"Solving" an inequality means determining precisely which real numbers give a true statement when you substitute them for the variable in the inequality. This is our second approach to solving inequalities.
 

Example: Solve the inequality x² - 3x + 3 <= 2x - 3.
                    (Netscape doesn't seem to recognize the "less than or equal to" symbol.)
 

Paper and Pencil Procedure:

• Graph the two equations y = x² - 3x + 3 and y = 2x - 3.
• Find the x-coordinates of the two points of intersection.
• Then looking at the picture should tell you which real numbers satisfy the inequality.

 

Calculator Procedure:

• On the HOME screen, start a new problem.

• Go to the Y= screen.
• Clear the screen of any leftover functions. (F1,8 works.)
• Set y1(x) equal to x² - 3x + 3.
• Set y2(x) equal to 2x - 3.
• Now graph.

There are lots of ways to do this.

• You could use pencil and paper to solve the equation x² - 3x + 3 = 2x - 3. (This is not much different from using pencil and paper to solve the original inequality .)
• You could use the calculator HOME screen to step through the procedure for solving the same equation.
• You could ask the calculator to solve the equation for you.
• On the GRAPH screen, you could use the Trace feature (F3) to locate intersection points visually.
• The Math menu (F5) on the GRAPH screen includes an intersection utility.

 

You don't actually need the graph.

• Use one of the non-graphical methods for solving the equation x² - 3x + 3 = 2x - 3.
• Plot the solutions on a number line; they divide the line into three pieces (five if you count the points themselves as pieces).
• Pick one number in each piece of the number line and check whether the inequality is satisfied at that point. (On the calculator, type in the inequality and follow it with |x=3 or the like. The vertical line is the with symbol; it temporarily substitutes 3 for x. The calculator should respond either true or false.)
• Checking one point in each piece tells the story for the whole piece. (This follows from the Intermediate Value Theorem and the fact that both formulas define continuous functions,)
• Modification for rational functions: Any number which makes some denominator 0 needs to be included in your collection of dividing points.

 

1. Solve x² - 3x + 3 => 3x - 3.

2. Solve |2x - 5| < 3. (You get absolute value on the calculator by typing abs( ).)

3. Solve 1/x <= 4.

4. Solve |2x - 5| <= |5x - 2|.

5. Solve 3.217x² - 9.402x + 8.306 <= 7.841x - 6,669.

6. Solve 3.1x³ + 2.4x² - 1.3x - 0.7 < 0.

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