Inequalities - GMAT Quantitative
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Which of the following is equivalent to
?
Which of the following is equivalent to ?
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To solve this problem we need to isolate our variable
.
We do this by subtracting
from both sides and subtracting
from both sides as follows:



Now by dividing by 3 we get our solution.
or 
To solve this problem we need to isolate our variable .
We do this by subtracting from both sides and subtracting
from both sides as follows:
Now by dividing by 3 we get our solution.
or
How many integers
satisfy the following inequality:

How many integers satisfy the following inequality:
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There are two integers between 2.25 and 4.5, which are 3 and 4.
There are two integers between 2.25 and 4.5, which are 3 and 4.
Solve the following inequality:

Solve the following inequality:
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Like any other equation, we solve the inequality by first grouping like terms. Grouping the
terms on the left side of the equation and the constants on the right side of the equation, we have:





Like any other equation, we solve the inequality by first grouping like terms. Grouping the terms on the left side of the equation and the constants on the right side of the equation, we have:
How many integers (x) can complete this inequality?
7< 2x-3 <15
How many integers (x) can complete this inequality?
7< 2x-3 <15
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7< 2x-3 <15
3 is added to each side to isolate the x term:
10< 2x <18
Then each side is divided by 2 to find the range of x:
5< x <9
The only integers that are between 5 and 9 are 6, 7, and 8.
The answer is 3 integers.
7< 2x-3 <15
3 is added to each side to isolate the x term:
10< 2x <18
Then each side is divided by 2 to find the range of x:
5< x <9
The only integers that are between 5 and 9 are 6, 7, and 8.
The answer is 3 integers.
Solve 5 < 3x + 10 leq 16.
Solve 5 < 3x + 10 leq 16.
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5 < 3x + 10 leq 16
Subtract 10: -5 < 3x leq 6
Divide by 3: -5/3 < x leq 2
We must carefully check the endpoints.
is greater than
and cannot equal
, yet
CAN equal 2. Therefore
should have a parentheses around it, and 2 should have a bracket:
is in
![\left ( -$\frac{5}{3}$, 2 \right ]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/84009/gif.latex)
5 < 3x + 10 leq 16
Subtract 10: -5 < 3x leq 6
Divide by 3: -5/3 < x leq 2
We must carefully check the endpoints. is greater than
and cannot equal
, yet
CAN equal 2. Therefore
should have a parentheses around it, and 2 should have a bracket:
is in
Solve
.
Solve .
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Subtract 3 from both sides: 
Divide both sides by
: 
Remember: when dividing by a negative number, reverse the inequality sign!
Now we need to decide if our numbers should have parentheses or brackets.
is strictly greater than
, so
should have a parentheses around it. Since there is no upper limit here,
is in (-2, infty ).
Note: Infinity should ALWAYS have a parentheses around it, NEVER a bracket.
Subtract 3 from both sides:
Divide both sides by :
Remember: when dividing by a negative number, reverse the inequality sign!
Now we need to decide if our numbers should have parentheses or brackets. is strictly greater than
, so
should have a parentheses around it. Since there is no upper limit here,
is in (-2, infty ).
Note: Infinity should ALWAYS have a parentheses around it, NEVER a bracket.
Solve $(x-1)^{2}$(x+4)<0.
Solve $(x-1)^{2}$(x+4)<0.
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$(x-1)^{2}$ must be positive, except when
. When
, $(x-1)^{2}$=0.
Then we know that the inequality is only satisfied when
, and xneq 1. Therefore
, which in interval notation is (-infty , -4).
Note: Infinity must always have parentheses, not brackets.
has a parentheses around it instead of a bracket because
is less than
, not less than or equal to
.
$(x-1)^{2}$ must be positive, except when . When
, $(x-1)^{2}$=0.
Then we know that the inequality is only satisfied when , and xneq 1. Therefore
, which in interval notation is (-infty , -4).
Note: Infinity must always have parentheses, not brackets. has a parentheses around it instead of a bracket because
is less than
, not less than or equal to
.
Solve
.
Solve .
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The roots we need to look at are 
:
Try 
, so
does not satisfy the inequality.
:
Try 

so
does satisfy the inequality.
:
Try 

so
does not satisfy the inequality.
:
Try
.

so
satisfies the inequality.
Therefore the answer is
and
.
The roots we need to look at are
:
Try
, so
does not satisfy the inequality.
:
Try
so does satisfy the inequality.
:
Try
so does not satisfy the inequality.
:
Try .
so satisfies the inequality.
Therefore the answer is and
.
Find the domain of $y=$\sqrt{x^{2}$$-4}.
Find the domain of $y=$\sqrt{x^{2}$$-4}.
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We want to see what values of x satisfy the equation. $x^{2}$-4 is under a radical, so it must be positive.
$x^{2}$-4geq 0
$x^{2}$geq 4
xgeq 2, xleq -2
We want to see what values of x satisfy the equation. $x^{2}$-4 is under a radical, so it must be positive.
$x^{2}$-4geq 0
$x^{2}$geq 4
xgeq 2, xleq -2
Solve the inequality:

Solve the inequality:
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When multiplying or dividing by a negative number on both sides of an inequality, the direction of the inequality changes.

When multiplying or dividing by a negative number on both sides of an inequality, the direction of the inequality changes.
Find the solution set for
:

Find the solution set for :
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Subtract 7:


Divide by -1. Don't forget to switch the direction of the inequality signs since we're dividing by a negative number:

Simplify:
or in interval form,
.
Subtract 7:
Divide by -1. Don't forget to switch the direction of the inequality signs since we're dividing by a negative number:
Simplify:
or in interval form,
.
What is the lowest value the integer
can take?

