All GRE Math Resources
Example Questions
Example Question #4 : How To Find The Solution To An Inequality With Addition
Find all solutions of the inequality .
All .
All .
All .
All .
All .
All .
Start by subtracting 13 from each side. This gives us . Then subtract from each side. This gives us . Divide both sides by 2 to get . Therefore all values of where will satisfy the original inequality.
Example Question #21 : Inequalities
What values of x make the following statement true?
|x – 3| < 9
x < 12
–3 < x < 9
–6 < x < 12
6 < x < 12
–12 < x < 6
–6 < x < 12
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Example Question #22 : Inequalities
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
w2
3w/2
|w|0.5
|w|
w/2
3w/2
3w/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
Example Question #23 : Inequalities
Solve for .
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
Example Question #24 : Inequalities
If and , then which of the following could be the value of ?
To solve this problem, add the two equations together:
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
Example Question #6 : How To Find The Solution To An Inequality With Addition
What values of make the statement true?
First, solve the inequality :
Since we are dealing with absolute value, must also be true; therefore:
Example Question #1 : Algebraic Fractions
Which of the following are answers to the equation below?
I. -3
II. -2
III. 2
II and III
III only
II only
I, II, and III
I only
III only
Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept. Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,
The expression is a difference of squares that can be factored as
Solving this for gives either or . That means either of these values will make our numerator equal zero. We might be tempted to conclude that both are valid answers. However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself. That means that any values that make our denominator zero must be rejected. Therefore we must also look at the denominator.
The left side factors as follows
This means that if is or , we end up dividing by zero. That means that cannot be a valid solution, leaving as the only valid answer. Therefore only #3 is correct.
Example Question #1 : How To Find Excluded Values
Which of the following provides the complete solution set for ?
No solutions
The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as . Thus all values except for 2 will work.
Example Question #3 : How To Find Excluded Values
If then which values of cannot exist?
The denominator of a fraction can never be equal to 0.
Therefore, to find out what x cannot be equal to, we must factor the denominator, and determine what values of x would make it equal to 0.
Therefore, and .
Example Question #1 : Algebraic Fractions
If then which cannot be an value?
You cannot take the square root of a negative number.
Setting up the inequality we get:
Solving for we get:
Therefore any value less than four will not work, .
Another approach is to plug in each of the possible values.
When plugged into all of the answers give us a value greater than or equal to 0, except for , which gives us .