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Lesson 2 for Algebra 2 Review

Page history last edited by Math in a Box - Susan Johnsey gm 10 years, 5 months ago

Lesson 2  Solving Inequalities  

 

 

Inequalities are two or more expressions that might be equal but may not be!!

 

Let us look at just two simple expressions first.

 

x>2  is read x is greater than 2 if we read from left to right.

x>2 is read 2 is less than x if we read from right to left.     You need to understand that these do mean the same thing, and that you do need to be able to read them both ways.

 

The symbol < or the  the symbol >   will always point toward the lesser expression.

 

   x< 4 is read "x is less than 4"  or "4 is greater than x".   Do you see that the symbol is pointing to the x, thus x is the lesser expression.   So you can state that x is less than 4  or you can state that 4 is greater than x.   Either way we know x is the lesser expression.


 

Can you list all the numbers less than 4?    That could take a long time.   There are many fractions and many whole numbers (positive and negative) that are less than 4.    We do not want to LIST them so we use the symbols < and > and some special notation instead. 

 

We write {x| x<4}   to mean all REAL numbers less than 4.   Did you know our numbers that we use daily have a name?   There name is REAL numbers.   There are other numbers and they are refered to as IMAGINARY!   When we use these together we call them the COMPLEX numbers.    In this lesson and for several lessons we will use only the REAL numbers.    They are the fractions, decimals, whole numbers, counting numbers, integers (positive and negative whole numbers; zero is an integer).  

 


 

 

Do you recall that we have two types of decimals?

The type that comes from fractions:  1/4 = .25   or 2/3 = .66666 forever  or 1/6 = .166666 forever  or 5/2 = 2.5.  But also the type that do not repeat and do not end.   These are pi ( we round it off to 3.14 most of the time)  or  square root of 5, or cube root of 9 and many others.   The REAL numbers are all fractions, all decimals, all integers (whole numbers -positive and negative and ZERO).  They are all the numbers that you have probably heard of all your life.    The others are IMAGINARY and use the letter "i" in them.  You may have studied those if you have taken Algebra 2 before now.


 

Recall we use the properties of equality to solve equations.  We can use the ones for adding and subtracting here too but we have to learn another property about multiplication and division that is specially for inequalities.

 

LOOK   -2 < 5   is this true?   yes    In math we want TRUE statements.    And we can use properties or laws on the statements only if they keep the statement TRUE.        WE did this in Lesson 1 every time we divided on both sides or we multiplied on both sides or ADDED on both sides or Subtracted on both sides!!!

 

 But watch:   -2< 5   is true   Does it stay true if I add -2 to both sides?  -2+(-2) < 5+(-2)     yes or no?__1

  Does it stay true if I subtract  -2 on  both sides? -2-(-2) < 5-(-2)  yes or no?      _____2

  Does it stay true if I multiply  -2 on  both sides? -2(-2) < 5(-2)  yes or no?      _____ 3

  Does it stay true if I divided by  -2 on  both sides? -2/(-2) < 5/(-2)  yes or no?      _____4.

 

Two of the answers above are NO .   Send me an email to tell me which numbered statements are false.

sojohnsey@gmail.com   Please add Subject :  "Inequality" to the email.

 

What if I use a positive 2 rather than the -2.    Will the 4 statements remain TRUE?     Try that and  send me another email telling me what you found out?

 


 

Now let us solve inequalities.

 

Example 2A

3x+4 >6x-20

3x+4 -6x > 6x-6x -20   I subtracted 6x from both sides.

-3x +4 > -20     NOw subtract 4 from both sides and we get??

-3x > -24    Now divide both sides by -3.   do you know what you learned above?  

 

If we divide both sides by a negative then the symbol must be reversed (flipped about) other wise

 the new statement is false.   You cannot "see" that is correct when we have a variable but what

we learned above is still CORRECT.

 

We get  x<8.    To be sure you get this try letting the x = 7, I chose that because it is less than 8.  

Replace all the x with (7). 

               3x + 4 >6x - 20

              3(7)+ 4 >6(7) - 20   becomes  21+4 >42 -20     Is this true?  25> 22       YES.   If we had fogotten to change the "greater than" to a "less than" in our last step (division by -3) then we would have had x>8.  Try 9 in the original statement.  It will be false!         

 

The answer or SOLUTION as it is called in Algebra is all REAL numbers less than 8  and in notation is written

       {x| x<8}    The  " x|"   just means "all x where".  

         {x| x<8} is read " the set of REAL numbers x where x is less than 8". 

 

Example 2B

 

   3(x-5) +2 > 5x+ 6/5   Hope you recall fractions and simplifying.  

                                      Do the simplfiying first.

  3x- 15  +2 > 5x + 6/5     Now let use the LCM and multiply all by +(5) .

 

 3x(5) - 15(5)   +2(5)  > 5x(5)  + (5) 6/5     That is a multiplying by positive (5)

                                                                             so we do NOT change the symbol to <.

  15x - 75 + 10 > 25x +6                           Simplify more.   Do not solve yet.

  15x -65  > 25x  + 6

       -10x -65 > 6

       -10x > 71    Ok we are needing to divide and it is dividing by -10.  

                             We must change the symbol.

             x < -7.1      Solution is all REAL numbers less that -7.1  or  {x | x<-7.1}

 

Example 2C

 

  3x/5  - 1/4 <= 1/5       that means "less than OR equal".   I can not write it here on the web as a math book does.

      We have fractions and no simplifying so let us multiply by the LCM.  It is 20.  

        I hope you knew that from Lesson 1.

 

   3x/5  - 1/4 <= 1/5           Multiply all by (+20).

 

    (20) 3x/5  - (20)1/4 <= (20)1/5  now recall we do not multiply until after we divide out. 

 

      Let us divide.

     ( 4)3x  - (5)1 <= (4)1  BE SURE YOU know where I got the 4 and 5 and 4

 

  12x  - 5  <= 4           Now add 5 on both sides.

    12x <= 9                 Divide by 12.  That is a positive 12 so we do not change the symbol.

       x<= 9/12   or x< =3/4.    

 

Solution is {x| x<=3/4}      TO CHECK let x= 0.   3(0)/5  - 1/4 <= 1/5 Is this true?

 

 

Example 2D

 

  -3x/5  - 1/4 <= 1              Almost the same!   LCM is still 20.

     (20)( -3x/5)  - (20)1/4 <= (20)1       YES we multiply all terms even the 1 .

         Let us divide.

     ( 4)(-3x)  - (5)1 <= (20)1

       -12x - 5 <= 20

 

           -12x<=  25  Now divide by the -12 and flip the symbol.

 

       Thus x>= -25/12 is our solution.      Choose a number larger than -25/12 and test it.   Does it make the original inequality TRUE?   I hope so.  

Try x=0 since 0 is larger than -25/12 we should have a true inequality.  But if you want to try x =-5 then you should have a false inequality since the -5 is not in our solution!    Do you know why I suggested 0 and -5?  It is because we have 3x/5 in our inequality and the x/5 means I must divide x by 5.   Both 0 and -5 are divisible by 5!   You can pick other numbers, but I wanted an easy division!!

 

Complete Assignment 2AD.   

    Look in Navigator for your Assignment folder.