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How to solve $frac{x-1}>fracx+12x+1$?
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up vote
4
down vote
favorite
I am working on the following problem:
$$fracx-1>fracx+12x+1$$
Here's what I have done so far:
$$|x+2|>fracx+12x+1times(x-1)$$
$$-left(frac(x+1)(x-1)(2x+1)right)<x+2<frac(x+1)(x-1)(2x+1)$$
This is where I stopped. I am not entirely sure how to go about solving this type of inequality. Am I using the correct approach?
algebra-precalculus inequality absolute-value
edited Aug 6 at 18:27
user1551
66.9k564122
asked Aug 6 at 18:14
AugieJavax98
308210
add a comment |Â
up vote
4
down vote
favorite
I am working on the following problem:
$$fracx-1>fracx+12x+1$$
Here's what I have done so far:
$$|x+2|>fracx+12x+1times(x-1)$$
$$-left(frac(x+1)(x-1)(2x+1)right)<x+2<frac(x+1)(x-1)(2x+1)$$
This is where I stopped. I am not entirely sure how to go about solving this type of inequality. Am I using the correct approach?
algebra-precalculus inequality absolute-value
edited Aug 6 at 18:27
user1551
66.9k564122
asked Aug 6 at 18:14
AugieJavax98
308210
4
Multiplying both sides by $x-1$ and leaving the "greater-than" sign intact is valid ONLY if you know $x-1$ is positive.
– Michael Hardy
Aug 6 at 18:20
From $|A|> B$ you cannot conclude $-B<A<B;$ rather, that follows from $|A|<B. qquad$
– Michael Hardy
Aug 6 at 18:22
I'd start by finding a common denominator and moving everything except $0$ to the other side.
– Michael Hardy
Aug 6 at 18:26
The problem doesn't make sense if $x=1$ or if $x=-1/2$ You probably need to break it up into cases: $1)x<-1/2; 2)-1/2<x<1; 3) x>1$
– saulspatz
Aug 6 at 18:26
You have written an inequality, but an inequality on its own is not a problem. What are you being asked to do with the inequality? What is the source of the inequality - what book or course?
– Carl Mummert
Aug 6 at 19:18
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am working on the following problem:
$$fracx-1>fracx+12x+1$$
Here's what I have done so far:
$$|x+2|>fracx+12x+1times(x-1)$$
$$-left(frac(x+1)(x-1)(2x+1)right)<x+2<frac(x+1)(x-1)(2x+1)$$
This is where I stopped. I am not entirely sure how to go about solving this type of inequality. Am I using the correct approach?
algebra-precalculus inequality absolute-value
edited Aug 6 at 18:27
user1551
66.9k564122
asked Aug 6 at 18:14
AugieJavax98
308210
I am working on the following problem:
$$fracx-1>fracx+12x+1$$
Here's what I have done so far:
$$|x+2|>fracx+12x+1times(x-1)$$
$$-left(frac(x+1)(x-1)(2x+1)right)<x+2<frac(x+1)(x-1)(2x+1)$$
This is where I stopped. I am not entirely sure how to go about solving this type of inequality. Am I using the correct approach?
algebra-precalculus inequality absolute-value
edited Aug 6 at 18:27
user1551
66.9k564122
asked Aug 6 at 18:14
AugieJavax98
308210
edited Aug 6 at 18:27
user1551
66.9k564122
edited Aug 6 at 18:27
user1551
66.9k564122
edited Aug 6 at 18:27
user1551
66.9k564122
66.9k564122
asked Aug 6 at 18:14
AugieJavax98
308210
asked Aug 6 at 18:14
AugieJavax98
308210
asked Aug 6 at 18:14
AugieJavax98
308210
308210
4
Multiplying both sides by $x-1$ and leaving the "greater-than" sign intact is valid ONLY if you know $x-1$ is positive.
– Michael Hardy
Aug 6 at 18:20
From $|A|> B$ you cannot conclude $-B<A<B;$ rather, that follows from $|A|<B. qquad$
– Michael Hardy
Aug 6 at 18:22
I'd start by finding a common denominator and moving everything except $0$ to the other side.
