Absolute Value - Math

To find the -intercepts, we must set .

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

Now we must set up our two scenarios:

and

and

and

To solve, we must first isolate the absolute value on one side:

Next, we must remember that the equation holds true for both the positive and negative of what is inside the absolute value sign, because the absolute value makes whatever is inside of it positive:

Solving for x, we get

One can plug the solutions back into the original equation to confirm.

The inequality compares an absolute value function with a negative integer. Since the absolute value of any real number is greater than or equal to 0, it can never be less than a negative number. Therefore, can never happen. There is no solution.

Solve for positive values by ignoring the absolute value. Solve for negative values by switching the inequality and adding a negative sign to 7.

First, subtract 5 from both sides to get the absolute value expression alone.

Split this into two linear equations:

or

The solution set is

Divide both sides by 3.

Consider both the negative and positive values for the absolute value term.

Subtract 2 from both sides to solve both scenarios for .

The absolute value gives two problems to solve. Remember to switch the "less than" to "greater than" when comparing the negative term.

or

Solve each inequality separately by adding to all sides.

or

This can be simplified to the format .

Remove the absolute value by setting the term equal to either or . Remember to flip the inequality for the negative term!

Solve each scenario independently by subtracting from both sides.

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

and

This gives us:

and

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as cannot equal a negative number.

Our final solution is then

The absolute value of any number is nonnegative, so must always be greater than . Therefore, any value of makes this a true statement.

To solve this inequality, it is best to break it up into two separate inequalities to eliminate the absolute value function:

or .

Then, solve each one separately:

Combining these solutions gives:

We start by finding the midpoint of the interval, which is enclosed by 60% of 204 and 80% of 204.

We find the midpoint, or average, of these endpoints by adding them and dividing by two:

142.5 is exactly 20.5 units away from both endpoints, 122 and 163. Since we are looking for the range of numbers between 122 and 163, all possible values have to be within 20.5 units of 142.5. If a number is greater than 20.5 units away from 142.5, either in the positive or negative direction, it will be outside of the \[122, 163\] interval. We can express this using absolute value in the following way:

We start by finding the midpoint of the interval, which is enclosed by 90 and 210. We find the midpoint, or average, of these two endpoints by adding them and dividing by two:

150 is exactly 60 units away from both endpoints, 90 and 210. Since we are looking for the range of numbers that fall in between 90 and 210, this means that any possible value can't be more than 60 units away from 150. If a number is more than 60 units away from 150, in either the increasing or decreasing direction, it will be outside of the \[90, 210\] interval. We can express this using absolute value in the following way:

We start problems like these by finding the midpoint of the two endpoints, which in this case are 21 and 65. We find the midpoint, or average, by adding them and dividing by two:

43 is right in between 21 and 65, and is exactly 22 units away from each endpoint. Since we are looking for all numbers which fall outside of the the \[21, 65\] interval, we are looking for values which are further than 22 units away from 43, in both the positive and negative directions. Using absolute value, we express this as:

When you subtract 43 from any number larger than 65, the absolute value of the result will be greater than 22. Similarly, when you subtract 43 from any number less than 21, the absolute value of the result will be greater than 22.

Using the "greater than or equal to" sign is necessary in order to include the endpoints, 21 and 65, in our set of allowed ages.

When solving for absolute values, remember there's an equation for positive value and another equation for negative value.

. S

ubtract on both sides of all equations.

Solve the following equation for b:

Let's begin by subtracting 6 from both sides:

Next, we can get rid of the absolute value signs and make our two equations:

Next, add 13 to both sides

And divide by 5:

or

Next, add 13 to both sides

And divide by 5:

So our answers are

When solving for absolute values, remember there's an equation for positive value and another equation for negative value.

Add to both sides of both equations.