# Inequations and Absolute Values

## 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,

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

### Examples:

Solve:

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

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

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

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

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

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?