
a - c > b - d
PROP 5: Taking Square Root on both sides of an inequality DOES NOT Change the sign of the inequality (provided it is possible to take square root on both the sides and get real values).
Ex 1: \(a^2\) > \(b^2\) [given that a and b are positive numbers]
Taking square root on both the sides we will get
a > b
PROP 6: Square of a number is always non-negative
Ex 1: \(a^2\) ≥ 0 [ this is true for all real values of a ]
this will be equal to 0 only when a itself is zero
Types of Inequality Problems
Type 1: x * y > 0
If product of two variables > 0 that means that the two variables have SAME SIGN
Either Both are Positive => x > 0 and y > 0
Or Both are Negative => x < 0 and y < 0
Type 2: x / y > 0
If division of two variables > 0 that means that the two variables have SAME SIGN
Either Both are Positive => x > 0 and y > 0
Or Both are Negative => x < 0 and y < 0
Type 3: x * y < 0
If product of two variables < 0 that means that the two variables have DIFFERENT SIGN
Either x > 0 and y < 0
Or x < 0 and y > 0
Type 4: x / y < 0
If division of two variables < 0 that means that the two variables have DIFFERENT SIGN
Either x > 0 and y < 0
Or x < 0 and y > 0
Basic Problems on Inequalities
Q1. Is bd > 0 ?
A. ab > 0
B. cd > 0
Sol:
Stat A: ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: cd > 0
There are two cases
c>0 and d>0
c<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT Sufficient
Combining both the statements we will have four cases
(1) a>0 b>0 c> 0 d>0 (2) a>0 b>0 c< 0 d<0
(3) a<0 b<0 c>0 d>0 (4) a<0 b<0 c<0 d<0
In case 1 and 4 bd > 0 and in case 2 and 3 bd < 0
So, Together also NOT sufficient. So, Answer will be E
Q2. Is bd>0 ?
A. ab > 0
B. ad > 0
Sol:
Stat A: ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: ad > 0
There are two cases
a>0 and d>0
a<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT Sufficient
Combining both the statements we will have two cases
(Since we have a common variable “a” in both the statements so we will combine the two statements based on the sign of the common variable
First case of STAT A will be combined with the first case of Stat B and
Second case of STAT A will be combined with the second case of Stat B
(1) a>0 b>0 d>0 (2) a<0 b<0 d<0
In both the cases bd > 0
So, Together the two statements are sufficient. So, Answer will be C
Q3. Given that y = 4 + \((1-x)2\). Find yMin( Minimum Value of y) and find the value of x for which y = yMin
Sol:
y = 4 + \((1-x)^2\)
We know that square of a number can never be negative, so Min value of \((1-x)^2\) = 0 when 1-x = 0 or when x = 1
So, yMin = 4 when x = 1
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
Now we will start marking + and - as mentioned below:
Now, get your answer as below:
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
Problems:
1. Problem Link
If \(x^5 + x^2 < 0\), then which one of the following must be true?
