Unizor - Algebra - Linear Inequalities and Absolute Value скачать в хорошем качестве

Unizor - Algebra - Linear Inequalities and Absolute Value

Consider a more complicated case of linear inequalies that involves absolute values. As an example, let's start with an inequality |X + 2| - |X - 1| 'is less than' 0. This is not exactly a linear inequality in its original sense, but a few logical steps will lead us to its solution in a manner equivalent to a "pure" linear case. When dealing with absolute values of expressions that contain unknowns we have to consider separately all cases when each absolute value equals to zero and all values of unknown in-between these key points. The reason for this is that, by definition, |Z| equals to Z for non-negative Z and equals to -Z for negative Z. The key points for an unknown X in the example above are X=-2 and X=1. These points divide all the values of X into three separate intervals: -Infinity 'is less than' X 'is less than' -2, -2 = X 'is less than' 1, 1 = X 'is less than' +Infinity. Now we will consider our original inequality on each of these intervals and in each case we can get rid of the absolute value using its definition mentioned above. Case A: -infinity 'is less than' X 'is less than' -2 In this case both X+2 and X-1 are negative, therefore |X+2| = -(X+2) and |X-1| = -(X-1). Therefore, the left part of our inequality becomes -(X+2) - (-(X-1)) = -X - 2 + X - 1 = -3. Since -3 'is less than' 0, all values of X we consider in this case are solutions to the original inequality: -infinity 'is less than' X 'is less than' -2. Case B: -2 'is less than or equal to' X 'is less than' 1 In this case X+2 is non-negative and X-1 is negative, therefore |X+2| = (X+2) and |X-1| = -(X-1). Therefore, the left part of our inequality becomes (X+2) - (-(X-1)) = X + 2 + X - 1 = 2·X + 1. We can solve now a linear inequality 2·X + 1 'is less than' 0 getting the condition X 'is less than' -1/2, which we have to consider in conjunction with the case condition -2 'is less than or equal to' X 'is less than' 1. This results in the following solution to the original inequality: -2 'is less than or equal to' X 'is less than' -1/2. Case C: 1 'is less than or equal to' X 'is less than' +infinity In this case X+2 is non-negative and X-1 is non-negative, therefore |X+2| = (X+2) and |X-1| = (X-1). Therefore, the left part of our inequality becomes (X+2) - (X-1) = 3 Since 3 'is greater than' 0, no values of X we consider in this case are solutions to the original inequality. Combining all three cases, we conclude that the solution to our inequality can be described as two conditions on the unknown variable X: -infinity 'is less than' X 'is less than' -2 and -2 'is less than or equal to' X 'is less than' -1/2. Or, since these intervals are adjacent to each other, we can combine them into one interval: -infinity 'is less than' X 'is less than' -1/2.