In this week of precalc 11, we learned about solving inequalities. This was partly a review from grade 9, with new skills added on. Linear equations make a line on a graph, while quadratic equations create a parabola (a curved symetrical shape). Both of these equations can be inequalities as well.

Here are some examples of linear inequalities:

6x > 18

4x + 1 < 9

3x – 3 ≥ 4x + 1

x² + 3x + 2 ≤ 0

2x² > -3x – 7

x² – 7x + 9 ≥ 9

To solve an inequality, you must get the variable on its own. For example:

For this inequality: 6x > 18, you divide both sides by 6, so the x is by itself. Therefore, x would be greater than 3 (x > 3)

To solve this inequality: 3x – 3 ≥ 4x + 1, start by subtracting the 4x from both sides. Then add the 3 from the left side to the right. This would create the equation:  – 1x ≥ 4. To make x positive, divide both sides by negative 1. This would make x less than or equal to 4 (x ≤ -4), because when you multiply or divide by a negative, the sign changes.

Here is one final example using the quadratic inequality: x² – 4x > 5

This is done a bit differently than the linear inequalities because you need to factor. First, move the 5 to the left side so the inequality has the zero on the right: x² – 4x – 5 > 0

Then, factor the right side: (x – 5) (x + 1) > 0

Next, find the roots. This is the two numbers, 5 and 1, but with opposite signs: +5, -1.

Finally, write out the solution: x < -1, and x > 5. This is the answer because the original question asked where the inequality was greater than 0. Because this equation would open up and have a minimum vertex (the x² was positive), it is greater than zero to the left (less than -1) and to the right (greater than 5) of the parabola. In between the parabola, x is less than 0, so it wouldn’t satisfy the inequality.