This week in Pre-calculus 11 we learned all about inequalities. Something that I learned that really stuck with was how to solve linear inequalities and graph them.

Signs and symbols to know:

• > greater than / < less than
• ≥ greater than or equal to / ≤ less than or equal to
• An open dot is used to indicate that the boundary point isn’t apart of the solution / A closed dot is used to indicate that the boundary point is a part of the solution
• A dotted line is used to is used when the values along the line aren’t included in the solution [> or <] / A solid line is used when the values along the line are included in the solution [≥ or ≤]

Example 1:

• First, you need to make it so that one side is equal to zero.
• Next, you factor the inequality and find its zeros. You do this using the zero product law. In this case the zeros would be: -5 and 1.
• Next you would write these on a number line and pick three test numbers. You pick one number from each of the three sections of the number line. (In this case you would pick a number less than -5, one in between -5 and 1, and one greater than 1).
• after doing that you test the points by putting them into the inequality. This will help you write your solution. Out of the three test points there will be some that prove to be true and others that will not. You take the one(s) that are true and use them to write your solution. (In this case our test point in between -5 and 1 proved to be true).
• Write your solution. Since we know that the value of x is in between -5 and 1 we would write it as {-5<x<1}.

Example 2:

*Note: another method used to solve is a sign chart*

• First you need to factor the expression. In order to do this you need to make sure that one side is equal to zero. To do this you would just rearrange the inequality and move things from one side to the other.
• After factoring you find the zeros of the inequality. (in this case they would be -2 and 6).
• Next you would create a sign chart. To do this you create two number lines with the zeros written on them. You would then write one of the binomials beside one of the number lines and the one beside the other and circle the zero that corresponds with that binomial. For the first number line we would write in each section of the line if the solution would be + or -. If we know that putting -2 into the binomial (x+2) = 0, then we know that anything less than it would result in a negative and anything greater than it would result in a positive. For the second line we do the same thing. If we know that when we put 6 into the binomial (x-6) = 0, then we also know that any value less than 6 would result in a negative and anything greater than 6 would result in a positive.
• Next you multiply the signs that are across from each other. We know: [(-)(-)= +/ (+)(-)= -/ (+)(+)= +]. You then compare your inequality to the new values and use the sections that prove the inequality to be true to write your solution. (in this case you would compare the inequality to + – +)
• Then you write the solution. We know that the solution is -2≤x≤6 because the only time that zero is greater than or equal to x is when x is in the middle section (in between -2 and 6).