## GRE: Solving Inequalities

Inequalities may be written with the symbols Shown here:-

 < Less than > Greater than ≤ Less than or equal to ≥ Greater than or equal to

A range of values is often expressed on a number line.

Two ranges are shown below:-

a) Represents the set of all numbers between -4 and 0 excluding the endpoints -4 and 0 or -4< x < 0

b) Represents the set of all numbers greater than -1, upto and including 3, or -1 < x ≤ 3

Rules for working with inequalities:-

• Treat inequalities like equations when adding or subtracting terms or when multiplying/Dividing by a positive number on both sides of the inequality.
• Flip the inequality sign if you multiply or divide both sides of an equality by a negative number.,

Problem 1) Solve for x and represent the solution set on a number line.

$$3-\frac { x }{ 4 } \ge 2$$

Sol.     Multiply both sides by 4                              12-x  ≥ 8

Subtract 12 from both sides                        -x ≥ -4

Multiply both sides by -1 and flip the inequality symbol      x ≤ 4

Problem 2) Solve the inequality

$$\frac { 3x }{ 2 } -\frac { 3x }{ 4 } <\frac { 7x }{ 4 } -1$$

Sol.

$$\frac { 3x }{ 2 } -\frac { 3x }{ 4 } -\frac { 7x }{ 4 } <-1\quad (Subtract\quad \frac { -7x }{ 4 } \quad from\quad both\quad sides)\\ \frac { 3x }{ 4 } -\frac { 7x }{ 4 } <-1\\ \frac { -4x }{ 4 } <-1\\ -x<-1\quad (Multiply\quad both\quad sides\quad by\quad -1)\\ x>1$$

Absolute Value: The absolute value of a number describes how far that number is away from 0 on a number line. The symbol of absolute value is |number|.

STEPS TO SOLVE ABSOLUTE VALUE EQUATION:-

• Take what’s inside the absolute value sign and set up two equations.First sets the positive value equal to the other side of the equation, and the second sites the negative value equal to the other side.
• Solve both equations:

Problem 3)        $$|2x+4|=30$$

Sol.

$$2x+4=30\quad or\quad -(2x+4)=30\\ x=13\quad \quad \quad \quad \quad \quad \quad +2x+4=-30\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=-17\\ x=13\quad or\quad (-17)$$

Inequalities and absolute value:-

### Steps for solving questions involving both inequalities and absolute value

• Set up Two Equations:- The first inequality replaces the absolute value with the positive of what’s inside and the second replaces the absolute value with the negative of what’s inside.

Problem 4)  | x | ≥ 4

Sol.

+( x ) ≥ 4 or –x ≥ 4

x ≥ 4 or x ≤ -4

Problem 5) | x+3 | ≤ 5

Sol.

+( x+3) < 5 and        -( x+3 ) < 5

x < 2                        x > -8

The only numbers that make the original inequality true are those that are true for both inequalities. X should be greater than -8 and less than 2.

NOTE:-   In conclusion, for solving absolute value expressions that are greater than some quantity set up two equations & after sol; having two equations.We have to take real numbers that are satisfied by either equation. However, for solving absolute value expressions that are less than some quantity set up two equations and after solving two equations, We have to take real numbers that are satisfied by both the equations.

Problem 6) If |3x + 7|  ≥ 2x + 12 , then which of the following is TRUE?

a) $$x\le \frac { -19 }{ 5 }$$

b) $$x\ge \frac { -19 }{ 5 }$$

c) $$x\ge 5$$

d) $$x\le \frac { -19 }{ 5 } \quad or\quad x\ge 5$$

Ans:- ( 3x + 7 ) ≥ 2x +12                               -( 3x + 7 ) ≥ 2x + 12

x ≥ 5                                                    -3x ≥ 2x +19

-5x ≥ 19

x ≤  -19/5

$${ (x\le \frac { -19 }{ 5 } \quad or\quad x\ge 5) }^{ Ans }$$

a), c) and d) are correct answers.

Using extreme values: Extreme values are helpful where the questions involve the potential range of value for variables in the problem.

Problem 7) If  -7  ≤ a  ≤ 6   and  -7  ≤ b  ≤ 8, What is the maximum possible value for ab?

Sol. Let us consider the different extreme value scenarios for a, b and ab.

abab
Min -7Min -749
Min -7Max 8-56
Max 6Min -7-42
Max 6Max 848

We can easily reckon that ab is maximized when we take the negative extreme values for both a and b. Therefore maximum value of  ab = 49

Problem 8) If $$|y|\le -4x\quad and\quad |3x-4|=2x+6$$,what is the value of x?

Sol.      $$|y|\le -4x$$

Any absolute value cannot be negative. |y| must be positive & positive value cannot be lesser than any negative value. Therefore, -4x must be positive.

For -4x to be positive, x must be negative.

