GRE Quantitative Reasoning - How to find the solution to an inequality with division

$|3x+4|\leq29$

$|2y+6|\leq18$

Quantity A:

The smallest possible value for

Quantity B:

The smallest possible value for

Which of the following is true?

Quantity A is larger.

Quantity B is larger.

The two quantities are equal.

A comparison cannot be detemined from the given information.

Explanation

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

$|3x+4|\leq29$ ,

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

$-29 \leq 3x+4 \leq 29$

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

First, subtract :

$-33 \leq 3x \leq 25$

Then, divide by :

$-11\leq x \leq \frac{25}{3}$

Next, do the same for the other equation.

$|2y+6|\leq18$

becomes...

$-18 \leq 2y+6 \leq18$

Then, subtract :

$-24 \leq 2y \leq 12$

Then, divide by :

$-12 \leq y \leq 6$

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

What is the solution set of the inequality $\dpi{100} \small 3x+8<35$ ?

$\dpi{100} \small x<9$

$\dpi{100} \small x>9$

$\dpi{100} \small x<27$

$\dpi{100} \small x>27$

$\dpi{100} \small x<35$

Explanation

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

$\dpi{100} \small 3x+8-8<35-8$

$\dpi{100} \small \frac{3x}{3}<\frac{27}{3}$

Thus $\dpi{100} \small x<9$

For how many positive integers, , is it true that

None

More than

Explanation

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

$x^4<27x\Rightarrow x^3<27\Rightarrow x<3\Rightarrow x= 1$ or .

Each of the following is equivalent to

xy/z * (5(x + y)) EXCEPT:

xy(5x + 5y)/z

xy(5y + 5x)/z

5x²y + 5xy²/z

5x² + y²/z

Explanation

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.

Solve for the -intercept:

$3y+11\geq 5y+6x-1$

Explanation

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

$3y+11\geq 5y+6x-1$

$-2y+11\geq6x-1$

$-2y\geq6x-12$

$-\frac{1}{2}(-2y\geq 6x-12)$

$y\leq -3x+6$

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

Solve for .

$x=\frac{3}{4}$

Explanation

For the second equation, solve for in terms of .

Plug this value of y into the first equation.

Solve for :

$x > -\frac{27}{4}$

$x < \frac{27}{4}$

$x > \frac{15}{4}$

$x<\frac{15}{4}$

Explanation

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

Or...

Next, divide both sides by :

$x > -\frac{27}{4}$

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

Let S be the set of numbers that contains all of values of x such that 2x + 4 < 8. Let T contain all of the values of x such that -2x +3 < 8. What is the sum of all of the integer values that belong to the intersection of S and T?

-3

-2

-7

Explanation

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.

What is a solution set of the inequality ?