Solving Absolute Number Inequalities

[Mathdabbler]

Hello,
I am working through some inequalities of absolute numbers. I’ve worked through the one in the attachment and arrive at an answer of [-2,4/3]. However, the text book ( Schaum’s Intermediate Algebra) gives the answer as 0 ie an empty set. I do not see how they get that result. Can someone point me in the right direction? Am I making a fundamental error somewhere? I am getting most of the exercises correct, but there’s one or two like thIs that are problematic.

 

[MarkFL]

We are given to solve:

[MATH]|3s+1|\le-5[/MATH]
Right away, we should observe that there is no real value of \(s\) for which the absolute value of a function of \(s\) can be negative. Thus, there is no solution.

 

[Dr.Peterson]

Hello,
I am working through some inequalities of absolute numbers. I’ve worked through the one in the attachment and arrive at an answer of [-2,4/3]. However, the text book ( Schaum’s Intermediate Algebra) gives the answer as 0 ie an empty set. I do not see how they get that result. Can someone point me in the right direction? Am I making a fundamental error somewhere? I am getting most of the exercises correct, but there’s one or two like thIs that are problematic.

Given that you are following the process, rather than observing at the start that there can be no solution, your error is in not seeing what you have actually found.

When you start with an inequality of the form [MATH]|x| < c[/MATH], you transform it into [MATH]x < c[/MATH] AND [MATH]x > -c[/MATH]. That "and" is important. So is the reversal of the inequality in the second part. [It's also important that this is valid only when c > 0; but I'm stipulating that you are ignoring that.]

You showed that [MATH]s\le -2[/MATH] and [MATH]s\ge 4/3[/MATH], after correcting the second inequality.

But there is no number that is both less than -2 and greater than 4/3. So there is no solution. Your bottom line has the directions reversed, probably because you assumed everything was as usual!

There are several points along the way where you can notice that your mechanical process will not yield a valid solution.

 

[Mathdabbler]

Thanks for your feedback. I’m familiar with Boolean logic, so understand your answer.

I have not seen the use of the AND operator in the book I am using (or it was not explicit enough), so that’s why I’ve missed it. The process I’m following is to change the sign of the RHS of the equation and then solve for the unknown variable to arrive at two solutions. From what you are saying, if I understand it correctly, I should always apply the AND operator between the two solutions?

 

[Mathdabbler]

We are given to solve:

[MATH]|3s+1|\le-5[/MATH]
Right away, we should observe that there is no real value of \(s\) for which the absolute value of a function of \(s\) can be negative. Thus, there is no solution.

Thanks for your feedback.

 

[Dr.Peterson]

Thanks for your feedback. I’m familiar with Boolean logic, so understand your answer.

I have not seen the use of the AND operator in the book I am using (or it was not explicit enough), so that’s why I’ve missed it. The process I’m following is to change the sign of the RHS of the equation and then solve for the unknown variable to arrive at two solutions. From what you are saying, if I understand it correctly, I should always apply the AND operator between the two solutions?

What you have read must be incomplete; or perhaps it is really about solving absolute value equations, rather than inequalities. Inequalities require considerably more care.

See this explanation of the topic: https://www.purplemath.com/modules/absineq.htm

When an inequality has the form [MATH]|x| < c[/MATH], it is equivalent to [MATH]x < c[/MATH] AND [MATH]x > -c[/MATH], which can also be written as [MATH]-c < x < c[/MATH]. (Note that in your example, this latter form would be [MATH]4/3 < x < -2[/MATH], which has no solutions, as 4/3 is not less than -2.)

But when it has the form [MATH]|x| > c[/MATH], it is equivalent to [MATH]x > c[/MATH] OR [MATH]x < -c[/MATH]. In this case, you do not use "and".

 

[Mathdabbler]

Thanks - I’ve looked at the link you provided and that cleared it up