(A) \(x < –1\)
(B) \(x < 0\)
(C) \(x > 0\)
(D) \(x > 1\)
(E) \(x^4 < x^2\)
Solution:
\(x^5 + x^2 < 0\)
Taking \(x^2\) common on the left hand side we have
\(x^2(x^3 + 1) < 0\)
Now, we know that \(x^2(x^3 + 1) < 0\) that means that product of two terms \(x^2\) and \((x^3 + 1)\) is < 0 so one of them is positive and other is negative
We also know that \(x^2\) is a square of a number so it can never be < 0 => \((x^3 + 1)\) < 0
or \(x^3\) < -1
That means x is a negative number less than 1 as \(x^3\)=-1 when x=-1 and for any value of x >-1 \(x^3\)>-1
So, Answer will be A
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
=> 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
\(\frac{1}{y}\) - 1 < 0
or, \(\frac{(1-y)}{y}\) < 0
or \(\frac{(y-1)}{y}\) > 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
4. If y>0>x, and \(\frac{(3+5y)}{(x−1)}\) < −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:
\(\frac{3+5y}{x−1}\) < −7
=> \(\frac{3+5y}{x−1}\) + 7 < 0
=> \(\frac{{(3+5y) + 7*(x-1)}}{x-1}\) < 0
=> \(\frac{7x + 5y -4}{x-1}\) < 0
Now, we know that x < 0 so, x- 1 < 0
in \(\frac{7x + 5y -4}{x-1} \)< 0
we know that x - 1 < 0
=> (7x + 5y -4 ) > 0
So, Answer will be B
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
Show: :: DOWNLOAD PDF: How to Solve Inequalities
Attachment:
⸙ What is Inequality and Types of Inequalities
⸙ Graphing Inequalities
⸙ Properties of Inequalities
⸙ Types of Inequality Problems
● x*y > 0
● x/y > 0
● x*y < 0
● x/y < 0
⸙ Basic Problems on Inequalities
⸙ Combining Inequalities
⸙ Four types of Inequality Problems solved using two methods
● Method 1: Algebra
● Method 2: Sine Wave Method / Wave Method / Wavy method
What is Inequality and Types of Inequalities
Usually we are given discrete values of variables like x=2, y=3 etc. In case of inequalities we are given a range of values. Let's take some examples to understand this:
x > 3 => x can take all real values which are greater than 3, i.e. 3.001, 4, 5, 6, 8, 100, etc...
So, instead of giving a single value in case of inequalities we are given a set of values for the variables.
Let's understand various types of inequalities now:
Greater Than Inequality ( > ) : Ex: x > 3 (We have seen above)
Less Than Inequality ( < ) : Ex: y < 2 => Y can take all read values which are less than 2
Greater Than or Equal to Inequality ( ≥ ): Ex: x ≥ 5 (x can take all real values greater than or equal to 5)
Less Than or Equal to Inequality ( ≤ ): Ex: y ≤ 3 (y can take all real values less than or equal to 3)
In-Between Inequality ( -2 ≤ x < 5 ): Ex: x can take all values which are greater than or equal to -2 and less than 5
Graphing Inequalities
Now let's talk about how to plot and inequality on the number line
Ex 1: Graph a ≤ 2 and b > -3 on a number line
Sol:
To plot a ≤ 2 we need to draw a line starting at 2 and extending till -∞ on the left hand side (Refer Orange line in below image). Note that we need to darken point 2 because it is included (as a ≤ 2, so 2 is included)
To plot b > -3 we need to draw a line starting at -3 and extending till +∞ on the right hand hand. (Refer Green line in below image). Note that we DO NOT darken point -3 as it is excluded (As b > -3 and not ≥ -3 )
Properties of Inequalities
PROP 1: Adding or Subtracting the same number from both the sides of the inequality DOES NOT change the sign of the inequality.
Ex 1: 7 > 3
Add 4 on both the sides we get
7 + 4 > 3 + 4 => 11 > 7 [Which is True and Note that sign of inequality which was > is still > ]
Ex 2: 8 > 4
Subtract 9 from both the sides we get
8 -9 > 4 -9 => -1 > -5 [Which is True and Note that sign of inequality which was < is still < ]
Ex 3: a > b
Add k on both the sides we get
a + k > b + k [ which is true and sign of inequality did not change ]
PROP 2: Multiplying / Dividing an inequality equation with a positive number DOES NOT change the sign of the inequality
Ex 1: 7 > 3
Multiply both the sides by +2 we get
7 * 2 > 3 * 2 => 14 > 6 [ which is true and sign of inequality did not change ]
Ex 2: 8 > 4
Divide both the sides by +2 we get
\(\frac{8}{2}\) > \(\frac{4}{2}\) => 4 > 2 [ which is true and sign of inequality did not change ]
Ex 3: a > b
Multiply both the sides with a positive variable k we get
ak > bk
PROP 3: Multiplying / Dividing an inequality equation with a negative number REVERSES the sign of the inequality
Ex 1: 7 > 3
Multiply both the sides by -2 we get
7 * -2 < 3 * -2 => -14 < -6 [ note the sign of inequality has changed from > to < ]
Ex 2: 8 > 4
Divide both the sides by -2 we get
\(\frac{8}{-2}\) < \(\frac{4}{-2}\) => -4 < -2 [ note the sign of inequality has changed from > to < ]
Ex 3: a > b
Multiply both the sides with a negative variable t we get
at < bt [ note the sign of inequality has changed from > to < ]
PROP 4: We can add two inequalities which have the same inequality sign
Ex 1:
7 > 3 and
8 > 2, Since the two inequalities have same sign of (>) so we can add both of them to get
7 + 8 > 3 + 2 => 15 > 5
Ex 2:
a > b and
c > d
Since the two inequalities have same sign of (>) so we can add both of them to get
a + c > b + d [Note that this is true irrespective of the signs of a, b, c and d]
Ex 3:
If two inequalities have different signs then we can multiply one of them to make the signs same and then add them
a > b
c < d
we can multiple c < d with -1 to get
-c > -d and now we can add a > b and -c > -d to get
a - c > b - d
PROP 5: Taking Square Root on both sides of an inequality DOES NOT Change the sign of the inequality (provided it is possible to take square root on both the sides and get real values).
