What are the different methods for solving an inequality?

[EngWiPy]

Hello,

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

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

 

[Hurkyl]

Hello,

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

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?

 

[HallsofIvy]

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 [tex]\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 ">".[/tex]

 

[Дьявол]

Hello,

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

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

And why don't you try:

$$\frac{2x-5}{x-2} - 1 < 0$$

?

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.

 

[arildno]

Hello,

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

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

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.

 

[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

 

[EngWiPy]

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

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

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

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

$$\begin{align*}<br /> 2x-5&<x-2\\<br /> x&<3<br /> \end{align}$$

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

$$\begin{align*}<br /> 2x-5&>x-2\\<br /> x&>3<br /> \end{align}$$

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

 

[Дьявол]

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

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

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

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

Because 3<x<2 is not valid,

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

Regards.

 

[EngWiPy]

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

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

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

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

Because 3<x<2 is not valid,

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

Regards.

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