Problem in solving inequality.
$begingroup$
In my book I found this following inequality :
$frac{2x}{x-1}>x$
In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?
inequality
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
$endgroup$
add a comment |
$begingroup$
In my book I found this following inequality :
$frac{2x}{x-1}>x$
In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?
inequality
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
$endgroup$
add a comment |
$begingroup$
In my book I found this following inequality :
$frac{2x}{x-1}>x$
In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?
inequality
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
$endgroup$
In my book I found this following inequality :
$frac{2x}{x-1}>x$
In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?
inequality
inequality
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
asked Jan 2 at 10:38
Asif IqubalAsif Iqubal
1397
1397
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can not divide by $x$ because $x$ can be positive or $x$ can be negative.
It's $$frac{2x}{x-1}-x>0$$ or
$$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
$$(-infty,0)cup(1,3).$$
I used the intervals method.
We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.
Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.
Thus, the sign on $(1,3)$ is $+$.
If we pass a point $1$ then the sign changes again.
Thus, the sign on $(0,1)$ is $-$.
And if we pass a point $0$ then the sign still changes.
Thus, the sign on $(-infty,0)$ is $+$.
Now, we can write the answer.
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
add a comment |
$begingroup$
You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.
In this example:
$$
frac{2x}{x-1}-x>0
$$
$$
frac{3x-x^2}{x-1}>0
$$
$$
frac{x(3-x)}{x-1}>0
$$
A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.
Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.
Once you take care about these points, you should find easily the full solution.
answered Jan 2 at 10:43
SergeDSergeD
214
$endgroup$
add a comment |
$begingroup$
We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can not divide by $x$ because $x$ can be positive or $x$ can be negative.
It's $$frac{2x}{x-1}-x>0$$ or
$$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
$$(-infty,0)cup(1,3).$$
I used the intervals method.
We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.
Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.
Thus, the sign on $(1,3)$ is $+$.
If we pass a point $1$ then the sign changes again.
Thus, the sign on $(0,1)$ is $-$.
And if we pass a point $0$ then the sign still changes.
Thus, the sign on $(-infty,0)$ is $+$.
Now, we can write the answer.
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
add a comment |
$begingroup$
You can not divide by $x$ because $x$ can be positive or $x$ can be negative.
It's $$frac{2x}{x-1}-x>0$$ or
$$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
$$(-infty,0)cup(1,3).$$
I used the intervals method.
We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.
Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.
Thus, the sign on $(1,3)$ is $+$.
If we pass a point $1$ then the sign changes again.
Thus, the sign on $(0,1)$ is $-$.
And if we pass a point $0$ then the sign still changes.
Thus, the sign on $(-infty,0)$ is $+$.
Now, we can write the answer.
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
add a comment |
$begingroup$
You can not divide by $x$ because $x$ can be positive or $x$ can be negative.
It's $$frac{2x}{x-1}-x>0$$ or
$$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
$$(-infty,0)cup(1,3).$$
I used the intervals method.
We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.
Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.
Thus, the sign on $(1,3)$ is $+$.
If we pass a point $1$ then the sign changes again.
Thus, the sign on $(0,1)$ is $-$.
And if we pass a point $0$ then the sign still changes.
Thus, the sign on $(-infty,0)$ is $+$.
Now, we can write the answer.
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
You can not divide by $x$ because $x$ can be positive or $x$ can be negative.
It's $$frac{2x}{x-1}-x>0$$ or
$$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
$$(-infty,0)cup(1,3).$$
I used the intervals method.
We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.
Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.
Thus, the sign on $(1,3)$ is $+$.
If we pass a point $1$ then the sign changes again.
Thus, the sign on $(0,1)$ is $-$.
And if we pass a point $0$ then the sign still changes.
Thus, the sign on $(-infty,0)$ is $+$.
Now, we can write the answer.
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
edited Jan 2 at 11:16
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 2 at 10:45
Michael RozenbergMichael Rozenberg
106k1893198
106k1893198
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
add a comment |
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
$endgroup$
– Asif Iqubal
Jan 2 at 10:47
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
@Asif Iqubal It's the intervals method.
$endgroup$
– Michael Rozenberg
Jan 2 at 10:49
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
Can you be kind to elaborate please?
$endgroup$
– Asif Iqubal
Jan 2 at 10:50
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
$begingroup$
@Asif Iqubal I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 2 at 11:00
add a comment |
$begingroup$
You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.
In this example:
$$
frac{2x}{x-1}-x>0
$$
$$
frac{3x-x^2}{x-1}>0
$$
$$
frac{x(3-x)}{x-1}>0
$$
A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.
In this example:
$$
frac{2x}{x-1}-x>0
$$
$$
frac{3x-x^2}{x-1}>0
$$
$$
frac{x(3-x)}{x-1}>0
$$
A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.
In this example:
$$
frac{2x}{x-1}-x>0
$$
$$
frac{3x-x^2}{x-1}>0
$$
$$
frac{x(3-x)}{x-1}>0
$$
A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.
In this example:
$$
frac{2x}{x-1}-x>0
$$
$$
frac{3x-x^2}{x-1}>0
$$
$$
frac{x(3-x)}{x-1}>0
$$
A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
edited Jan 2 at 10:50
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
answered Jan 2 at 10:43
A.Γ.A.Γ.
22.8k32656
22.8k32656
add a comment |
add a comment |
$begingroup$
Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.
Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.
Once you take care about these points, you should find easily the full solution.
answered Jan 2 at 10:43
SergeDSergeD
214
$endgroup$
add a comment |
$begingroup$
Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.
Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.
Once you take care about these points, you should find easily the full solution.
answered Jan 2 at 10:43
SergeDSergeD
214
$endgroup$
add a comment |
$begingroup$
Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.
Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.
Once you take care about these points, you should find easily the full solution.
answered Jan 2 at 10:43
SergeDSergeD
214
$endgroup$
Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.
Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.
Once you take care about these points, you should find easily the full solution.
answered Jan 2 at 10:43
SergeDSergeD
214
answered Jan 2 at 10:43
SergeDSergeD
214
answered Jan 2 at 10:43
SergeDSergeD
214
answered Jan 2 at 10:43
SergeDSergeD
214
214
add a comment |
add a comment |
$begingroup$
We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
$endgroup$
We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
answered Jan 2 at 10:48
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
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