Ex 1: \(a^2\) > \(b^2\) [given that a and b are positive numbers]
Taking square root on both the sides we will get
a > b
PROP 6: Square of a number is always non-negative
Ex 1: \(a^2\) ≥ 0 [ this is true for all real values of a ]
this will be equal to 0 only when a itself is zero
Types of Inequality Problems
Type 1: x * y > 0
If product of two variables > 0 that means that the two variables have SAME SIGN
Either Both are Positive => x > 0 and y > 0
Or Both are Negative => x < 0 and y < 0
Type 2: x / y > 0
If division of two variables > 0 that means that the two variables have SAME SIGN
Either Both are Positive => x > 0 and y > 0
Or Both are Negative => x < 0 and y < 0
Type 3: x * y < 0
If product of two variables < 0 that means that the two variables have DIFFERENT SIGN
Either x > 0 and y < 0
Or x < 0 and y > 0
Type 4: x / y < 0
If division of two variables < 0 that means that the two variables have DIFFERENT SIGN
Either x > 0 and y < 0
Or x < 0 and y > 0
Basic Problems on Inequalities
Q1. Is bd > 0 ?
A. ab > 0
B. cd > 0
Sol:
Stat A: ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: cd > 0
There are two cases
c>0 and d>0
c<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT Sufficient
Combining both the statements we will have four cases
(1) a>0 b>0 c> 0 d>0 (2) a>0 b>0 c< 0 d<0
(3) a<0 b<0 c>0 d>0 (4) a<0 b<0 c<0 d<0
In case 1 and 4 bd > 0 and in case 2 and 3 bd < 0
So, Together also NOT sufficient. So, Answer will be E
Q2. Is bd>0 ?
A. ab > 0
B. ad > 0
Sol:
Stat A: ab > 0
There are two cases
a>0 and b>0
a<0 and b<0
In both the cases we don’t know anything about the sign of d so NOT sufficient
Stat B: ad > 0
There are two cases
a>0 and d>0
a<0 and d<0
In both the cases we don’t know anything about the sign of b so NOT Sufficient
Combining both the statements we will have two cases
(Since we have a common variable “a” in both the statements so we will combine the two statements based on the sign of the common variable
First case of STAT A will be combined with the first case of Stat B and
Second case of STAT A will be combined with the second case of Stat B
(1) a>0 b>0 d>0 (2) a<0 b<0 d<0
In both the cases bd > 0
So, Together the two statements are sufficient. So, Answer will be C
Q3. Given that y = 4 + \((1-x)2\). Find yMin( Minimum Value of y) and find the value of x for which y = yMin
Sol:
y = 4 + \((1-x)^2\)
We know that square of a number can never be negative, so Min value of \((1-x)^2\) = 0 when 1-x = 0 or when x = 1
So, yMin = 4 when x = 1
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
Problems:
1. Problem Link
If \(x^5 + x^2 < 0\), then which one of the following must be true?
