- #1
xzi86
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Homework Statement
(3x+1)/(x+4)>=1
Homework Equations
The Attempt at a Solution
(3x+1)>=(x+4)
2x>=3
x>=3/2
But this is wrong?? Why?
Inequality rational polymonial
- Thread starter xzi86
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- Inequality Rational
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Homework Help Overview
The problem involves solving the inequality (3x+1)/(x+4) ≥ 1, which falls under the subject area of rational inequalities. Participants are exploring the conditions under which the inequality holds true and the implications of the signs of the expressions involved.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- The original poster attempts to simplify the inequality directly but questions the validity of their approach. Some participants suggest breaking the problem into cases based on the sign of (x+4) and exploring the implications of multiplying by negative numbers. Others propose an alternative method involving squaring the expression to avoid sign issues.
Discussion Status
The discussion is active, with participants providing different perspectives on how to approach the inequality. There is a recognition of the need to consider the sign of (x+4) and the implications for the inequality, but no explicit consensus has been reached on a single method or solution.
Contextual Notes
Participants note the importance of considering the conditions under which (x+4) is positive or negative, as well as the implications of these conditions on the inequality. There is also mention of a resource that may provide further clarification on solving rational inequalities.
(3x+1)/(x+4)>=1
(3x+1)>=(x+4)
2x>=3
x>=3/2
But this is wrong?? Why?
- #2
HallsofIvy
Science Advisor
Homework Helper
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- 983
If you multiply both sides of an inequality by a negative number, you reverse the direction of the inequality. But you don't know whether x+ 4 is positive or negative.
Instead break it into two parts. x+ 4 is positive, that is x+ 4> 0 for x> -4. If that is the case, then 3x+1> x+ 4 so that 2x> 3, x> 3/2. Because any value of x that is larger than 3/2 certainly is larger than -4, x>= 3/2 is part of the solution.
But if x< -4, then x+ 4 is negative. And then 3x+1< x+ 4 so that 2x< 3 so that x< 3/2. Now we argue the other way: any number that is less than -4 certainly is less than 3/2 so x< -4 is the other part of the solution. (x= -4 is not because the fraction is not defined if x= -4.)
You can check: if x= -5, then 3x+1= -15+1= -14, x+ 4= -1 so that (3x+1)/(x+4)= -14/-1= 14 which is certainly greater than 1.
That is, the solution set is all values of x less than -4 and all values of x greater than 3/2.
Instead break it into two parts. x+ 4 is positive, that is x+ 4> 0 for x> -4. If that is the case, then 3x+1> x+ 4 so that 2x> 3, x> 3/2. Because any value of x that is larger than 3/2 certainly is larger than -4, x>= 3/2 is part of the solution.
But if x< -4, then x+ 4 is negative. And then 3x+1< x+ 4 so that 2x< 3 so that x< 3/2. Now we argue the other way: any number that is less than -4 certainly is less than 3/2 so x< -4 is the other part of the solution. (x= -4 is not because the fraction is not defined if x= -4.)
You can check: if x= -5, then 3x+1= -15+1= -14, x+ 4= -1 so that (3x+1)/(x+4)= -14/-1= 14 which is certainly greater than 1.
That is, the solution set is all values of x less than -4 and all values of x greater than 3/2.
- #3
xzi86
- 7
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so my method was good except i forgot to consider x is negative
- #4
Mentallic
Homework Helper
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xzi86 said:
Not x is negative, but x+4 is negative, or x+4<0, x<-4.
Another way to solve this problem without considering cases is to multiply by (x+4)2 because any number squared is always positive, so there is no need to switch the inequality sign. You have to solve a quadratic and you'll get the same answer.
- #5
BloodyFrozen
- 353
- 1
Read this, it should help.
http://tutorial.math.lamar.edu/Classes/Alg/SolveRationalInequalities.aspx
In your case,
(3x+1)/(x+4)≥1
You want to get one side zero. Here is a small part...
(3x+1)/(x+4)≥1
(3x+1)/(x+4)-1≥0
Now, what can you do to make the LHS into one fraction?
If you read carefully, you should know.
http://tutorial.math.lamar.edu/Classes/Alg/SolveRationalInequalities.aspx
In your case,
(3x+1)/(x+4)≥1
You want to get one side zero. Here is a small part...
(3x+1)/(x+4)≥1
(3x+1)/(x+4)-1≥0
Now, what can you do to make the LHS into one fraction?
If you read carefully, you should know.
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