Solving Inequalities

There are a number of different methods used for solving inequalities. These methods are outlined in different math textbooks. The method used in this presentation demonstrates one of these methods.

Concept & Philosophy: To solve an inequality, find out where equality occurs. If one can determine where equality occurs ... then you also can determine where inequality occurs. The only issue becomes "which region/interval" of inequality is that being sought by/in the original problem.

An example: Imagine you need to solve the inequality ...

Step 1: Change the inequality to an equality and solve this equality. (You should drop the inequality symbol and replace it with an equality symbol ... and then solve the resulting equation using whatever methods seem appropriate.)

Step 2:  Make an "x" number line locating the points of equality found above (in step 1). Put a "hard dot" (filled-in dot) on the points if the original problem allowed equality, Put an "open circle" around the points if the original problem was a "pure" inequality (allowing for no potential of equality). In this case ... locate the points and put open circles around them.

Step 3:  Examine the number line from step 2. It will be divided into intervals ... each of which represents a region of inequality. In this instance, there are three regions of inequality: all values to the left of and less than -5, the region between -5 and 2, and then the region comprised of all values to the right of and greater than 2. The issue you face: which of these three regions represent the one(s) required by the original problem ... the inequality stated in the original problem.

To determine which (if any) of the regions work ... pick a value from each of the regions and test this value back in the original inequality. If the value makes the original inequality true, then ALL values in the interval work. If the test value makes the original inequality false, then NO value in the interval work. In this case:

From left most interval try x = -10:  (-10)²  +  3*(-10)  -  10  =  60 which is NOT less than "0". NO VALUE IN LEFT MOST INTERVAL WORK!!!

From the middle interval try x = 1:  1²  +  3*1  -  10  = -6 which IS less than "0". ALL VALUES IN THE MIDDLE INTERVAL SATISFY THE ORIGINAL & WORK!!!

From the right most interval try x = 4:  4²  +  3*4  - 10  = 18 which is NOT less than "0". NO VALUE IN RIGHT MOST INTERVAL WORK!!!

Step 4: Shade the region (interval) of the number line that makes the original inequality true ... then express this solution as an algebraic statement and as an interval. In this case, the algebraic solution is -5 < x < 2 and the interval solution is (-5,2).

Other:  If dealing with a rational (fractional) expression, one must use all values which cause a zero in the denominator. In general, any values that cause a function to be undefined are critical values ... and must be considered and graphed in step 2 above.

Return to Index page