(A) \(x < –1\)
(B) \(x < 0\)
(C) \(x > 0\)
(D) \(x > 1\)
(E) \(x^4 < x^2\)
Solution:
\(x^5 + x^2 < 0\)
Taking \(x^2\) common on the left hand side we have
\(x^2(x^3 + 1) < 0\)
Now, we know that \(x^2(x^3 + 1) < 0\) that means that product of two terms \(x^2\) and \((x^3 + 1)\) is < 0 so one of them is positive and other is negative
We also know that \(x^2\) is a square of a number so it can never be < 0 => \((x^3 + 1)\) < 0
or \(x^3\) < -1
That means x is a negative number less than 1 as \(x^3\)=-1 when x=-1 and for any value of x >-1 \(x^3\)>-1
So, Answer will be A
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
=> 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
\(\frac{1}{y}\) - 1 < 0
or, \(\frac{(1-y)}{y}\) < 0
or \(\frac{(y-1)}{y}\) > 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
4. If y>0>x, and \(\frac{(3+5y)}{(x−1)}\) < −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:
\(\frac{3+5y}{x−1}\) < −7
=> \(\frac{3+5y}{x−1}\) + 7 < 0
=> \(\frac{{(3+5y) + 7*(x-1)}}{x-1}\) < 0
=> \(\frac{7x + 5y -4}{x-1}\) < 0
Now, we know that x < 0 so, x- 1 < 0
in \(\frac{7x + 5y -4}{x-1} \)< 0
we know that x - 1 < 0
=> (7x + 5y -4 ) > 0
So, Answer will be B
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
Re: GRE Quant - Solving Linear Inequalities Theory
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10 Jun 2020, 12:06
10 Jun 2020, 12:06
Expert Reply
1
Bookmarks
Solving Linear Inequalities
Rules of Operations in an Inequality
1. Multiplication/Division Properties for Inequalities: when multiplying/dividing by a positive value
If a < b AND c is positive, then \(ac < bc\)
If a < b AND c is positive, then \(a/c < b/c\)
2. Multiplication/Division Properties for Inequalities: when multiplying/dividing by a negative value
If a < b AND c is negative, then ac > bc
If a < b AND c is negative, then a/c > b/c
When solving inequalities, if you multiply or divide through by a negative, you must also flip the inequality sign.
3. Compound Inequality
Solve 10 < 3x + 4 < 19.
This is what is called a "compound inequality". It works just like regular inequalities, except that it has three "sides". So, for instance, when I go to subtract the 4, I will have to subtract it from all three "sides".
Three Step Solving Strategy:
Step 1: Simplify each side, if needed.
This would involve things like removing ( ), removing fractions, adding like terms, etc.
Step 2: Use Add./Sub.
Properties to move the variable term on one side and all other terms to the other side.
Step 3: Use Mult./Div.
Properties to remove any values that are in front of the variable.
Example:
\(2(y + 1) <= y - 4\)
or \(2y + 2 <= y - 4\) Step 1
or \(2y + 2 -y <= y - 4 -y\) Step 2
or \(y + 2 <= - 4\)
or \(y + 2 - 2 <= - 4 - 2\) Step 2
or \(y <= - 6\)
Absolute value equations and inequalities
An absolute value of a number \(a\) is represented as, \(|a|\). An absolute equation, \(|x| = a\) consists of two solutions, \(x = a\) and \(x = -a\). Hence, care should be taken while solving an equation consisting of an absolute value.
For example, the equation, \(|x - 17| = 45\) is solved in two different parts.
Part I: \(x - 17 = 45\) making \(x = 62\).
Part II: \(x - 17 = -45\) making \(x = -28\).
An inequality equation with an absolute value, \(|x| > a\) means \(x > a\) and \(x < -a\).
For example,
\(5|3x + 18| < 45\)
In order to solve this equation we first divide it by 5, \(|3x + 18| < 9\).
Now comes the most important step of removing absolute sign and rewriting the equation as, \(-9 < 3x + 18 < 9\).
This equation is finally solved as, \(-27 < 3x < -9\). Hence, \(-9 < x < 3\) is the solution.
Re: GRE Quant - Solving Linear Inequalities Theory
[#permalink]
10 Jun 2020, 12:11
10 Jun 2020, 12:11
Expert Reply
Inequalities Tips
ADDING/SUBTRACTING INEQUALITIES
1. You can only add inequalities when their signs are in the same direction:
If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).
