Q5) Inequalities and absolute values

Question 1

Multiplying or dividing an inequality by a NEGATIVE NUMBER

Answer 1

While we use many of the same rules when adding or subtracting numbers as we do in solving regular EQUATIONS….

> When an inequality is multiplied or divided BY A NEGATIVE NUMBER, the inequality sign must also REVERSE its orientation

> aka, do the operation, THEN “flip the inequality sign”

Please note:
> Be careful when MULTIPLYING or DIVIDING UNKNOWN VARIABLES (unless you know the sign!!!!)

Question 2

Example DS involving inequalities

Example 1:
Is 2d > 4c?

(1) 3c/10 = 10 d
(2) d^3 < 0

Example 2: If x > y > 0, is x > 3?

(1) 2/x + 1/y = 11/15
(2) 2x + 3y = 16

Answer 2

Inequalities in DS often involve working with inequalities AND EQUATIONS
> need to SUB in known variables INTO the inequality to answer the question stem AND MOVE variables to ONE SIDE to compare to 0 (easiest method)

Simplify the question stem to is d > 2c?

** (1) Simplifies to d = 3c/100
—-> PLUG THIS INTO the question stem

Is 3c/100 > 2c ? –> MOVE terms to one side

Is 197c/100 < 0 ? —> UNKNOWN, depends on the value of c

NS (even simple test values could have HELPED)

(2) d^3 < 0 means that d < 0

Test case 1: -1 > 2? –> NO

Test case 2: -1 > - 2 –> YES

NS

(3) We know the SIGN OF d

Re-express question stem as is -197d/3 > 0 and since d < 0, we know that -197d/3 must always be positive

S (ans C)

You can also sub in known variables into CONDITIONS that are inequalities

e.g., we know that x > y > 0

(1) equation simplifies into 30y + 15x = 11xy

ISOLATE for y
y = 15x/(11x-30)

Then, sub expression for y into inequality condition x > y > 0

x > 15x/(11x-30) > 0

Move all x variables to one side and see if we can conclude that x > 3: (also note that since y > 0, we know that 11x-30 must also be greater than 0)

x > 45/11 —> greater than 3, so SUFFICIENT

(2) Rearrange for y
y = (16-2x)/3

Then, sub expression for y into inequality condition x > y > 0

x > (16-2x)/3 > 0

Move all x variables to one side and see if we can conclude that x > 3:

x > 16/5 = —> greater than 3, so SUFFICIENT

D

Question 3

Adding inequalities

e.g., 5a > 2b and 2b > 4a

Answer 3

In order to ADD inequalities, the SIGN OF EACH inequality must be facing the SAME DIRECTION

> then just add the Left side together, and separately add the Right side together (same way as adding multiple equations - one side at a time)

(only true for addition)

e.g., 5a + 2b > 2b + 4a
a > 0

Question 4

DS involving simple inequalities

e.g., Is p < 6? given -p > 1

Answer 4

Sometimes it’s easier to evaluate when the variable is isolated on one side and the number is on the other side (vs moving all the terms to one side and having the other side = 0)

Question 5

What are compound inequalities?

Answer 5

Refer to three-part inequality with TWO inequality signs
> you can create a compound inequality by COMBINING two individual inequalities
> set the signs to the same direction
> make sure the middle part is common to the other two parts

e.g., x<=5 and x>=-4 can be combined into -4<=x<=5

When you have a compound inequality, what we do to ONE PART of the compound inequality we must do to ALL PARTS of the compound inequality
> when a compound inequality is MULTIPLIED or DIVIDED by a negative number, BOTH INEQUALITY SIGNS must be reversed

Question 6

Substitution involving equations and inequalities

Answer 6

We can substitute FROM AN EQUATION INTO an inequality –> result is you can understand one UNKNOWN’s possible range of numbers

You can even perform substitution from an equation INTO a compound inequality (substitute for one variable and you can understand the other unknown’s possible range of numbers)

Question 7

Comparing the relative size of variables expressed in multiple inequalities

e.g., a > b, c > a, d < b which of the following is true?

Answer 7

Strategy for solving:
> Set up a NUMBER LINE to compare values graphically

Question 8

Solving inequalities containing SQUARE VARIABLES

Answer 8

Recall that sqrt(x^2) = |x|

To unwind the absolute sign, think about the two possible conditions - positive and negative
> keep the other side the SAME
> don’t flip the inequality sign until after you unwind the absolute sign
> “unwinding” refers to REWRITING the absolute value term without it and sticking a positive or negative sign so the equation still equates

For a given expression |##| –>
Positive = |##| = ##
Negative = |##| = -##

IN GENERAL:

x^2 > b (where b > 0) …
x > sqrt(b) or x < -sqrt(b)

x^2 < b (where b > 0) … (“less than”)
-b < x < b

Question 9

Finding the minimum or maximum value of the PRODUCT OF TWO+ UNKNOWNS, given a range of values for each unknown

e.g., If a <= x <= b and c <= y <= d, what is the min (or max) of xy?

