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Linear Inequalites with Absolute Value

Linear Equations present many problems to students, and by the time a student finally grasps the basics, teachers start throwing inequalities into the mix. However, students soon adapt and follow along, sometimes with understanding and sometimes without understanding. But one common topic that confuses most students is Absolute Value. The absolute value of a negative number is a positive number and the absolute value of a positive number is a positive number. Therefore, absolute value is always positive.

Let’s start with a regular equation to demonstrate an important concept.

I x I > 15

Start with the Positive Case. And remove the absolute value symbols to obtain the ordinary equation.

x > 15

Graphing on a number line gives

Remember to keep the open circle to indicate that it begins x begins at 15 but does not include 15. Very simple and easy to understand this, just act like it’s a normal inequality and remove the absolute value bars. However, what about when x is negative?

Next we have to explore the negative case. But first I want us to think through what is happening. When x = -3, the absolute value bars turn the negative number into a positive number which gives the following results

I x I = I -3 I = 3 Equation 1

In other words, if x is negative we get a positive number. So what happens if we continue the pattern by decreasing the value of x by one digit at a time?

I x I = I -5 I = 5 Equation 2

I x I = I -8 I = 8 Equation 3

I x I = I -10 I = 10 Equation 4

Notice that as the value of x decreases negatively, the absolute value of x increases positively. This means that if

x < -15 then

I x I > 15

Now that the concept makes sense, let’s do the actual math to develop our understanding of the negative case.

If x is less than zero (x < 0) then -1 times x (-x) is positive and therefore equal to the absolute value of x.

This gets into what the negative case is all about, substituting –x into the original equation and then solving for x which gives us

-x > 15 divide both sides by -1 and switch the inequality around

x < -15

So, looking at our number line and remembering the solution for the positive case we can graph the negative case and observe our results.

This is actually quite easy to solve one step equations like this, simply substitute –x for x and solve for x. With a linear equation, you would solve for y and define it in terms of x on opposite sides of an inequality sign. Simple example being:

y > I x I First take the positive case where x > 0 and obtain

y > x For all x greater than 0. Graphing the equation we obtain

Remember, because y is greater than but not equal to x, we have to use a dotted line, other than that, let’s break this down and understand what just happened. First thing we notice is that y = x is just a line set at a 45 degree angle. Every time x goes over one, y goes up one – plain and simple. Notice that also, when exploring the positive case of this equation we are only considering x > 0 which is why the graph takes place in the 1st Quadrant and not in any other. Since y > x, we shaded the region above the line y = x.

Now let’s explore the negative case, where x < 0. Remember, everything inside the Absolute value sign gets multiplied by -1. And since this case only involves 1 variable x, we can isolate x by only considering x < 0. Remember what happens when a negative number gets multiplied by a negative? It becomes positive. Therefore the inequality remains the same and the result is y > -x for all x < 0 giving us the following result:

As with the positive case, we are simply looking at the line y = -x and noting that y is greater than –x for all values of x is less than 0. Because we are only considering values where x is less than 0, the graph takes place in the second quadrant. Connecting the two graphs gives us the overall graph of y > I x I (y is greater than the absolute value of x. Again, we used a dotted line because y is greater than but not equal to –x. Thus we have the answer to our basic question. If you feel you have a good grasp of the basics, try the problem in my next post and see if you can solve it.