The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Verifying the Solution to an Inequality ... by Substituting? by Checking Boundaries?

Date: 04/28/2010 at 22:02:02
From: J.C. 
Subject: I'm confused with inequalities. 4 - 2/3b < 3 - 1/3b verify?

O.K. I am totally confused.

I have to solve this inequality, show my work, and then verify it:

   4 - (2/3)b < 3 - (1/3)b

I think I somewhat understand how to solve it ... but I have no idea how
to verify it.

Here's what I did:

   4 - (2/3)b < 3 - (1/3)b      Multiplied the fractions (2/3)b and (1/3)b
                                by their lowest common denominator (3) below

      3(2/3b) = (6/3)b          Simplified 6/3b = 2b 
      3(1/3b) = (3/3)b          Simplified = 3(1/3b) = 1b
       4 - 2b < 3 - 1b          Substituted those simplified fractions
      2b + 1b < 3 - 4           Transposed, by switching the sides of the      
                                expression 1b and the quantity 4, and 
                                then changing their signs 

           3b < -1 
       3b / 3 < -1 / 3          Divided both sides by 3
            b < -1/3  

Normally with equations, to verify the answer I just replace the variable
with what I found out the variable to be. For example, if I had to solve
for x in some equation, and found out x = -4, I would replace all of the x
variables in that equation with -4.

But with this inequality, I got "b" as

            b < -1/3

So now I don't know if I have to replace "b" with something less than

Date: 04/28/2010 at 23:14:40
From: Doctor Peterson
Subject: Re: I'm confused with inequalities.  4 - 2/3b < 3 - 1/3b verify?

Hi, J.C.

First, I want to look carefully at your work to see that it looks right;
I'll write just the main steps:

  4 - (2/3)b < 3 - (1/3)b 
      4 - 2b < 3 - 1b     This is wrong: you multiplied only one term on
                          each side by 3, which is illegal. You have to
                          multiply each ENTIRE SIDE (every term) by 3.
     2b + 1b < 3 - 4      Here you lost a negative sign; -2b is staying
                          where it was, so its sign doesn't change.
          3b < -1 
           b < -1/3  

So your work is wrong. The solution is probably wrong, too.

But I want to show you how to check; presumably, checking will also reveal
the error.

How do you check a solution to an inequality? Ideally, you have to check
that EVERY number in the solution set (in this case, every number less
than -1/3) makes the original inequality true, and that EVERY number NOT
in the solution set (-1/3 and every number greater than that) makes the
original inequality false. You can't do that, of course.

But we know that a solution to a linear inequality will be an interval, so
we just have to check that we have the right interval. To do that, we can
check the BOUNDARY -- the -1/3 that you determined -- and then check which
SIDE of it is correct.

So I would first take b = -1/3, and see if that is on the boundary of the
solution, which means that it should make the two sides of the original
inequality EQUAL. Let's do that:

       4 - (2/3)b =? 3 - (1/3)b      let b = -1/3

  4 - (2/3)(-1/3) =? 3 - (1/3)(-1/3)

       4 - (-2/9) =? 3 - (-1/9)

            4 2/9 =? 3 1/9

They are not equal, so this is not the correct boundary. Nevertheless, b =
-1/3 is not in your claimed solution set (since only numbers LESS than
this are in it), and it turns out not to be a solution (since the left
side is not less than the right). So the problem is not that it's not a
solution, only that it is not the correct boundary to the solution set.
I'm going to continue, just to show more clearly that the solution is

If it HAD worked out, my next step would have been to choose some point IN
the claimed solution set, and see if it satisfies the inequality. Let's
take an easy number; try b = -1, which is less than -1/3:

     4 - (2/3)b <? 3 - (1/3)b      let b = -1
  4 - (2/3)(-1) <? 3 - (1/3)(-1)
     4 - (-2/3) <? 3 - (-1/3)
          4 2/3 <? 3 1/3

This value is supposed to be a solution, but it isn't. In this case, we
didn't have to do this, because we already knew the solution was wrong;
but if we had had the right boundary, this might have shown that we had
the wrong direction for the inequality.

Now correct the errors in your solution, and do the check again. If you
have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 04/28/2010 at 23:54:48
From: J.C. 
Subject: Thank you (I'm confused with inequalities.  4 - 2/3b < 3 - 1/3b verify?)

Thank you so much. I finally understand it now. You and everybody else at
Dr. Math are just the BEST!

I greatly appreciate that you took time to answer my question.

Once again, thank you! :D
Associated Topics:
Middle School Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.