Inequalities - GRE Subject Test: Math

Since the absolute value with x in it is alone on one side of the inequality, you set the expression inside the absolute value equal to both the positive and negative value of the other side, 11 and -11 in this case. For the negative value -11, you must also flip the inequality from less than to a greater than. You should have two inequalities looking like this.

and

Add 5 to both sides in each inequality.

and

Divide by -4 to both sides of the inequality. Remember, dividing by a negative will flip both inequality symbols and you should have this.

and

The correct answer is and

The correct answer is and

Before expanding the quantity within absolute value brackets, it is best to simplify the "actual values" in the problem. Thus becomes:

From there, note that the absolute value means that one of two things is true: or . You can therefore solve for each possibility to get all possible solutions. Beginning with the first:

means that:

For the second:

means that:

Note that the two solutions can be connected by putting the inequality signs in the same order:

Because Absolute Value must be a non-negative number, there is no solution to this Absolute Value inequality.

The following absolute value inequality can be used to assess the bowling balls that are tolerable:

The correct answer is and

At this point, you've isolated the absolute value and can solve this problems for both cases, and . Beginning with the first case:

Then for the second case:

The correct answer is and

The Absolute Value Inequality that can assess which cell phones are tolerable is:

Step 1: Separate the equation into two equations:

First Equation:
Second Equation:

Step 2: Solve the first equation

Step 3: Solve the second equation

The solution is

Method 1:

1. Multiply-out the left side then rewrite the inequality as an equation:

1. Now rewrite as a quadratic equation and solve the equation:

1. Next set up intervals using the solutions and test the original inequality to see where it holds true by using values for on each interval.

2. The interval between and holds true for the original inequality.

3. Solution:

Method 2:

Using a graphing calculator, find the graph. The function is below the x-axis (less than ) for the x-values . Using interval notation for , .

Method 3:

For the inequality , the variable expression in terms of is less than , and an inequality has a range of values that the solution is composed of. This means that each of the solution values for are strictly between the two solutions of . 'Between' is for a 'less than' case, 'Outside of' is for a 'greater than' case.

We begin by moving all of our terms to the left side of the inequality.

We then factor.

That means our left side will equal 0 when . However, we also want to know the values when the left side is less than zero. We can do this using test regions. We begin by drawing a number line with our two numbers labeled.

We notice that our two numbers divide our line into three regions. We simply need to try a test value in each region. We begin with our leftmost region by selecting a number less than . We then plug that value into the left side of our inequality to see if the result is positive or negative. Any value (such as ) will give us a positive value.

We then repeat this process with the center region by selecting a value between our two numbers. Any value (such as ) will result in a negative outcome.

Finally we complete the process with the rightmost region by selecting a value larger than . Any value (such as ) will result in a positive value.

We then label our regions accordingly.

Since we want the result to be less than zero, we want the values between our two numbers. However, since our left side can be less than or equal to zero, we can also include the two numbers themselves. We can express this as

Before we get started, read the question carefully. We need to find x squared, not x. Don't call it quits too early!

So, we start here:

Get the x's on one side and the constants on the other.

When we divide by a negative number in an inequality, remember that we need to switch the direction of the sign.

We begin by solving the equation for its zeros. This is done by changing the sign into an sign.

Since we know the zeros of the equation, we can then check the areas around the zeros since we naturally have split up the real line into three sections :

First we check

Therefore, the first interval can be included in our answer. Additionally, we know that satisfies the equation, therefore we can say with certainty that the interval is part of the answer.

Next we check something in the second interval. Let , then

Therefore the second interval cannot be included in the answer.

Lastly, we check the third interval. Let , then

Which does satisfy the original equation. Therefore the third interval can also be included in the answer. Since we know that satisfies the equation as well, we can include it in the interval as such:

Therefore,

The first thing we must do is solve the given equation for :

Since we are looking for values when , we can set up our equation as follows:

Solve.

So, is the largest integer of x which makes the statement true.

We can either divide the other side of the inequality by or distribute it. We'll go ahead and distribute it here:

Now we just solve:

Our first step will be to solve the given equation for :

Since we want to know the smallest integer of for which , we can set up our equation as

This problem requires first combining like terms that are on opposite sides of the inequality.

First add 3x to both sides of the inequality to get

.

Then subtract 2 from both sides.

Now, divide 7 from both sides to get x alone.

which you should switch the order to be

.