Now solve the absolute value equation by using the identity that |a|=a when a is positive or zero & |a|=-a when a is negative.

$$|3x-4|=2x+6\\ 3x-4=2x+6\quad \quad or\quad -(3x-4)=2x+6\\ 3x-4-2x-6=0\quad \quad \quad \quad -3x+4=2x+6\\ x=10\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -5x=2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=(\frac { -2 }{ 5 } )$$

x must be negative. Therefore, x can only be $$\frac { -2 }{ 5 } [latex]. Problem 9) If [latex]2{ (x-1) }^{ 3 }+3\le 19$$, which of the following must be true?

a) $$x\ge 3$$

b) $$x\le 3$$

c) $$x\ge -3$$

d) $$x\le -3$$

e) $$x<-3\quad or\quad x>3$$

Sol.

$$2{ (x-1) }^{ 3 }+3\le 19\\ 2{ (x-1) }^{ 3 }\le 16\\ { (x-1) }^{ 3 }\le 8$$

Taking the cube root of an inequality is permissible here, because cubing a number, while squaring it, does not change its sign.

$$x-1\le 2\\ { (x\le 3) }^{ Ans }$$

Problem 10) Which of the following could be the graph of all values of $$x$$ that satisfy the inequality

$$4-11x\ge (\frac { -2x+3 }{ 2 } )$$

Sol.First solve the inequality,

$$4-11x\ge \frac { -2x+3 }{ 2 } \\ 8-22x\ge 20x\\ \frac { 1 }{ 4 } \ge x\\ x\le \frac { 1 }{ 4 }$$

Thus, the correct choice should show the gray line beginning to the right of zero (in the positive zone), & continuing indefinitely into the negative zero.

Problem 11) $$p+|k|>|p|+k$$

Quantity A:         p

Quantity B:         k

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  In general, there are four cases for the signs of p & k, some of which can be ruled out by the constraints of this equation:

 p k p+|k|>|p|+k + + In this case, both sides should be positive. Therefore, this case is not true. + - This case is true. (positive value)+absolute value>(positive value)+(negative value) - - It may between. Both sides are positive plus a negative. For this condition to the true (|k|>|p|) - + This case is not true. k+(negative value)

Additionally, check whether p or k could be zero.

If p=0, p+|k|>|p|+k is equivalent to |k|>k.

This is true when k is negative.

If k=0, p+|k|>|p|+k (when k=0)

p>|p|

This is not true for any p value. So, there are three possible cases for p & k.

Let’s interpret this more: –

Use identity that |a|=-a, when a is negative.

 p k Interpret + - p=positive >k=negative - - p+|k|>|p|+k p-k>-p+k 2p>2k Therefore, p>k 0 - p=0>k=negative Therefore, p>k

In all cases, p is greater than k.

Therefore, Quantity A is greater.

Problem 12) $$|x|y>x|y|$$

Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  In general, there are four cases for the signs of x & y, some of which can be ruled out by the constraint in the question using identity that |a|=-a when a is negative.

 x y |x|y>x|y| is equivalent to True or False? + + xy>xy False: xy=xy + - xy>x(-y) False: xy is negative & (-xy) is negative. - + (-x)y>(-xy) True: (-xy) is positive & xy is negative - - (-x)y>(-xy) False: -xy=-xy

Note that if either x or y equals 0, that case would also fail the constraint.

Quantity A: $${ (x+y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy$$

Quantity B: $${ (x-y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-2xy$$

X is negative & y is positive.

Therefore, xy is negative.

$${ x }^{ 2 }+{ y }^{ 2 }-2xy>{ x }^{ 2 }+{ y }^{ 2 }+2xy\\ { (x-y) }^{ 2 }>{ (x+y) }^{ 2 }$$

Therefore, Quantity B is greater.

Problem 13) If y<0 & 4x>y, which of the following could be equal to

$$\frac { x }{ y }$$  ?

a) 0

b) $$\frac { 1 }{ 4 }$$

c) $$\frac { 1 }{ 2 }$$

d) 1

e) 4

Sol.  4x>y (divide both sides by y)

$$\frac { 4x }{ y } <1$$

Remember to switch the direction of the inequality sign when multiplying or dividing by a negative Quantity.

$$\frac { 4x }{ y } <1$$ (Now divide both sides by 4)

$$\frac { x }{ y } <\frac { 1 }{ 4 }$$

The only answer choice less than $$\frac { 1 }{ 4 }$$ is 0.

Problem 14) If |1-x|=6 & |2y-6|=10, which of the following could be the value of xy?

Indicate all such value?

a) -40

b) -10

c) -14

d) 56

Sol. Remember that to solve the absolute value equation,

First set the positive value equal to the other side of the equation & the second set the negative value equal to the other side of the equation.

$$|1-x|=6,\quad -(1-x)=6\\ -1+x=6\\ x=7\\ \\ |1-x|=6,\quad +(1-x)=6\\ 1-x=6\\ x=-5\\ Therefore,\quad x=-5\quad or\quad 7\\ \\ |2y-6|=10,\quad 2y-6=10\\ 2y=16,\quad y=8\\ \\ 2y-6=-10,\quad 2y=-10+6=-4\\ y=-2\\ Therefore,\quad y=8\quad or\quad y=-2$$

Now calculate all four possible combinations for xy:

(-5) (8) =-40

(-5) (-2) =10

7(8) =56

7(-2) =-14

The correct answer are -40, -14 & 56 only.

Problem 15) If & x<0, which of the following could be the value of x?

Indicate all such values:

a) -6

b) -14

c) -18

d) -22

e) 36

Sol.  Remember that for solving absolute value expressions that are greater than some quantity, set up two equations (the first inequality replaces the absolute value with the positive of what’s inside & the second replaces the absolute value with the negative of what’s inside) & we have to take real numbers that are satisfied by either equation.

$$\frac { |x+4| }{ 2 } >5,\quad |x+5|>10\\ x+4>10\quad \quad \quad -(x+4)>10\\ x>6\quad \quad \quad \quad \quad \quad -x>14\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x<-14$$

x>6 is not a valid solution range, as the other inequality indicates that x is negative.

x<-14, note that this fits the other inequality which states that x<0.

If x<-14, only -18 & -22 are correct answers.