Solving a Fractional, Single-Variable, Inequality

[nicksbyman]

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

 

[Mentallic]

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 x\neq -2 so let's look at just x&gt;-2. 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

 

[nicksbyman]

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 x\neq -2 so let's look at just x&gt;-2. 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.

 

[eumyang]

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:
\frac{3x-7}{x+2} &lt; 1

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

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.
3x - 7 &lt; x + 2

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.
3x - 7 &gt; x + 2

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

 

[nicksbyman]

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:
\frac{3x-7}{x+2} &lt; 1

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

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.
3x - 7 &lt; x + 2

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.
3x - 7 &gt; x + 2

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.

 

[eumyang]

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).

 

[Mark44]

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.

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.