Inequalities - GRE Quantitative Reasoning

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.

3_w_/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.

The question does not give us any specifics about the variables x and y.

If we substitute the same numbers for x and y (say, x = 1 and y = 1), the two expressions are equal.

If we substitute different number in for x and y (say, x = 2 and y = 1), the two expressions are not equal.

If there are two possible outcomes, then we need more information to determine which quantity is greater. Don't be afraid to pick "The relationship cannot be determined from the information given" as an answer choice on the GRE!

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.

The expression can be rewritten as .

The only integer that satisfies the inequality is 0.

Thus, Quantity A and Quantity B are equal.

First, solve the inequality :

Since we are dealing with absolute value, must also be true; therefore:

Start by subtracting 3 from each side of the inequality. That gives us . Divide both sides by 2 to get . Therefore every value for where is a solution to the original inequality.

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.

Choice a is equivalent because we can say that technically we are multiplying two fractions together: (xy)/z and (5(x + y))/1. We multiply the numerators together and the denominators together and end up with xy (5x + 5y)/z. xy (5y + 5x)/z is also equivalent because it is only simplifying what is inside the parentheses and switching the order- the commutative property tells us this is still the same expression. 5x²y + 5xy²/z is equivalent as it is just a simplified version when the numerators are multiplied out. Choice 5x² + y²/z is not equivalent because it does not account for all the variables that were in the given expression and it does not use FOIL correctly.

First, we need to find all of the values that are in the set S, and then we need to find the values in T. Once we do this, we must find the numbers in the intersection of S and T, which means we must find the values contained in BOTH sets S and T.

S contains all of the values of x such that 2x + 4 < 8. We need to solve this inequality.

2x + 4 < 8

Subtract 4 from both sides.

2x < 4

Divide by 2.

x < 2

Thus, S contains all of the values of x that are less than (but not equal to) 2.

Now, we need to do the same thing to find the values contained in T.

-2x + 3 < 8

Subtract 3 from both sides.

-2x < 5

Divide both sides by -2. Remember, when multiplying or dividing an inequality by a negative number, we must switch the sign.

x > -5/2

Therefore, T contains all of the values of x that are greater than -5/2, or -2.5.

Next, we must find the values that are contained in both S and T. In order to be in both sets, these numbers must be less than 2, but also greater than -2.5. Thus, the intersection of S and T consists of all numbers between -2.5 and 2.

The question asks us to find the sum of the integers in the intersection of S and T. This means we must find all of the integers between -2.5 and 2.

The integers between -2.5 and 2 are the following: -2, -1, 0, and 1. We cannot include 2, because the values in S are LESS than but not equal to 2.

Lastly, we add up the values -2, -1, 0, and 1. The sum of these is -2.

The answer is -2.

For the second equation, solve for in terms of .

Plug this value of y into the first equation.

We simplify this inequality similarly to how we would simplify an equation

Thus

Don't forget to switch the inequality direction if you multiply or divide by a negative.

Now that we have the equation in slope-intercept form, we can see that the y-intercept is 6.

Since is positive, we can divide both sides of the inequality by :

or .

In order to find the solution set, we solve as we would an equation:

Therefore, the solution set is any value of .

Begin by adding to both sides, this will get the variable isolated:

Or...

Next, divide both sides by :

Notice that when you divide by a negative number, you need to flip the inequality sign!

Recall that when you have an absolute value and an inequality like

,

this is the same as saying that must be between and . You can rewrite it:

To solve this, you just apply your modifications to each and every part of the inequality.

First, subtract :

Then, divide by :

Next, do the same for the other equation.

becomes...

Then, subtract :

Then, divide by :

Thus, the smallest value for is , while the smallest value for is . Therefore, quantity A is greater.

For quantitative comparison questions involving a shared variable between quantities, the best approach is to test a positive integer, a negative integer, and a fraction. Half of our work is eliminated, however, because the question stipulates that x > 0. We only need to check a positive integer and a positive fraction between 0 and 1. Plugging in 2, we see that quantity A is greater than quantity B. Checking 1/2, however, we find that quantity B is greater than quantity A. Thus the relationship cannot be determined.