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Finding a Set, Solving a System


Date: 06/24/98 at 12:28:21
From: Monika Bailey
Subject: Find the solution set

I actually have two questions. The first is:

   Find the solution set:  3-1/4x <_ 2+ 3/8 x

The other problem is:

   Solve the system
  
       3x-6y = 1
           x = 2y+3

Thanks for all your help.  I couldn't do this without all of you.

Monika


Date: 06/25/98 at 12:27:25
From: Doctor Peterson
Subject: Re: Find the solution set

Hi, Monika. 

I'll try doing a problem similar to each of yours so you'll have 
a chance to practice on your own problems.

The first is an inequality. I'll assume when you write "<_" you mean 
"less than or equal to"; I'll write that as "<=". Here's my own 
problem:

         1            3
    4 + --- x <= 1 + --- x
         4            8

Your goal, as in solving an equation, is to get x alone on one side. 
The only thing really different is that you have to be careful about 
the direction of the inequality, because if you multiply both sides 
by a negative number, it will reverse the comparison from <= to >=. 
I prefer just to avoid that, by making sure I never have to multiply 
or divide by a negative number. 

Since the 3/8 is bigger than the 1/4, I'll move all the x's to the 
right side, where they will be positive, by subtracting 1/4 x from 
both sides:

              1
    4 <= 1 + --- x
              8

Now I can subtract 1 from both sides to get the constants together:

          1
    3 <= --- x
          8

and then multiply by 8 (which, as I planned, is positive and safe to 
use!)

    24 <= x

Now we can turn the whole thing around to get

    x >= 24

So the answer is that the inequality is true for all values of x 
greater than or equal to 24. As a check, you can plug in x = 24 and 
see that it is equal, and then try, say, 25, to see that it is 
greater.
------------------------------------------------------------

Your second problem is a pair of simultaneous equations with a little 
twist. I'll do this one, which has the same twist:

    3x - 9y = 3
          x = 3y + 4

First you have to put them in the same form, by moving the 3y to the 
left:

    3x - 9y = 3
     x - 3y = 4

There are several ways to solve these; I'll use the method in which we 
make the coefficients of x the same and subtract the resulting 
equations. Just multiply the second equation by 3:

    3x - 9y = 3
    3x - 9y = 12

Now you would normally subtract these, but you would get 0x + 0y = -9, 
which can never be true. How can that be? It means that these two 
equations represent lines that never intersect, because they are 
parallel. Whenever you end up with an impossible equation, it means 
there are no solutions.

Does that help?

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


Date: 06/25/98 at 12:33:57
From: Anonymous
Subject: Re: Find the solution set

Thank you Dr. Peterson.  Dr. Math came to the rescue once again.
    
Associated Topics:
High School Basic Algebra
Middle School Algebra

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