Solution Of Systems Of Linear Inequalities In One Variable - QuickMath

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:

6x+3>=2x-5 and 3x-7<5x-9

Solution  

We first find the solution set of each inequality.

6x+3>=2x-5	3x-7<5x-9
4x>=-8	92x<-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)
12-4x<7+6-3x	3x-3<4-1+x
-x<1	2x<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 A∈R, 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 a∈R.

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-1>=0

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-3>=0 that is, x>=3

|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

x>=3 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 |2x+3| = 9.

Solution First:

When 2x+3>=0 that is x>=-(3/2)

|2x+3| = 2x+3

The equation now becomes

|2x+3| = 2x+3 = 9, or x=3

The solution set is the intersection of the solution sets of

x>=-(3/2) and x=3

The solution set (Figure 5.11) is {3}.

  

Figure 5.11

Second: When 2x+3<0, that is, x<-3/2

|2x+3| = -(-2x+3) = -2x-3

The equation now becomes

|2x+3| = -2x-3 = 9, or x=-6

The solution set is the intersection of the solution sets of

x<-3/2 and x=-6

The solution set (Figure 5.12) is {-6}.

  

Figure 5.12

The solution set of |2x+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 |3x+5|=-4 is Φ

Find the solution set of |2x-5|=x+3.

First

When 2x-5>=0 that is x>=5/2 (1)

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

Thus |2x-5| = x+3 becomes

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

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

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

then |2x-5| = -(2x-5) = -2x+5

Thus, |2x-5| = x+3 becomes

-2x+5 = x+3, or x=2/3 (4)

From (3) and (4) the solution set is {2/3}.

The solution set of |2x-5| = x+3 is the union of the solution sets in the tow case

Hence the solution set is {2/3, 8}.

EXAMPLE Find the solution set of |4-3x| = 3x-4.

Solution First

When 4-3x>=0 that is, x>=4/3 (1)

then |4-3x| = 4-3x

Thus |4-3x| = 3x-4 becomes

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

From (1) and (2) the solution set is {4/3}.

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

then |4-3x| = -(4-3x) = -4+3x

Thus |4-3x| = 3x-4 becomes

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

Which is true for all x ∈ R (4)

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

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

Hence the solution set is

 

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