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General Strategies on Solving Linear Algebraic Equations

Date: 04/04/2005 at 02:55:16
From: Ernie
Subject: How do I solve equations with variables on each side

How do I solve the following equations?

12k + 15 = 35 + 2k
3b + 8 = 10b
3k + 10 = 2k - 21
3x + 2 = 4x - 1
3(a + 22) = 12a + 30
n + 4 = -n + 10

What are the rules to follow in solving these equations?  They all 
appear to be so different.  Is there a standard rule to follow?

Help, I need to teach this to my 7th graders.



Date: 04/04/2005 at 15:03:10
From: Doctor Ian
Subject: Re: How do I solve equations with variables on each side

Hi Ernie,

The standard rules are these:

  1) You can change the values on both sides of an equation in
     the same way.

  2) You can change the appearance of one side of an equation
     so long as you don't change the value. 

  3) You can't ever divide by zero.

How you apply those rules is something of an art; but the general
"rule for applying the rules" is this:  You're always looking for a
way to change what you have into something simpler.  Ideally, you want
to change it into something that you know you recognize as already
knowing how to solve.

The simplest thing might look like

  x = 4

Can I solve that?  Well, it's already solved!  So if I can start with
some other equation, and turn it into something like this, I'm in fat
city, right?

Next, suppose I have something like this:

   x + 3 = 7

If I add -3 to each side (Rule 1), I get

   x + 3 + -3 = 7 + -3

which I can simplify (Rule 2) to get

        x + 0 = 4

            x = 4

And this is something I know how to solve.  So I want to learn to
recognize

  x + constant = constant

as an equation that I like, because I know how to deal with it.  Which
is to say, if I see an equation like

  2x + 5 = x + 6

my first thought should be:  "Can I turn it into something that looks like

  x + constant = constant ?

And if so, how could I do that?"

A simple way to do that would be to add -x to each side:

  -x + 2x + 5 = -x + x + 6 

        x + 5 = 6

So now I'm not really done yet, but I know I can get home from here,
because this is a form I like.

Another form I like is 

   constant * x = constant

e.g.,


   3x = 5

Why?  Because I can multiply both sides by the reciprocal of the
coefficient of the variable, to get

  (1/3) * 3 * x = (1/3) * 5

              x = 5/3

Which is to say, when I have

  constant * x = constant

I'm only one step away from a solution. 

So, what if I have something like 

  3x - 4 = 9

What then?  Well, there are a couple of ways I might go.  If I add 4
to both sides, I can get

  3x -4 + 4 = 9 + 4

         3x = 13

which is one of the forms I like.  Or, if I divide both sides by 3, I
can get

  3x   4   13
  -- - - = --
   3   3    3

       4   13
   x - - = --
       3    3

which is also one of the forms I like (even though it's a little 
messier, because it involves fractions).  

So now I'm building up a little map of "escape routes":

[Note--the meanings of constants like 'a', 'b', and so on aren't 
preserved as you move through the map.  Just think of each equation as
representing something with variables and constants in some particular
form.]

                 ax + b = cx + d

                        |
                        | add -cx 
                        |

                   ax + b = c

                 /            \
   divide by a  /              \ add -b 
               /                \
              /                  \

       x + a = b               ax = b 
 
              \                  /
               \                /
       add -a   \              /  divide by a
                 \            /
               
                      x = a 

                     (Home!)
 

As equations get more complicated, I add more escape routes to my map.
For example, given something like

   3(a + 22) = 12a + 30

I don't like parentheses, so I'll distribute the multiplication:

  3a + 66 = 12a + 30

I'd like to have a's on only one side, so I'll add -3a to each side:

       66 = 9a + 30

and this is something I have on my map... so from here, I know how to
get home. 

So now I can add this to my map:

              a(bx + c) = dx + e

                        |
                        |  distribute multiplication
                        |

                 ax + b = cx + d

If I have something like 

     2
     -(5x - 6) = 7x - 4
     3

I can multiply by the denominator of the fraction to get something 
like I just saw, so that's something else that can go on my map:

              a
              -(cx + d) = ex + f
              b

                        |
                        |  multiply by b
                        | 

              a(bx + c) = dx + e

                        |
                        |  distribute multiplication
                        |

                 ax + b = cx + d

So as you can see, the number of rules is very small, but they can be
combined in lots of ways.  A big part of succeeding at this level of
algebra is building a good map, 


              a
              -(cx + d) = ex + f
              b

                        |
                        |  multiply by b
                        | 

              a(bx + c) = dx + e

                        |
                        |  distribute multiplication
                        |

                 ax + b = cx + d

                        |
                        | add -cx 
                        |

                   ax + b = c

                 /            \
   divide by a  /              \ add -b 
               /                \
              /                  \

       x + a = b               ax = b 
 
              \                  /
               \                /
       add -a   \              /  divide by a
                 \            /
               
                      x = a 

                     (Home!)


which means learning to recognize particular forms of equations that
can be reduced to simpler forms. 

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 04/05/2005 at 01:47:29
From: Ernie
Subject: Thank you (How do I solve equations with variables on each side)

Thank you for all your help.  I have been agonizing over this.  I can 
not begin to tell you how valuable your help has been to me.  I teach 
7th grade multiple subjects (including math), and 8th grade math.  I 
do not think of myself as a math teacher, but rather a teacher 
teaching math at grade level.  We have upgraded our math books this 
year, so I have had to continue with my math studying in order to help 
my students.  Thank you! (-:
Associated Topics:
Middle School Equations

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