


Solution of systems of linear inequalities in one variable & linear equations with absolute values
Sometimes it is necessary to ﬁnd the common solution, or solution set, of two or more inequalities, called a system of inequalities. The solution set of a system of inequality is thus the intersection of the solution set of each inequality in the system.
EXAMPLE Find the solution set of the following system: 6 x+3≥2 x5 and 3 x7<5 x9 Solution We first ﬁnd the solution set of each inequality.
The solution set of the system (Figure 5.7) is
FIGURE 5.7 Let’s see how our Linear EXAMPLE Find the solution set of the following system: 4 (3x)<7+3 (2x) and 3 (x1)<4(1x) Solution We ﬁrst ﬁnd the solution set of each inequality
The solution set of the system (Figure 5.8) is
FIGURE 5.8 Let’s see how our Linear
The absolute value of a number A∈R, denoted by a, is either + a That is,
EXAMPLES 1. 6=6 2. − 4=− (− 4)=4 Note that the absolute value of any real number is either zero or a positive number, never a negative number. That is, a≥0 for all a∈R. When we have the absolute value of a quantity involving a variable such as x1 that quantity, x1. could be 1. greater than or equal to zero, or 2. less than zero. When x1 is greater than or equal to zero, that is, x1≥0 then x1 = x1 When x1 is less than zero, that is, x1<0, then x1 = − (x1) = − x+1 The following examples illustrate how to solve a linear equation in one variable involving absolute value.
EXAMPLE Solve the equation x3 = 5 Solution To ﬁnd the solution set of this equation, we have to consider two cases First When x3≥0 that is, x≥3 x3 = x3 The equation now becomes x3 = x3 = 5 or x=8 The solution set is the intersection of the solution sets of x≥3 and x=8 The solution set (Figure 5.9) is {8}.
Figure 5.9 Second When x3<0 that is, x<3 x3 = − (x3) = − x+3 The equation now becomes x3 = − x+3 = 5 or x=− 2 The solution set is the intersection of the solution sets of x<3 and x=− 2 The solution set (Figure 5.10) is {− 2}.
Figure 5.10 The solution set of x3 = 5 is the union of the solution sets in the tow case Hence the solution set is {− 2,8}.
EXAMPLE Find the solution of 2 x+3 = 9. Solution First: When 2 x+3≥0 that is x≥− (32) 2 x+3 = 2 x+3 The equation now becomes 2 x+3 = 2 x+3 = 9, or x=3 The solution set is the intersection of the solution sets of x≥− (32) and x=3 The solution set (Figure 5.11) is {3}.
Figure 5.11 Second: When 2 x+3<0, that is, x<− 32 2 x+3 = − (− 2 x+3) = − 2 x3 The equation now becomes 2 x+3 = − 2 x3 = 9, or x=− 6 The solution set is the intersection of the solution sets of x<− 32 and x=− 6 The solution set (Figure 5.12) is {− 6}.
Figure 5.12 The solution set of 2 x+3 = 6 is the union of the solution sets in the tow case Hence the solution set is {− 6,3}. Note Since the absolute value of any real number is never negative. the solution set of the equation 3 x+5=− 4 is Φ Find the solution set of 2 x5=x+3. First When 2 x5≥0 that is x≥52 (1) Then 2 x5 = 2 x5 Thus 2 x5 = x+3 becomes 2 x5=x+3, or x=8 (2) From (1) and (2) the solution set is {8}. Second When 2 x5<0, that is, x<52; (3) then 2 x5 = − (2 x5) = − 2 x+5 Thus, 2 x5 = x+3 becomes − 2 x+5=x+3, or x=23 (4) From (3) and (4) the solution set is {23}. The solution set of 2 x5 = x+3 is the union of the solution sets in the tow case Hence the solution set is {23,8}.
EXAMPLE Find the solution set of 43 x = 3 x4. Solution First When 43 x≥0 that is, x≥43 (1) then 43 x = 43 x Thus 43 x = 3 x4 becomes 43 x=3 x4, or x=43 (2) From (1) and (2) the solution set is {43}. Second When 43 x<0, that is x>43; (3) then 43 x = − (43 x) = − 4+3 x Thus 43 x = 3 x4 becomes − 4+3 x = 3 x4 or 0 x=0 Which is true for all x ∈ R (4) From (3) and (4) the solution set is {x x}>{43}. The solution set of 43 x = 3 x4 is the union of the solution sets in the tow case Hence the solution set is
Let’s see how our Linear