Answer 9

To solve, we need to evaluate all the possible PRODUCTS of the END POINTS

e.g., ac, ad, bc, and bd

The LARGEST of these quantities = maximum value value of xy

The SMALLEST of these quantities = minimum value of xy

Please note:
> the maximum product is not necessarily the product of the two largest positive ranges (could also be the product of the smallest negative ranges)!

Question 10

Equations with absolute value

Answer 10

Need to solve the equation THREE TIMES
> once for the condition in which the value inside the absolute value bars is POSITIVE
> second for the condition in which the value inside the absolute value bars is NEGATIVE
> THIRD time for the condition in which the value inside the absolute value bars is ZERO (pretty easy to rule this one out)
> “unwind” the absolute value sign

Please note:
> There could be 1+ terms within the absolute value bars

Question 11

** Under what conditions would two absolute values EQUAL EACH OTHER

e.g., when would |x| = |y|?

e.g., what are all the possible values of x if |16x + 14| = |8x + 6|?

Answer 11

If two absolute values are EQUAL, it must be true that the expressions within the absolute value bars are EITHER EQUAL OR OPPOSITE

(1) when x = y (numbers are equal)
e.g., x = y = 5

(2) x = -y (opposite signs)
e.g., x = 5 and y = -5

This works because we are just simplifying the testing of 4 different cases:
> both positive
> x positive, y negative
> x negative, y positive (equal to statement above)
> both negative (equal to both positive)

Please note:
> This is TRUE regardless of the EXPRESSIONS represented by x and y
> It is only necessary to calculate ONE equation with opposite absolute values
> STILL NEED TO CHECK your solutions by plugging back into the original equation with absolute values

Question 12

Keep in mind when you are solving equations and inequalities containing ABSOLUTE VALUES

Answer 12

> Keep track of your conditions and make sure to ALWAYS CHECK whether the answer and the condition work out (you need to strike out invalid answers)

Becomes crucial when:
> both sides of an absolute-value equation contain variable
> one side of absolute value equation is negative (never true)

Question 13

If |2x + 8| = |4x - 4|, what is the value of x?

(1) |x| > x^2
(2) -|x| < -x

Answer 13

Absolute value and inequalities
> When you see two absolute values set equal to each other, you know it’s ONLY TRUE IF the inner expressions are equal OR opposite values

Simplify question stem:
EQUAL expressions -
2x+8 = 4x-4
x = 6

OPPOSITE expressions -
2x+8 = -4x+4
x = -2/3

So we need to now determine whether x = 6 or x = -2/3

(1) |x| > x^2

Case 1: x > 0
x > x^2 (you can DIVIDE both sides by x)
1 > x –> CHECK CONDITION and COMBINE WITH THE CONDITION, you get
0 < x < 1 —> neither 6 nor -2/3 fit this criteria, NOT POSSIBLE, x cannot be greater than 0 and x =/ 6 –> Sufficient already, but let’s check others quickly

Case 2: x < 0
-x > x^2 (you can divide both sides by -x = positive number, so sign stays the same direction)
1 > -x
-1 < x –> CHECK CONDITION, you get -1 < x < 0 —> x must be -2/3 if x < 0

Case 3: x = 0
NOT possible case because we are told |x| > x^2

Sufficient

*** FASTER WAY: ALTERNATIVELY, sub in possible values of x into each statement and see if it’s true or not

(2) -|x| < -x —> divide both sides by -1 and change direction of sign

|x| > x

Case 1: x > 0
x > x –> NOT POSSIBLE, so x is not greater than 0 (x =/ 6)

Sufficient

Case 2: x < 0
-x > x
0 > 2x
0 > x –> Same thing as the condition, x = -2/3

Case 3: x = 0, we know this is not possible

Sufficient

Question 14

Adding absolute values - what is the rule here?

e.g., |a| + |b| vs |a+b|

Answer 14

|a+b|<= |a| + |b|

> this is true for ALL value of p and q (so multiple possible answers are possible)

> this is because a and b could be negative

|a+b| equals |a| +
b| WHEN:
> one or both quantities = 0
> or both quantities have the SAME sign (if a and b are non-zero numbers)

aka ab > 0

Question 15

Subtracting absolute values - what is the rule here?

e.g., |a| - |b| vs |a-b|

Answer 15

|a-b|>= |a| - |b|

> this is because b could be negative (when a is positive) and a could be negative (when b is positive)

|a-b| equals |a| -
b| WHEN:
> one or both quantities = 0
> or both quantities have the SAME sign (if a and b are non-zero numbers) **AND the absolute value of first quantity is GREATER THAN OR EQUAL TO the absolute value of the second quantity

aka ab >0 AND|a| >= |b|

Question 16

What are the similarities between x^2 and |x|? What are the differences?

