Mistake in solving inequality

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In summary: Best regards.In summary, the conversation discusses different methods for solving a given inequality and identifies the mistake in a previous solution. The methods include multiplying both sides by a squared term, splitting the analysis into cases, and solving the corresponding equation. The final solution is (2,3).
  • #1
EngWiPy
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Hello,

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

[tex]\begin{align*}
&\frac{2x-5}{x-2}<1\\
&2x-5<x-2\\
&x<3, x\ne 2
\end{align}[/tex]
 
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  • #2
S_David said:
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?
 
  • #3
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 ">".
 
  • #4
S_David said:
Hello,

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

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

And why don't you try:

[tex]\frac{2x-5}{x-2} - 1 < 0[/tex]

?

Then consider these cases

[tex]a/b <0 [/tex]

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

Regards.
 
  • #5
S_David said:
Hello,

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

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

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
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
uart said:
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:

[tex]\begin{align}
x-2&>0\\
x-2&<0
\end{align}[/tex]

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

[tex]\begin{align*}
2x-5&<x-2\\
x&<3
\end{align}[/tex]

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

[tex]\begin{align*}
2x-5&>x-2\\
x&>3
\end{align}[/tex]

But [tex]x<2[/tex], 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 [tex]x[/tex].

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
You could solve it this way:
[tex]
\frac{2x-5}{x-2} - 1 < 0
[/tex]

[tex]\frac{x-3}{x-2}<0[/tex]

[tex]\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}
[/tex]

[tex]\begin{bmatrix}
\left\{\begin{matrix}
x<3\\
x>2
\end{matrix}\right.
\\
\\
\left\{\begin{matrix}
x>3\\
x<2
\end{matrix}\right.
\end{matrix}[/tex]

Because 3<x<2 is not valid,

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

Regards.
 
  • #9
Дьявол said:
You could solve it this way:
[tex]
\frac{2x-5}{x-2} - 1 < 0
[/tex]

[tex]\frac{x-3}{x-2}<0[/tex]

[tex]\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}
[/tex]

[tex]\begin{bmatrix}
\left\{\begin{matrix}
x<3\\
x>2
\end{matrix}\right.
\\
\\
\left\{\begin{matrix}
x>3\\
x<2
\end{matrix}\right.
\end{matrix}[/tex]

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
 

What is an inequality?

An inequality is a mathematical statement that compares two quantities using <, >, ≤, or ≥ symbols. These symbols represent less than, greater than, less than or equal to, and greater than or equal to, respectively.

How do you solve an inequality?

To solve an inequality, you must isolate the variable on one side of the inequality symbol. This can be done by using inverse operations, such as adding, subtracting, multiplying, or dividing both sides of the inequality by the same number. Remember to flip the inequality symbol if you multiply or divide by a negative number.

What is the solution set of an inequality?

The solution set of an inequality is the set of all possible values that make the inequality true. This set can be represented using interval notation or set builder notation. For example, if the solution set is all real numbers greater than 3, it can be written as (3, ∞) or {x | x > 3}.

What is the difference between solving an inequality and solving an equation?

The main difference between solving an inequality and solving an equation is that solving an inequality can result in a range of values, while solving an equation results in a single value. Additionally, when solving an inequality, the inequality symbol must be maintained throughout the solving process, while in equations, the equal sign can be replaced with an equivalent expression.

When is it necessary to flip the inequality symbol when solving an inequality?

It is necessary to flip the inequality symbol when multiplying or dividing both sides of an inequality by a negative number. This is because multiplying or dividing by a negative number changes the direction of the inequality. For example, if you have x < 3 and you multiply both sides by -2, the new inequality is -2x > -6, which can be rewritten as x < 3. Therefore, you must flip the inequality symbol to maintain the correct direction of the inequality.

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