Solving an Inequality

[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}

[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 ">".

[Дьявол]

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}

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

[S_David]

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}
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

[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.

[S_David]

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.

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