Recall that the absolute value sign will convert any value to a positive sign. There will be no occurrences of that will evaluate into a negative one as a final solution.

There are no solutions for this equation.

The answer is:

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

Separate the absolute value and solve both the positive and negative components of the absolute value.

Solve the first equation. Add on both sides.

Add two on both sides.

Divide by five on both sides.

One of the solutions is after substitution is valid.

Evaluate the second equation. Distribute the negative sign in the binomial.

Subtract on both sides.

Add two on both sides.

If we substitute this value back to the original equation, the equation becomes invalid.

The answer is:

The expression inside the absolute value brackets could be either positive or negative.

Positive:

subtract 5 from both sides

subtract 2x from both sides

Negative:

subtract 5 from both sides

add 2x to both sides

divide both sides by 5

Now see if either actually works:

Neither actually works.

Since absolute values are concerened with the distance from zero, you need two equations to show both possible solutions.

FIRST

SECOND

Recall that when dealing with absolute value, you have to write the inequality two different ways to solve it:

1. 

2. 

Now, solve each one

Put those answers together to get your final answer:

When dealing with absolute value equations, we need to deal with negative values as well.

Subtract on both sides.

Divide on both sides.

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.

In order to solve the equation, we have to isolate the variable. We do this by performing the same operation to either side of the equation.

First, divide both sides by 3.

The absolute value function makes whatever is inside positive, so this has two potential solutions. could either be positive or negative 9, so we have to solve for both:

or

First, divide both sides by -3:

Whatever is inside the absolute value brackets could be positive or negative.

Positive:

Negative:

Both of those answers may not work, however, so test both:

The only answer that works is -1.