What is the lowest value the integer can take?
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The lowest value n can take is 6.
The lowest value n can take is 6.
What value of
will make the following expression negative:

What value of will make the following expression negative:
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Our first step is to simplify the expression. We need to remember our order of operations or PEMDAS.
First distribute the 0.4 to the binomial.


Now distribute the 10 to the binomial.

Now multiply 0.6 by 5



Remember to flip the sign of the inequation when multiplying or dividing by a negative number.

220 will make the expression negative.
Our first step is to simplify the expression. We need to remember our order of operations or PEMDAS.
First distribute the 0.4 to the binomial.
Now distribute the 10 to the binomial.
Now multiply 0.6 by 5
Remember to flip the sign of the inequation when multiplying or dividing by a negative number.
220 will make the expression negative.
Solve for
.

Solve for .
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To solve this problem all we need to do is solve for
:



To solve this problem all we need to do is solve for :
What value of
satisfies both of the following inequalities?


What value of satisfies both of the following inequalities?
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Solve for
in both inequalities:
(1)

Subtract 9 from each side of the inequality:

Then, divide by
. Remember to switch the direction of the sign from "less than or equal to" to "greater than or equal to."

(2)

Add
to both sides of the inequality:

Subtract 7 from each side of the inequality:

Divide each side of the inequality by 3:

From solving both inequalities, we find
such that:


Only
is in that interval
.
and
are less than 1.33, so they are not in the interval.
and
are more than 2, so they are not in the interval.
Solve for in both inequalities:
(1)
Subtract 9 from each side of the inequality:
Then, divide by . Remember to switch the direction of the sign from "less than or equal to" to "greater than or equal to."
(2)
Add to both sides of the inequality:
Subtract 7 from each side of the inequality:
Divide each side of the inequality by 3:
From solving both inequalities, we find such that:
Only is in that interval
.
and
are less than 1.33, so they are not in the interval.
and
are more than 2, so they are not in the interval.
Which of the following could be a value of
, given the following inequality?

Which of the following could be a value of , given the following inequality?
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The inequality that is presented in the problem is:

Start by moving your variables to one side of the inequality and all other numbers to the other side:


Divide both sides of the equation by
. Remember to flip the direction of the inequality's sign since you are dividing by a negative number!

Reduce:

The only answer choice with a value greater than
is
.
The inequality that is presented in the problem is:
Start by moving your variables to one side of the inequality and all other numbers to the other side:
Divide both sides of the equation by . Remember to flip the direction of the inequality's sign since you are dividing by a negative number!
Reduce:
The only answer choice with a value greater than is
.


If
and
are two integers, which of the following inequalities would be true?
If and
are two integers, which of the following inequalities would be true?
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First let's solve each of the inequalities:





Don't forget to flip the direction of the sign when dividing by a negative number:

is the correct answer.
is an integer greater than 3 and
is greater than 9. Therefore, the sum of
and
is greater than 12.
is not true.
and
are two positive integers as
is greater than 3 and
is greater than 9. The sum of two positive integers cannot be a negative number.
is not true.
and
are two positive integers as
is greater than 3 and
is greater than 9. The division of two positive numbers is positive and therefore cannot be less than 0.
is not true.
is greater than 3 and
is greater than 9. The product of
and
cannot be less than 3.
is not true.
and
are positive. Therefore, the product of
and
is negative and cannot be greater than 0.
First let's solve each of the inequalities:
Don't forget to flip the direction of the sign when dividing by a negative number:
is the correct answer.
is an integer greater than 3 and
is greater than 9. Therefore, the sum of
and
is greater than 12.
is not true.
and
are two positive integers as
is greater than 3 and
is greater than 9. The sum of two positive integers cannot be a negative number.
is not true.
and
are two positive integers as
is greater than 3 and
is greater than 9. The division of two positive numbers is positive and therefore cannot be less than 0.
is not true.
is greater than 3 and
is greater than 9. The product of
and
cannot be less than 3.
is not true.
and
are positive. Therefore, the product of
and
is negative and cannot be greater than 0.


If an integer
satisfies both of the above inequalities, which of the following is true about
?
If an integer satisfies both of the above inequalities, which of the following is true about
?
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First, we solve both inequalities:





If
satisfies both inequalities, then
is greater than 5 AND
is greater than 11. Therefore
is greater than 11.
is the correct answer.
First, we solve both inequalities:
If satisfies both inequalities, then
is greater than 5 AND
is greater than 11. Therefore
is greater than 11.
is the correct answer.
Solve the following inequality:

Solve the following inequality:
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We start by simplifying our inequality like any other equation:


Now we must remember that when we divide by a negative, the inequality is flipped, so we obtain:

We start by simplifying our inequality like any other equation:
Now we must remember that when we divide by a negative, the inequality is flipped, so we obtain:
Solve the following inequality:

Solve the following inequality:
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Solving inequalities is very similar to solving equations, but we need to remember an important rule:
If we multiply or divide by a negative number, we must switch the direction of the inequality. So a "greater than" sign will become a "less than" sign and vice versa.
We are given

Start by moving the
and the
over:

Simplify to get the following:

Then, we will divide both sides of the equation by
. Remember to switch the direction of the inequality sign!


So,

Solving inequalities is very similar to solving equations, but we need to remember an important rule:
If we multiply or divide by a negative number, we must switch the direction of the inequality. So a "greater than" sign will become a "less than" sign and vice versa.
We are given
Start by moving the and the
over:
Simplify to get the following:
Then, we will divide both sides of the equation by . Remember to switch the direction of the inequality sign!
So,