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Chapter 1.4: Solving Absolute Value Inequalities Let’s use the definition of Absolute Value to discuss the meaning of the following inequalities: a. ! < 3 According to the definition of absolute value, the definition of this inequality is a is less than 3 units from zero on the number line”. So, lets think about this. Where are we, if we are less than 3 units from zero on the number line? Well, all of the values from 0 to 3 (but not equal to 3) are less than 3 units from zero on the number line, and all of the values from -3 to 0 (but not equal to -3) are less than 3 units from zero. In number line notation, it looks like the green shaded area between -3 and 3, below: 3 0 3 If we would express this number line using inequality notation, without the absolute value, we would rewrite this as 3 < a < 3. b. ! 5 Again, using the definition of absolute value, the definition of this inequality is n is greater than or equal to 5 units from zero on the number line”. So, lets think about this one as well. If we are 5 units, or more from zero on the number line, then we are hitting all of the values of 5, up to positive infinity. On the other side of the number line, we are starting at 5, then going down to negative infinity. In number line notation, it looks like the following description below: 5 0 5 If we would express this number line using inequality notation, without the absolute value, we would rewrite this as: x 5 and x 5.

# Chapter 1.4: Solving Absolute Value Inequalities · 2018. 1. 15. · Toexpress(absolute(value(inequalities(without(using(absolute(values,(we(can(rewrite(as(the(following:(WHENSOLVINGABSOLUTEVALUEINEQUALITIES,(1

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Chapter 1.4: Solving Absolute Value Inequalities

Let’s  use  the  definition  of  Absolute  Value  to  discuss  the  meaning  of  the  following  inequalities:

a. ! < 3

According  to  the  definition  of  absolute  value,  the  definition  of  this  inequality  is  “a  is  less  than  3  units  from  zero  on  the  number  line”.

So,  lets  think  about  this.    Where  are  we,  if  we  are  less  than  3  units  from  zero  on  the  number  line?    Well,  all  of  the  values  from  0  to  3  (but  not  equal  to  3)  are  less  than  3  units  from  zero  on  the  number  line,  and  all  of  the  values  from  -3  to  0  (but  not  equal  to  -3)  are  less  than  3  units  from  zero.    In  number  line  notation,  it  looks  like  the  green shaded area between -3 and 3,  below:

-­‐3 0                 3

If  we  would  express  this  number  line  using  inequality  notation,  without  the  absolute  value,  we  would  rewrite  this  as  -­‐3  <  a  <  3.

b. !  ≥ 5

Again,  using  the  definition  of  absolute  value,  the  definition  of  this  inequality  is  “n  is  greater  than  or  equal  to  5  units  from  zero  on  the  number  line”.

So,  lets  think  about  this  one  as  well.    If  we  are  5  units,  or  more  from  zero  on  the  number  line,  then  we  are  hitting  all  of  the  values  of  5,  up  to  positive  infinity.    On  the  other  side  of  the  number  line,  we  are  starting  at  -­‐5,  then  going  down  to  negative  infinity.    In  number  line  notation,  it  looks  like  the  following  description  below:

-­‐5 0 5

If  we  would  express  this  number  line  using  inequality  notation,  without  the  absolute  value,  we  would  rewrite  this  as:      x  ≤  -­‐5    and  x  ≥ 5.

To  express  absolute  value  inequalities  without  using  absolute  values,  we  can  rewrite  as  the  following:

WHEN  SOLVING  ABSOLUTE  VALUE  INEQUALITIES,

1. Express  in  standard  form.    This  means  get  the  absolute  value  by  itself  on  one  side  ofthe  equation,  and  all  other  terms  on  the  other  side  of  the  equation.

2. Express  the  equation  in  terms  of  the  one  or  two  equations  that  do  not  involve  theabsolute  value,  and  solve.

Rewrite   !  <  a      to  the  compound (double)  inequality,    -a  <  x  <  a

Rewrite   ! > a    to  two  inequalities,    x  >  a    or      x  <  -­‐a

3. Graph  the  solution  set,  or  express  it  using  number  line  notation.

Lets  consider  the  following  examples:

Example:    Solve  the  following,  express  your  solution  set  using  number  line  for  p  and  q,  and  using  interval  notation  for  q,  r  and  s.

p.     2! − 3  ≤ 4   STEP  1:    It  is  already  in  standard  form

-­‐4    <  2s    -­‐  3    <    4   STEP  2:    Express  the  equation  in  terms  +3 +3          +3 of  an  equation  without  an  absolute  value.  -­‐1      <    2s    <    7    2                  2              2

So,  the  solution  set  to  this  absolute  value  inequality  is  -­‐  !!  <  s  <  !

