Solving a Fractional, Single-Variable, Inequality

In summary, when solving inequalities involving fractions, it is important to take into account the possible signs of the variables involved. In this case, we need to consider two cases: when x+2 is positive, and when it is negative. By multiplying both sides of the inequality by the denominator, we can solve for x in each case and find the intersection of the two sets of solutions to get the final answer. In this problem, the solution is (-2, 9/2).
  • #1
19
0

Homework Statement



Solve the Inequality:
(3x-7)/(x+2)<1

Homework Equations


The Attempt at a Solution



Cross Multiply: x+2>3x-7
Simplify: 9>2x
Simplify More: 9/2>x
My Answer: (-∞, 9/2)
I put this as my answer but the answer is really (-2, 9/2)

Can someone explain to me why this is? I know you can't divide by 0 and this has something to do with the answer but that is as far as I got.

Thanks
 
Physics news on Phys.org
  • #2
If you're trying to solve the inequality -x>1, then you would divide by -1 (or equivalently, multiply by -1) to give x<-1. Notice how we reversed the inequality sign because we divided/multiplied by a negative value. However, dividing by positive values doesn't reverse the inequality sign.

So what about when we divide or multiply through by a variable x? We don't know if x is positive or negative.

This is why we need to take two cases:

(3x-7)/(x+2)<1

1) Ok so we know that [tex]x\neq -2[/tex] so let's look at just [tex]x>-2[/tex]. Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption.

2) Now multiply through, assuming x<-2.

3) Now that you have two sets of solutions, obviously both need to be valid so the answer will be the intersection of these two sets of solutions. Say if we had 0<x<5 and 2<x<10, then our answer will be 2<x<5
 
  • #3
Mentallic said:
If you're trying to solve the inequality -x>1, then you would divide by -1 (or equivalently, multiply by -1) to give x<-1. Notice how we reversed the inequality sign because we divided/multiplied by a negative value. However, dividing by positive values doesn't reverse the inequality sign.

So what about when we divide or multiply through by a variable x? We don't know if x is positive or negative.

This is why we need to take two cases:

(3x-7)/(x+2)<1

1) Ok so we know that [tex]x\neq -2[/tex] so let's look at just [tex]x>-2[/tex]. Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption.

2) Now multiply through, assuming x<-2.

3) Now that you have two sets of solutions, obviously both need to be valid so the answer will be the intersection of these two sets of solutions. Say if we had 0<x<5 and 2<x<10, then our answer will be 2<x<5

Thank you for your reply Mentallic. I don't quite understand your solution. You said "Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption." I don't understand what you mean by solve the inequality, I thought I already did that, what can I do differently? Also why should I discard the values of x<-2, they don't satisfy which inequality? x<9/2 or x>-2. I'm quite confused.

Thanks.
 
  • #4
nicksbyman said:
Thank you for your reply Mentallic. I don't quite understand your solution. You said "Multiplying through won't change the inequality sign because we are multiplying through by a positive value (since our assumption is x>-2 or x+2>0). Solve the inequality now, and remember to discard any values of x<-2 because this doesn't satisfy our original assumption." I don't understand what you mean by solve the inequality, I thought I already did that, what can I do differently? Also why should I discard the values of x<-2, they don't satisfy which inequality? x<9/2 or x>-2. I'm quite confused.

Thanks.
You sort of solved the inequality, but you started off making the assumption (whether or not you realized it) that x + 2 is positive, because you did not change the inequality sign.

You started with this:
[tex]\frac{3x-7}{x+2} < 1[/tex]

I personally don't like cross multiplying with inequalities. Instead, I will multiply both sides by the denominator:
[tex]\frac{3x - 7}{x + 2} \cdot (x + 2) < 1 \cdot (x + 2)[/tex]

This is the problem: we don't know yet what the sign of x + 2 is. If x + 2 is positive, then we need to assume that x > -2. We leave the inequality sign alone.
[tex]3x - 7 < x + 2[/tex]

Solving for x, you get x < 9/2, as you said.

But wait! We assumed that x > -2 when we solved this. This means we have to throw out values less then or equal to -2. Test any of those values in the original inequality and you will end up with a false statement. So the solutions in this case is
(-2, 9/2)

Now consider the 2nd case, where x + 2 is negative. This means we assume that x < -2. Because we are multiplying both sides by a negative number, we must switch the inequality sign.
[tex]3x - 7 > x + 2[/tex]

Solve for x, and compare this with what we assumed in this 2nd case.
 
