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# Solution of systems of linear inequalities in one variable & linear equations with absolute values

Solution of Systems of Linear inequalities in One Variable

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+32 x-5 and 3 x-7<5 x-9

Solution

We first ﬁnd the solution set of each inequality.

 6 x+3≥2 x-5 3 x-7<5 x-9 4 x≥− 8 92 x<− 2 x≥− 2 x>1 The Solution set is The Solution set is

The solution set of the system (Figure 5.7) is

FIGURE 5.7

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EXAMPLE Find the solution set of the following system:

4 (3-x)<7+3 (2-x) and 3 (x-1)<4-(1-x)

Solution We ﬁrst ﬁnd the solution set of each inequality

 4 (3-x)<7+3 (2-x) 3 (x-1)<4-(1-x) 12-4 x<7+6-3 x 3 x-3<4-1+x − x<1 2 x<6 x>− 1 x<3 The Solution set is The Solution set is

The solution set of the system (Figure 5.8) is

FIGURE 5.8

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Solution of Linear Equations with Absolute Values

The absolute value of a number AR, denoted by |a|, is either + aor  a whichever is positive, and is zero if a=0

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 aR.

When we have the absolute value of a quantity involving a variable such as |x-1| that quantity, x-1. could be

1. greater than or equal to zero, or

2. less than zero.

When x-1 is greater than or equal to zero, that is,

x-10

then |x-1| = x-1

When x-1 is less than zero, that is, x-1<0, then

|x-1| = (x-1) = x+1

The following examples illustrate how to solve a linear equation in one variable involving absolute value.

EXAMPLE Solve the equation |x-3| = 5

Solution   To ﬁnd the solution set of this equation, we have to consider two cases

First

When x-30 that is, x3

|x-3| = x-3

The equation now becomes

|x-3| = x-3 = 5 or x=8

The solution set is the intersection of the solution sets of

x3 and x=8

The solution set (Figure 5.9) is {8}.

Figure 5.9

Second When x-3<0 that is, x<3

|x-3| = (x-3) = x+3

The equation now becomes

|x-3| = 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 |x-3| = 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+30 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 x-3

The equation now becomes

|2 x+3| = 2 x-3 = 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 x-5|=x+3.

First

When 2 x-50 that is x52 (1)

Then |2 x-5| = 2 x-5

Thus |2 x-5| = x+3 becomes

2 x-5=x+3, or x=8 (2)

From (1) and (2) the solution set is {8}.

Second   When 2 x-5<0, that is, x<52; (3)

then |2 x-5| = (2 x-5) = 2 x+5

Thus, |2 x-5| = 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 x-5| = 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 |4-3 x| = 3 x-4.

Solution   First

When 4-3 x0 that is, x43 (1)

then |4-3 x| = 4-3 x

Thus |4-3 x| = 3 x-4 becomes

4-3 x=3 x-4, or x=4-3 (2)

From (1) and (2) the solution set is {43}.

Second   When 4-3 x<0, that is x>43; (3)

then |4-3 x| = (4-3 x) = 4+3 x

Thus |4-3 x| = 3 x-4 becomes

4+3 x = 3 x-4 or 0 x=0

Which is true for all xR (4)

From (3) and (4) the solution set is {x |x|}>{43}.

The solution set of |4-3 x| = 3 x-4 is the union of the solution sets in the tow case

Hence the solution set is

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