– Michael Hardy
Aug 6 at 18:26
The problem doesn't make sense if $x=1$ or if $x=-1/2$ You probably need to break it up into cases: $1)x<-1/2; 2)-1/2<x<1; 3) x>1$
– saulspatz
Aug 6 at 18:26
You have written an inequality, but an inequality on its own is not a problem. What are you being asked to do with the inequality? What is the source of the inequality - what book or course?
– Carl Mummert
Aug 6 at 19:18
add a comment |Â
4
Multiplying both sides by $x-1$ and leaving the "greater-than" sign intact is valid ONLY if you know $x-1$ is positive.
– Michael Hardy
Aug 6 at 18:20
From $|A|> B$ you cannot conclude $-B<A<B;$ rather, that follows from $|A|<B. qquad$
– Michael Hardy
Aug 6 at 18:22
I'd start by finding a common denominator and moving everything except $0$ to the other side.
– Michael Hardy
Aug 6 at 18:26
The problem doesn't make sense if $x=1$ or if $x=-1/2$ You probably need to break it up into cases: $1)x<-1/2; 2)-1/2<x<1; 3) x>1$
– saulspatz
Aug 6 at 18:26
You have written an inequality, but an inequality on its own is not a problem. What are you being asked to do with the inequality? What is the source of the inequality - what book or course?
– Carl Mummert
Aug 6 at 19:18
4
4
Multiplying both sides by $x-1$ and leaving the "greater-than" sign intact is valid ONLY if you know $x-1$ is positive.
– Michael Hardy
Aug 6 at 18:20
Multiplying both sides by $x-1$ and leaving the "greater-than" sign intact is valid ONLY if you know $x-1$ is positive.
– Michael Hardy
Aug 6 at 18:20
From $|A|> B$ you cannot conclude $-B<A<B;$ rather, that follows from $|A|<B. qquad$
– Michael Hardy
Aug 6 at 18:22
From $|A|> B$ you cannot conclude $-B<A<B;$ rather, that follows from $|A|<B. qquad$
– Michael Hardy
Aug 6 at 18:22
I'd start by finding a common denominator and moving everything except $0$ to the other side.
– Michael Hardy
Aug 6 at 18:26
I'd start by finding a common denominator and moving everything except $0$ to the other side.
– Michael Hardy
Aug 6 at 18:26
The problem doesn't make sense if $x=1$ or if $x=-1/2$ You probably need to break it up into cases: $1)x<-1/2; 2)-1/2<x<1; 3) x>1$
– saulspatz
Aug 6 at 18:26
The problem doesn't make sense if $x=1$ or if $x=-1/2$ You probably need to break it up into cases: $1)x<-1/2; 2)-1/2<x<1; 3) x>1$
– saulspatz
Aug 6 at 18:26
You have written an inequality, but an inequality on its own is not a problem. What are you being asked to do with the inequality? What is the source of the inequality - what book or course?
– Carl Mummert
Aug 6 at 19:18
You have written an inequality, but an inequality on its own is not a problem. What are you being asked to do with the inequality? What is the source of the inequality - what book or course?
– Carl Mummert
Aug 6 at 19:18
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Hint: You must do case work:
Case 1: $$x>1$$ then we get $$2x+1>0$$ and $|x+2|=x+2$ and we have
$$(x+2)(2x+1)>x^2-1$$ so $2x^2+5x+2>x^2-1$.
Can you proceed?
For your Control:
The result is given by
$frac-52+frac12sqrt3<x<-frac12$ or $x>1$
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
add a comment |Â
up vote
1
down vote
Your last step is incorrect as we know nothing about the sign of $x+1$, and also if $|a|>b$, then $a<-b, a>b$, not $-b<a<b$.
What you need to do is consider each case between the points $x=-1, -frac12,1$ as they determine whether the inequality should be flipped over, and whether you could drop the absolute value.
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
add a comment |Â
up vote
1
down vote
We need to consider 2 cases
- for $x+2ge 0 implies xge -2$ we need to solve
$$fracx+2x-1>fracx+12x+1$$
- for $x+2< 0 implies x< -2$ we need to solve
$$frac-x-2x-1>fracx+12x+1$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$fracx+2x-1>fracx+12x+1iff fracx+2x-1-fracx+12x+1>0iff frac(x+2)(2x+1)-(x+1)(x-1)(x-1)(2x+1)>0$$
$$iff fracx^2+5x+3(x-1)(2x+1)>0iff fracleft(x-frac5+sqrt 132right)left(x-frac5-sqrt 132right)(x-1)(2x+1)>0$$
then we ca easily find the solutions under the condition $xge -2$.