Answer 16

Both are NON-NEGATIVE quantities regardless of the value of x (can never be negative)

The difference between the two is the WAY we ISOLATE (unwind) the variable

Question 17

What happens if you see an absolute value equal to a NEGATIVE value

e.g., |x+1| = -2

Answer 17

No solutions for this equation because the absolute value of anything must be NON-NEGATIVE

> DO NOT proceed with solving the equation using the typical method (you will get two solutions that appear to work but once you plug back into the original equation they in fact do not work)

Question 18

What does it mean when you see:

|p| + |q| = 0

Answer 18

Recall that the absolute value of any number will always be NON-NEGATIVE (>=0)

The absolute value of a NON-ZERO number will always be positive

The absolute value of ZERO will always be ZERO

Only way for SUM of two absolute values to equal 0 is for BOTH p and q to equal 0

Question 19

If x and y are integers and…
|x-4| < 6
y-7 > 23
y+2 < 50

What is the greatest possible value of xy?

Answer 19

(1) determine the possible values of x and y
(2) Watch out for the END RANGES (make the max and min values INCLUSIVE) –> we also know x and y are integers

Since -2 < x < 10, therefore -1 <= x <= 9
Since 30 < y < 48, therefore 31 <= y <= 47

Greatest possible value of xy is 947 = 423 (NOT 1048 = 480)

Question 20

If 5a > 2b, is the sum of a and b greater than zero?

(1) 2b > 4a
(2) b > 0

Answer 20

DS with many inequality statements
> keep track of them ALL (don’t just rely on the combined inequality)

is a + b > 0?

(1) We can add two inequalities as long as the inequality sign is facing the same direction
5a + 2b > 4a + 2b
a > 0 ——> It might not be obvious yet but we DO KNOW what b must be

if a > 0, then to make 2b > 4a true, b > 0 too

a + b always > 0, Sufficient

(2) b > 0, therefore to make 5a > 2b true, a must also > 0, Sufficient

D

Question 21

if x =/ y and x and y are non-zero integers, is (x+y)/(x-y) > 0?

(1) x > y
(2) |x| > |y|

Answer 21

DS involving DOUBLE absolute value signs AND inequality…
> think about ALL the possible cases (max 4)

Is (x+y)/(x-y) > 0 —-> are the signs of numerator and denominator the same?

ALSO we know x and y are non-zero integers

(1) x > y, means x - y > 0 —-> denominator is positive

Now question becomes is x + y > 0?

NS (knowing x-y cannot tell you about x+y and vice versa)

e.g., x = 10 and y = 5 —> x+y >0
But if x = 10 and y = -100 —> x+y <0

(2) |x| > |y|
(double absolute value sigs AND inequality…)
> tells me that x is farther from 0 than y on a number line
> draw this out and TEST ALL 4 possibilities of |x| > |y|

(a) x < 0 and y < 0 (e.g., x = -5, y = -1)
(b) x < 0 and y > 0 (e.g., x = -5, y = 1)
(c) x > 0 and y < 0 (e.g., x = 5 and y = -1)
(d) x > 0 and y > 0 (e.g., = 5 and y = 1)

In each of these 4 test cases (x+y)/(x-y) > 0
Sufficient

You can also think logically through this
> if we start at x, we have to move at least |x| units before we pass the origin and change the sign
> so only moving |y| units won’t reach the origin
> also x+y and x-y both have the same sign as x
> dividing two quantities with the same sign gives a positive quotient

Question 22

Is t = |r-s|?

(1) 3r > 3s
(2) t = r-s

Answer 22

Equations with absolute values and inequalities
> watch out for the SIGN OF t

(1) r > s, means r - s > 0
Careful… we don’t know anything about t, so this statement is not sufficient

(2) t = r-s
This is TEMPTING but if you replace t with r-s in the question stem you get: Is r-s = |r-s|?
> this is only true if r-s > 0
> if r-s < 0, then |r-s| = -(r-s)
> t could also be negative

Not sufficient

(3) we know r-s > 0 AND t = r-s
Sufficient

Question 23

Taking the reciprocal of an inequality (incl. compound inequality)

Answer 23

FLIP direction of sign when you do 1/x to each part of the inequality AS LONG AS you know the sign of a or b (+ or -)

IF YOU KNOW THE SIGNS: a and b both > 0
> Flip the inequality
e.g., 2 < 5 —> 1/2 > 1/5

IF YOU KNOW THE SIGNS: a and b both < 0
> Flip the inequality
e.g., -2 > -5 —> -1/2 < -1/5

IF YOU KNOW THE SIGNS: a and b are opposite signs
> DO NOT flip the inequality
e.g., -2 < 5 —> -1/2 < 1/5
e.g., 5 > -2 —> 1/5 > -1/2