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 /> &amp;\frac{2x-5}{x-2}&lt;1\\<br /> &amp;2x-5&lt;x-2\\<br /> &amp;x&lt;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 \frac{2x-5}{x-2}= 1[/itex]. The &quot;x&quot; that satisfies that and the &quot;x&quot; that makes the denominator 0 are the only places where the inequality can &quot;change&quot;. They divide the real line into three intervals- check one point in each interval to see which give &quot;&gt;&quot;.

 

[Дьявол]

Hello,

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

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

And why don't you try:

\frac{2x-5}{x-2} - 1 &lt; 0

?

Then consider these cases

a/b &lt;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 /> &amp;\frac{2x-5}{x-2}&lt;1\\<br /> &amp;2x-5&lt;x-2\\<br /> &amp;x&lt;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&amp;&gt;0\\<br /> x-2&amp;&lt;0<br /> \end{align}

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

\begin{align*}<br /> 2x-5&amp;&lt;x-2\\<br /> x&amp;&lt;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&amp;&gt;x-2\\<br /> x&amp;&gt;3<br /> \end{align}

But x&lt;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:
<br /> \frac{2x-5}{x-2} - 1 &lt; 0<br />

\frac{x-3}{x-2}&lt;0

\begin{bmatrix}<br /> \left\{\begin{matrix}<br /> x-3&lt;0\\ <br /> x-2&gt;0<br /> \end{matrix}\right.<br /> \\<br /> \\<br /> \left\{\begin{matrix}<br /> x-3&gt;0\\ <br /> x-2&lt;0<br /> \end{matrix}\right.<br /> <br /> \end{matrix}<br />

\begin{bmatrix}<br /> \left\{\begin{matrix}<br /> x&lt;3\\ <br /> x&gt;2<br /> \end{matrix}\right.<br /> \\<br /> \\<br /> \left\{\begin{matrix}<br /> x&gt;3\\ <br /> x&lt;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:
<br /> \frac{2x-5}{x-2} - 1 &lt; 0<br />

\frac{x-3}{x-2}&lt;0

\begin{bmatrix}<br /> \left\{\begin{matrix}<br /> x-3&lt;0\\ <br /> x-2&gt;0<br /> \end{matrix}\right.<br /> \\<br /> \\<br /> \left\{\begin{matrix}<br /> x-3&gt;0\\ <br /> x-2&lt;0<br /> \end{matrix}\right.<br /> <br /> \end{matrix}<br />

\begin{bmatrix}<br /> \left\{\begin{matrix}<br /> x&lt;3\\ <br /> x&gt;2<br /> \end{matrix}\right.<br /> \\<br /> \\<br /> \left\{\begin{matrix}<br /> x&gt;3\\ <br /> x&lt;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