In a similar way we can study case 2.
answered Aug 6 at 19:23
gimusi
65.5k73684
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: You must do case work:
Case 1: $$x>1$$ then we get $$2x+1>0$$ and $|x+2|=x+2$ and we have
$$(x+2)(2x+1)>x^2-1$$ so $2x^2+5x+2>x^2-1$.
Can you proceed?
For your Control:
The result is given by
$frac-52+frac12sqrt3<x<-frac12$ or $x>1$
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
add a comment |Â
up vote
2
down vote
Hint: You must do case work:
Case 1: $$x>1$$ then we get $$2x+1>0$$ and $|x+2|=x+2$ and we have
$$(x+2)(2x+1)>x^2-1$$ so $2x^2+5x+2>x^2-1$.
Can you proceed?
For your Control:
The result is given by
$frac-52+frac12sqrt3<x<-frac12$ or $x>1$
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: You must do case work:
Case 1: $$x>1$$ then we get $$2x+1>0$$ and $|x+2|=x+2$ and we have
$$(x+2)(2x+1)>x^2-1$$ so $2x^2+5x+2>x^2-1$.
Can you proceed?
For your Control:
The result is given by
$frac-52+frac12sqrt3<x<-frac12$ or $x>1$
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
Hint: You must do case work:
Case 1: $$x>1$$ then we get $$2x+1>0$$ and $|x+2|=x+2$ and we have
$$(x+2)(2x+1)>x^2-1$$ so $2x^2+5x+2>x^2-1$.
Can you proceed?
For your Control:
The result is given by
$frac-52+frac12sqrt3<x<-frac12$ or $x>1$
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
answered Aug 6 at 18:29
Dr. Sonnhard Graubner
67k32660
67k32660
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
add a comment |Â
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
Why did you start with x>1?
– AugieJavax98
Aug 7 at 13:02
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
The other cases are also to be considered, but the case $x>1$ seems to me easy for the start, and in this case all given terms are positive.
– Dr. Sonnhard Graubner
Aug 7 at 13:08
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
In the next case i would consider $-frac12<x<1$
– Dr. Sonnhard Graubner
Aug 7 at 13:10
add a comment |Â
up vote
1
down vote
Your last step is incorrect as we know nothing about the sign of $x+1$, and also if $|a|>b$, then $a<-b, a>b$, not $-b<a<b$.
What you need to do is consider each case between the points $x=-1, -frac12,1$ as they determine whether the inequality should be flipped over, and whether you could drop the absolute value.
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
add a comment |Â
up vote
1
down vote
Your last step is incorrect as we know nothing about the sign of $x+1$, and also if $|a|>b$, then $a<-b, a>b$, not $-b<a<b$.
What you need to do is consider each case between the points $x=-1, -frac12,1$ as they determine whether the inequality should be flipped over, and whether you could drop the absolute value.
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your last step is incorrect as we know nothing about the sign of $x+1$, and also if $|a|>b$, then $a<-b, a>b$, not $-b<a<b$.
What you need to do is consider each case between the points $x=-1, -frac12,1$ as they determine whether the inequality should be flipped over, and whether you could drop the absolute value.
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
Your last step is incorrect as we know nothing about the sign of $x+1$, and also if $|a|>b$, then $a<-b, a>b$, not $-b<a<b$.
What you need to do is consider each case between the points $x=-1, -frac12,1$ as they determine whether the inequality should be flipped over, and whether you could drop the absolute value.
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
answered Aug 6 at 18:32
TheSimpliFire
9,69261952
9,69261952
add a comment |Â
add a comment |Â
up vote
1
down vote
We need to consider 2 cases
- for $x+2ge 0 implies xge -2$ we need to solve
$$fracx+2x-1>fracx+12x+1$$
- for $x+2< 0 implies x< -2$ we need to solve
$$frac-x-2x-1>fracx+12x+1$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$fracx+2x-1>fracx+12x+1iff fracx+2x-1-fracx+12x+1>0iff frac(x+2)(2x+1)-(x+1)(x-1)(x-1)(2x+1)>0$$
$$iff fracx^2+5x+3(x-1)(2x+1)>0iff fracleft(x-frac5+sqrt 132right)left(x-frac5-sqrt 132right)(x-1)(2x+1)>0$$
then we ca easily find the solutions under the condition $xge -2$.
