# Solving an Inequality

1. Aug 26, 2009

### S_David

Hello,

Where is the mistake in the following solution of the inequality:

\begin{align*} &\frac{2x-5}{x-2}<1\\ &2x-5<x-2\\ &x<3, x\ne 2 \end{align}

2. Aug 26, 2009

### Hurkyl

Staff Emeritus
The following is not true:
If a<b, then ac<bc​

In fact you have three separate cases, depending on c -- do you know what they are?

3. Aug 26, 2009

### HallsofIvy

Staff Emeritus
Hurkyl's point is that you multiplied both sides of the inequality by x- 2 and, since you don't know what x is, you don't know if x- 2 is positive or negative. (Oops, I just gave you two of the three cases Hurkyl asked about!)

My preferred method for solving anything more than linear inequalities is to first solve the corresponding equation. Here, solve $$\frac{2x-5}{x-2}= 1[/itex]. The "x" that satisfies that and the "x" that makes the denominator 0 are the only places where the inequality can "change". They divide the real line into three intervals- check one point in each interval to see which give ">". 4. Aug 26, 2009 ### Дьявол And why don't you try: [tex]\frac{2x-5}{x-2} - 1 < 0$$

?

Then consider these cases

$$a/b <0$$

should "a" and /or "b" be positive or negative so that the the fraction a/b would be negative?

Regards.

5. Aug 26, 2009

### arildno

You may split your analysis into two cases:

1) x-2>0,
and

2) x-2<0

You will find that for 2), there are NO solutions, meaning that x must be greater than 2.

6. Aug 26, 2009

### uart

Another excelent method in this case is to multiply both sides by (x-2)^2, with the squared assuring that we haven't multiplied by a negative number, and then just re-arrange it to give a quadratic inequality : x^2 - 5x + 6 < 0

7. Aug 26, 2009

### S_David

Ok, I can see where I did mistake the inequality. So, we have two cases:

\begin{align} x-2&>0\\ x-2&<0 \end{align}

In the first case the multiplication does not change the direction of the inequality, so:

\begin{align*} 2x-5&<x-2\\ x&<3 \end{align}

Then the open interval $$(2,3)$$ is the solution set. In the second case, the direction of the inequality changed, so:

\begin{align*} 2x-5&>x-2\\ x&>3 \end{align}

But $$x<2$$, then there is no solution.

The solution provided by uart is a good one, too. Where he eliminated the problem of negative sign possibility in the unknown $$x$$.

Actually, I am reviewing the precalculus and calculus books, and such things I forgot because I don't practice it continousely.

Anyway, thank you all guys.

Best regards

8. Aug 26, 2009

### Дьявол

You could solve it this way:
$$\frac{2x-5}{x-2} - 1 < 0$$

$$\frac{x-3}{x-2}<0$$

$$\begin{bmatrix} \left\{\begin{matrix} x-3<0\\ x-2>0 \end{matrix}\right. \\ \\ \left\{\begin{matrix} x-3>0\\ x-2<0 \end{matrix}\right. \end{matrix}$$

$$\begin{bmatrix} \left\{\begin{matrix} x<3\\ x>2 \end{matrix}\right. \\ \\ \left\{\begin{matrix} x>3\\ x<2 \end{matrix}\right. \end{matrix}$$

Because 3<x<2 is not valid,

the only solution is 2<x<3 or (2,3).

Regards.

9. Aug 27, 2009

### S_David

Dear Дьявол,

This method is as described in my calculus book, and yes it is easier than the one I explained earlier. But I wanted to know the different methods to solve the inequality.

Thanks