Inequations and Absolute Values

Inequations

Inequations Signs and Operations

 

> means greater than

\ge  means greater than or equal to

< means less than

\le means less than or equal to

 

In order to solve inequations, we need to see what effect one operation applied to both sides has on the inequality sign.

For example,

3>2 \qquad  -> \qquad  3+1>2+1 \qquad  -> \qquad 4>3

2<3 \qquad  -> \qquad 2-1<3-1 \qquad -> \qquad 1<2

3>2 \qquad  -> \qquad 3 \times 2>2 \times 2 \qquad -> \qquad 6>4

3>2 \qquad  -> \qquad 3 \div 2>2 \div 2 \qquad ->\qquad 1.5>1

However, the inequality sign reverses when:

  • multiplying by a negative
  • dividing by a negative
  • taking the reciprocal of both sides

For example:

6>4 \qquad -> \qquad 6 \times -2 < 4 \times -2 \qquad -> \qquad -12<-8

6>4 \qquad -> \qquad 6 \div -2 < 4 \div -2 \qquad -> \qquad -3<-2

6>4 \qquad -> \qquad \cfrac { 1 }{ 6 } <\cfrac { 1 }{ 4 }

Solving Inequations

 

Inequations, just like equations are used to find unknown values of a pronumeral. For example,

5x+7\ge 17\\ \\ 5x+7-7\ge 17-7\\ \\ 5x\ge 10\\ \\ \cfrac { 5x }{ 5 } \ge \cfrac { 10 }{ 5 } \\ \\ x\ge 2

However, what does x\ge 2 mean?

This means that the solution is correct for all values of x greater or equal to 2.

If the solution is to be plotted on a number line, then the solution would like the following:

As seen above, the number 2 is circled fully (meaning that 2 is also part of a solution), as well as 3, 4, 5, 6, 7, and so on are calid solutions that satisfy the inequation.

Remember: fill the circle in if the solution is greater/less than or equal to (\ge or \le) but do NOT fill the circle in if the solution is only greater/less than (> or <).

Another kind of inequation as shown below:

1\le 2z+7\le 11\\ \\ 1-7\le 2z+7-7\le 11-7\\ \\ -6\le 2z\le 4\\ \\ \cfrac { -6 }{ 2 } \le\cfrac { 2z }{ 2 } \le \cfrac { 4 }{ 2 } \\ \\ -3\le z\le 2

The solutions are limited to a small region as shown in the number line below:

Absolute Values

 

An absolute value is the magnitude of a real number without regard to its sign. The absolute value of any number is given by:

\left| x \right| \quad =\quad \left| -x \right| \quad =\quad x

It could simply be understood as how far a number is from zero. For example, plot on a number line and evaluate x.

\left| x \right| =2

x=\pm 2

This means the distance of x from zero is 2 (in either direction).

However, what happens when \left| x \right| \le 2 or \left| x \right| \ge 2?

\left| x \right| \le 2 means the distance from zero to x is less than or equal to 2 (in either direction). The other way to write this as a statement is -2\le x\le 2.

On the other hand, \left| x \right| \ge 2 means that the distance from zero to x is greater than 2 (in either direction). The other way to write this statement is x\le -2\quad or\quad x\ge 2.

To conclude,

\left| x \right| =a\quad means\quad x=\pm a\\ \\ \left| x \right| <a\quad means\quad -a<x<a\\ \\ \left| x \right| >a\quad means\quad x<-a\quad or\quad x>a

 

 

Examples:

 

Solve:

1. \left| x+4 \right| =7

Show Answer

This means that the distance from zero to x+4 is 7 in either direction.

\left| x+4 \right| =7\\ \\ x+4=\pm 7\\ \\ x+4=7\qquad or\qquad x+4=-7\\ \\ x=3\qquad \qquad \qquad x=-11

 

2. \left| 2y-1 \right| \le 5

Show Answer

This means that the distance from zero to 2y-1 is less than or equal to 5 in either direction.

\left| 2y-1 \right| \le 5\\ \\ -5\le 2y-1\le 5\\ \\ -5+1\le 2y-1+1\le 5+1\\ \\ -4\le 2y\le 6\\ \\ \cfrac { -4 }{ 2 } \le \cfrac { 2y }{ 2 } \le \cfrac { 6 }{ 2 } \\ \\ -2\le y\le 3

 

3. \left| 5b-7 \right| >3

Show Answer

This means that the distance from 5b-7 to zero is greater than 3 in both direction.

\left| 5b-7 \right| >3\\ \\ 5b-7>\pm 3\\ \\ 5b-7<-3\qquad or\qquad 5b-7>3\\ \\ 5b<4\qquad \qquad or\qquad 5b>10\\ \\ b<\cfrac { 4 }{ 5 } \qquad \qquad or\qquad b>2

 

In several cases, it is better to check the solutions to make sure that they satisfy the inequations. Not all solutions are feasible.

 

4. \left| 2x+1 \right| =3x-2

Show Answer

2x+1=\pm (3x-2)\\ \\ 2x+1=3x-2\qquad or\qquad 2x+1=-(3x-2)\\ \\ 2x-3x=-2-1\qquad \qquad \quad 2x+1=-3x+2\\ \\ -x=-3\qquad \qquad \qquad \qquad 2x+3x=2-1\\ \\ x=3\qquad \qquad \qquad \qquad \qquad 5x=1 \qquad -> \qquad x=\cfrac { 1 }{ 5 }

Let's check whether both solutions satisfy the inequation.

(i) Substitute x=3 to thein equation, and check whether the LHS equals the RHS.

LHS=\left| 2x+1 \right| \qquad \qquad RHS=3x-2\\ \\ \qquad =\left| 2\times 3+1 \right| \qquad \qquad \qquad =3\times 3-2\\ \\ \qquad =\left| 7 \right| \qquad \qquad \qquad \qquad \quad =9-2\\ \\ \qquad =7\qquad \qquad \qquad \qquad \qquad =7

Since both sides have equal value, that means x=3 IS a solution.

(ii) Substitute x=1/5 and check.

LHS=\left| 2x+1 \right| \qquad \qquad RHS=3x-2\\ \\ \qquad =\left| 2\times \cfrac { 1 }{ 5 } +1 \right| \qquad \qquad =3\times \cfrac { 1 }{ 5 } -2\\ \\ \qquad =\left| \frac { 2 }{ 5 } +1 \right| \qquad \qquad \qquad =\cfrac { 3 }{ 5 } -2\\ \\ \qquad =\cfrac { 7 }{ 5 } \qquad \qquad \qquad \qquad =-\cfrac { 7 }{ 5 }

The opposite sides have unequal values and therefore x=1/5 IS NOT a solution.

 

Think: why couldn't all solutions be feasible?