!

|!| ≤ !      !"  !ℎ!  !"#\$  !ℎ!"#  !"  !"#\$%&  − !   ≤ !   ≤ !

|!| > !  !"  !ℎ!  !"#\$  !ℎ!"#  !"  !"#\$%&  ! > !        !"#        ! <  −!

(assuming  a  >  0  )

STEP  3:  Graph  the  solution  set.                            − !

!0 !

!

q.     5! + 2 − 6 >  2 STEP  1:    Express  in  standard  form  + 6        +  6

5! + 2 > 8 STEP  2:  Express  the  equation  in    terms  of  equations  without  absolute

5! + 2 >  8            or            5! + 2 <  −8 values.    -­‐2              -­‐2  -­‐2                    -­‐2

5a      >        6                    or                5a        <    -­‐10          5                    5 5                        5

! >   !  !        !"      !   <  −2

STEP  3:  Graph  the  solution  set,  and  Express  in  interval  notation.

-­‐2                              0   !!

If  we  were  to  express  this  in  interval  notation,  note  that  we  have  2  intervals  to  express  in  interval  notation.    First,  a  <  -­‐2  can  be  expressed  by  (  −∞,−2).    The  second  interval,  a  >   !

!  can

be  expressed  as  (  !  !, ∞).    Because  we  have  2  intervals,  we  need  a  connector  to  link  them

together.    In  this  case  we  use  a  ∪  (which  in  set  theory  means  ‘union’).

Putting  this  altogether,  we  can  express  the  solution  set  using  interval  notation  as:

(  −∞,−2)    ∪    (  !  !, ∞)

r.   5! − 2 > −4

STOP!    Before  you  being  solving  this  inequality,  note  what  happened  with  example  n.  from  above.    You  should  think  to  yourself,  this  is  strange!    Anytime  you  have  a  negative  solution  involving  absolute  values,  something  is  strange!    Let  think  about  what  this  means.  Lets  look  at  the  overall  picture,  which  is,  when  is  the  absolute  value  of  any  number  greater  than  negative  4?  Well,  recall  that  the  absolute  value  (the  distance  function)  of  any  number  is  going  to  be  positive.    And  when  is  a  positive  number  greater  than  -­‐4?    Always!    So,  the  solution  set  to  this

absolute  value  inequality  is  all  real  numbers!    Any  number  we  substitute  in  for  k,  multiply  by  5,  and  then  subtract  2  from  it  and  THEN  take  the  absolute  value-­‐  which  will  ALWAYS  make  that  number  greater  than  -­‐4.

Our  solution  set  is  all  Real  numbers.    In  terms  of  the  number  line,  the  whole  thing  would  be  shaded  as  below:

0

In  terms  of  interval  notation,  it  would  look  as  follows:

(  −∞, ∞  )

s.     −3! + 4 ≤ 7 STEP  1:  It  is  in  standard  form

−7   ≤ −3! + 4 ≤ 7        STEP  2:    Express  the  equation  in    -­‐4                                      -­‐4            -­‐4        terms  of  an  equation  without  an  -­‐11    ≤  −3!       ≤      3   absolute  value.      -­‐3          -­‐3              -­‐3

!!!  ≥ !   ≥  −1

(which  is  the  same  thing  as  saying)

-­‐1  ≤ !   ≤   !!!

[-­‐1  ,    !!!] STEP  3:    Express  the  solution  set  in

interval  notation.

Remember  when  you  divide  an  inequality  by  a  negative,    it

changes  direction!

t.   −3! + 4  ≥ 7 STEP  1:    It  is  in  standard  form

-­‐3x    +  4    ≥ 7            or          -­‐3x    +  4    ≤  −7   STEP  2:    Express  the  equation  in                  -­‐  4            -­‐4            -­‐4                -­‐  4   terms  of  an  equation  without  an    -­‐3x              ≥    3    or   -­‐3x            ≤  −11 absolute  value.

-­‐3                        -­‐3 -­‐3 -­‐3

x  ≤  −1              !"          !         ≥     !!!

−∞,−1  ∪     [!!!   ,∞)   STEP  3:    Note,  again  that  we  have  2

intervals  that  make  this  true.    This  will  typically  happen  when  we  have  an  absolute  value  GREATER  THAN.    So,  we  need  to  express  the  solution  set  in  terms  of  2  intervals.

AGAIN!    Remember  when  you  divide  an  inequality  by  a

negative,    it  changes  direction!

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