  • #5
eumyang said:
You sort of solved the inequality, but you started off making the assumption (whether or not you realized it) that x + 2 is positive, because you did not change the inequality sign.

You started with this:
[tex]\frac{3x-7}{x+2} < 1[/tex]

I personally don't like cross multiplying with inequalities. Instead, I will multiply both sides by the denominator:
[tex]\frac{3x - 7}{x + 2} \cdot (x + 2) < 1 \cdot (x + 2)[/tex]

This is the problem: we don't know yet what the sign of x + 2 is. If x + 2 is positive, then we need to assume that x > -2. We leave the inequality sign alone.
[tex]3x - 7 < x + 2[/tex]

Solving for x, you get x < 9/2, as you said.

But wait! We assumed that x > -2 when we solved this. This means we have to throw out values less then or equal to -2. Test any of those values in the original inequality and you will end up with a false statement. So the solutions in this case is
(-2, 9/2)

Now consider the 2nd case, where x + 2 is negative. This means we assume that x < -2. Because we are multiplying both sides by a negative number, we must switch the inequality sign.
[tex]3x - 7 > x + 2[/tex]

Solve for x, and compare this with what we assumed in this 2nd case.

But although the 2nd case works mathematically my answer sheet says that the only answer for this problem is (-2,9/2). Due to the 2nd case x can be greater than 9/2. When we solve for x we get x>9/2. In the 2nd case the answer would be (9/2, -2), which doesn't make any sense. In other words, it seems that my problem sheet is asking me to ignore the 2nd case, which doesn't make sense with what you guys are saying.
Thanks.
 
Last edited:
  • #6
nicksbyman said:
But although the 2nd case works mathematically my answer sheet says that the only answer for this problem is (-2,9/2). Due to the 2nd case x can be greater than 9/2. When we solve for x we get x>9/2. In the 2nd case the answer would be (9/2, -2), which doesn't make any sense. In other words, it seems that my problem sheet is asking me to ignore the 2nd case, which doesn't make sense with what you guys are saying.
Thanks.

"Ignor[ing] the 2nd case" makes perfect sense. In the 2nd case, we make the assumption that x < -2. We solve the inequality (switching the inequality sign because x + 2 < 0) and we get the answer x > 9/2. Since this contradicts the assumption that x < -2, the assumption is invalid. So the only solution set is (-2, 9/2).
 
  • #7
nicksbyman said:
But although the 2nd case works mathematically my answer sheet says that the only answer for this problem is (-2,9/2).
You're ignoring the basic assumption in this case. For the second case, you end up with x > 9/2 AND you assumed that x + 2 < 0. This is equivalent to x < - 2 AND x > 9/2, which is a contradiction, since x can't simultaneously be smaller than -2 and larger than 9/2.
nicksbyman said:
Due to the 2nd case x can be greater than 9/2. When we solve for x we get x>9/2. In the 2nd case the answer would be (9/2, -2), which doesn't make any sense. In other words, it seems that my problem sheet is asking me to ignore the 2nd case, which doesn't make sense with what you guys are saying.
Thanks.
The first case boils down to x + 2 > 0 AND x < 9/2, which is the same as saying -2 < x < 9/2, the same as your answer book.
 

What is an inequality?

An inequality is a mathematical statement that compares two values or expressions using symbols such as <, >, ≤, or ≥. It indicates that one value is either less than or greater than another value.

What is a fractional inequality?

A fractional inequality is an inequality that involves fractions or rational numbers. It can be solved by treating the inequality as a regular inequality and then multiplying both sides by the common denominator to eliminate the fractions.

How do you solve a single-variable inequality?

To solve a single-variable inequality, follow these steps:

  • Isolate the variable on one side of the inequality by using inverse operations.
  • If you multiply or divide by a negative number, flip the direction of the inequality.
  • Check the solution by plugging it back into the original inequality.

What is the solution set of an inequality?

The solution set of an inequality is the range of values that satisfy the inequality. It can be represented using interval notation or set-builder notation.

How can solving inequalities be applied in real-life situations?

Solving inequalities can be applied in various real-life situations such as budgeting, determining the best deals, and solving problems involving rates, proportions, and percentages. It can also be used to represent relationships between quantities in fields like economics, engineering, and science.

Suggested for: Solving a Fractional, Single-Variable, Inequality

Replies
2
Views
360
Replies
9
Views
655
Replies
11
Views
929
Replies
3
Views
706
Replies
7
Views
536
Replies
12
Views
1K
Back
Top