In a similar way we can study case 2.
answered Aug 6 at 19:23
gimusi
65.5k73684
add a comment |Â
up vote
1
down vote
We need to consider 2 cases
- for $x+2ge 0 implies xge -2$ we need to solve
$$fracx+2x-1>fracx+12x+1$$
- for $x+2< 0 implies x< -2$ we need to solve
$$frac-x-2x-1>fracx+12x+1$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$fracx+2x-1>fracx+12x+1iff fracx+2x-1-fracx+12x+1>0iff frac(x+2)(2x+1)-(x+1)(x-1)(x-1)(2x+1)>0$$
$$iff fracx^2+5x+3(x-1)(2x+1)>0iff fracleft(x-frac5+sqrt 132right)left(x-frac5-sqrt 132right)(x-1)(2x+1)>0$$
then we ca easily find the solutions under the condition $xge -2$.
In a similar way we can study case 2.
answered Aug 6 at 19:23
gimusi
65.5k73684
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We need to consider 2 cases
- for $x+2ge 0 implies xge -2$ we need to solve
$$fracx+2x-1>fracx+12x+1$$
- for $x+2< 0 implies x< -2$ we need to solve
$$frac-x-2x-1>fracx+12x+1$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$fracx+2x-1>fracx+12x+1iff fracx+2x-1-fracx+12x+1>0iff frac(x+2)(2x+1)-(x+1)(x-1)(x-1)(2x+1)>0$$
$$iff fracx^2+5x+3(x-1)(2x+1)>0iff fracleft(x-frac5+sqrt 132right)left(x-frac5-sqrt 132right)(x-1)(2x+1)>0$$
then we ca easily find the solutions under the condition $xge -2$.
In a similar way we can study case 2.
answered Aug 6 at 19:23
gimusi
65.5k73684
We need to consider 2 cases
- for $x+2ge 0 implies xge -2$ we need to solve
$$fracx+2x-1>fracx+12x+1$$
- for $x+2< 0 implies x< -2$ we need to solve
$$frac-x-2x-1>fracx+12x+1$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$fracx+2x-1>fracx+12x+1iff fracx+2x-1-fracx+12x+1>0iff frac(x+2)(2x+1)-(x+1)(x-1)(x-1)(2x+1)>0$$
$$iff fracx^2+5x+3(x-1)(2x+1)>0iff fracleft(x-frac5+sqrt 132right)left(x-frac5-sqrt 132right)(x-1)(2x+1)>0$$
then we ca easily find the solutions under the condition $xge -2$.
In a similar way we can study case 2.
answered Aug 6 at 19:23
gimusi
65.5k73684
edited Aug 6 at 19:39
answered Aug 6 at 19:23
gimusi
65.5k73684
answered Aug 6 at 19:23
gimusi
65.5k73684
answered Aug 6 at 19:23
gimusi
65.5k73684
65.5k73684
add a comment |Â
add a comment |Â
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4
Multiplying both sides by $x-1$ and leaving the "greater-than" sign intact is valid ONLY if you know $x-1$ is positive.
– Michael Hardy
Aug 6 at 18:20
From $|A|> B$ you cannot conclude $-B<A<B;$ rather, that follows from $|A|<B. qquad$
– Michael Hardy
Aug 6 at 18:22
I'd start by finding a common denominator and moving everything except $0$ to the other side.
– Michael Hardy
Aug 6 at 18:26
The problem doesn't make sense if $x=1$ or if $x=-1/2$ You probably need to break it up into cases: $1)x<-1/2; 2)-1/2<x<1; 3) x>1$
– saulspatz
Aug 6 at 18:26
You have written an inequality, but an inequality on its own is not a problem. What are you being asked to do with the inequality? What is the source of the inequality - what book or course?
– Carl Mummert
Aug 6 at 19:18