2. You can only apply subtraction when their signs are in the opposite directions:
If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).
RAISING INEQUALITIES TO EVEN/ODD POWER
1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);
But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.
2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).
MULTIPLYING/DIVIDING TWO INEQUALITIES
1. If both sides of both inequalities are positive and the inequalities have the same sign, you can multiply them.
For example, for positive \(x\), \(y\), \(a\), \(b\), if \(x < a\) and \(y < b\), then \(xy < ab\).
2. If both sides of both inequalities are positive and the signs of the inequality are opposite, then you can divide them.
For example, for positive \(x\), \(y\), \(a\), \(b\), if \(x < a\) and \(y > b\), then \(\frac{x}{y} < \frac{a}{b}\) (The final inequality takes the sign of the numerator).
MULTIPLYING/DIVIDING AN INEQUALITY BY A NUMBER
1. Whenever you multiply or divide an inequality by a positive number, you must keep the inequality sign.
2. Whenever you multiply or divide an inequality by a negative number, you must flip the inequality sign.
3. Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know the sign of it or are not certain that variable (or the expression with a variable) doesn't equal to zero.
SOLVING QUADRATIC INEQUALITIES: GRAPHIC APPROACH
Say we need to find the ranges of \(x\) for \(x^2-4x+3<0\). \(x^2-4x+3=0\) is the graph of a parabola and it look likes this:
Intersection points are the roots of the equation \(x^2-4x+3=0\), which are \(x_1=1\) and \(x_2=3\). "<" sign means in which range of \(x\) the graph is below x-axis. Answer is \(1<x<3\) (between the roots).
If the sign were ">": \(x^2-4x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above x-axis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
This approach works for any quadratic inequality. For example: \(-x^2-x+12>0\), first rewrite this as \(x^2+x-12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).
\(x^2+x-12<0\). Roots are \(x_1=-4\) and \(x_1=3\) --> below ("<") the x-axis is the range for \(-4<x<3\) (between the roots).
Again if it were \(x^2+x-12>0\), then the answer would be \(x<-4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
1. You can only add inequalities when their signs are in the same direction:
If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).
2. You can only apply subtraction when their signs are in the opposite directions:
If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).
RAISING INEQUALITIES TO EVEN/ODD POWER
1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);
But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.
2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).
MULTIPLYING/DIVIDING TWO INEQUALITIES
1. If both sides of both inequalities are positive and the inequalities have the same sign, you can multiply them.
For example, for positive \(x\), \(y\), \(a\), \(b\), if \(x < a\) and \(y < b\), then \(xy < ab\).
2. If both sides of both inequalities are positive and the signs of the inequality are opposite, then you can divide them.
For example, for positive \(x\), \(y\), \(a\), \(b\), if \(x < a\) and \(y > b\), then \(\frac{x}{y} < \frac{a}{b}\) (The final inequality takes the sign of the numerator).
MULTIPLYING/DIVIDING AN INEQUALITY BY A NUMBER
1. Whenever you multiply or divide an inequality by a positive number, you must keep the inequality sign.
2. Whenever you multiply or divide an inequality by a negative number, you must flip the inequality sign.
3. Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know the sign of it or are not certain that variable (or the expression with a variable) doesn't equal to zero.
SOLVING QUADRATIC INEQUALITIES: GRAPHIC APPROACH
Say we need to find the ranges of \(x\) for \(x^2-4x+3<0\). \(x^2-4x+3=0\) is the graph of a parabola and it look likes this:
Intersection points are the roots of the equation \(x^2-4x+3=0\), which are \(x_1=1\) and \(x_2=3\). "<" sign means in which range of \(x\) the graph is below x-axis. Answer is \(1<x<3\) (between the roots).
If the sign were ">": \(x^2-4x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above x-axis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
This approach works for any quadratic inequality. For example: \(-x^2-x+12>0\), first rewrite this as \(x^2+x-12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).
\(x^2+x-12<0\). Roots are \(x_1=-4\) and \(x_1=3\) --> below ("<") the x-axis is the range for \(-4<x<3\) (between the roots).
Again if it were \(x^2+x-12>0\), then the answer would be \(x<-4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
