No Solution, Infinite Solutions

Date: 01/03/2002 at 21:03:08
From: Corinne
Subject: Two algebra problems

(1)
     8(2x - 3) = 4(4x - 8)

      16x - 24 = 16x - 32

16x - 16x - 24 = 16x - 16x - 32

           -24 = -32

The 16x cancels so that there is no x left. How can this be solved?

(2)

    -3(x - 3) >= 5-3x

      -3x + 9 >= 5-3x

 -3x + 3x + 9 >= 5-3x+3x

            9 >= 5

The x's cancel and this seems unsolvable. Can it be solved and put on 
a number line?

Date: 01/03/2002 at 22:50:57
From: Doctor Ian
Subject: Re: Two algebra problems

Hi Corinne, 

Your first problem has no solution. One way to see that this must be 
true is that 

  y = 16x - 24

is the equation of a line with slope 16 that crosses the y-axis at 
(0,-24), and 

  y = 16x - 32

is the equation of a line the same slope that crosses the y-axis at 
(0,-32). 

Now, since these two lines have the same slope, they must either be 
the same line (in which case they intersect everywhere), or they must 
be parallel lines (in which case they intersect nowhere).  

Since they pass through the y-axis at different points, they must be 
different lines. So they're parallel. So they never intersect. So the 
equation has no solution. 

Does this make sense? 

For your second question, once again, try looking at each side of the 
inequality as the equation of a line. As before, the lines have the 
same slope, but they pass through the y-axis at two different 
locations, so they must be parallel. 

Now, in fact, the line 

  y = -3x +9 

is above the line

  y = -3x + 5

So in fact, this inequality is true _everywhere_... that is, for _all_ 
values of x. So whereas the first problem had no solution, this 
problem has an infinite number of solutions. 

Just to check that, let's pick some values for x and try them:

  x = 1      -3(1) + 9 >=  -3(1) + 5       True

  x = 5      -3(5) + 9 >=  -3(5) + 5       True

In fact, if you try _any_ value of x, you'll get a true inequality. 

These two problems make a point: when you have parallel lines, you 
have to do a little reasoning, instead of just jumping in and trying 
to apply the rules of algebra.

I hope this helps. Write back if you'd like to talk more about this, 
or anything else. 

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