Mixing
On Changing The Direction Of The Inequality Sign By Dividing By A Negative Number?
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When solving inequality equations, we have to change the sign of the inequality when we divide/multiply by a negative number. And I get why. Because your making one side positive and one side negative.
But let's say you were multiplying by say $(x^3+1)/x>1$ or something. When you multiply both sides by x, why don't you need to flip the sign? Because there's the possibility x is negative! Why don't you need to construct to equations, one where the sign is the same, and one where the sign is flipped.
As an extension, let's say you were multiplying both sides of $(x^3+1)/-x>1$ by -x. Do you need to change the sign too?
Thank you in advance.
algebra-precalculus inequality
asked Aug 2 at 13:48
Ethan Chan
591322
add a comment |Â
up vote
1
down vote
favorite
When solving inequality equations, we have to change the sign of the inequality when we divide/multiply by a negative number. And I get why. Because your making one side positive and one side negative.
But let's say you were multiplying by say $(x^3+1)/x>1$ or something. When you multiply both sides by x, why don't you need to flip the sign? Because there's the possibility x is negative! Why don't you need to construct to equations, one where the sign is the same, and one where the sign is flipped.
As an extension, let's say you were multiplying both sides of $(x^3+1)/-x>1$ by -x. Do you need to change the sign too?
Thank you in advance.
algebra-precalculus inequality
asked Aug 2 at 13:48
Ethan Chan
591322
1
When multiplying by an expression whose sign may vary (according to some variable, $x$ for instance), then one must separate the inequality into two different inequalities, one for each case, and solve each of them separately.
– Suzet
Aug 2 at 13:50
@Suzet So does one need to flip the sign in the 2 situations ($(x^3+1)/x>1$ etc) above? If so, why?
– Ethan Chan
Aug 2 at 13:52
See Inequalities.
– Mauro ALLEGRANZA
Aug 2 at 13:53
"And I get why. Because your making one side positive and one side negative." - not sure I understand what you are saying here, but it does not sound correct to me.
– NickD
Aug 2 at 15:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When solving inequality equations, we have to change the sign of the inequality when we divide/multiply by a negative number. And I get why. Because your making one side positive and one side negative.
But let's say you were multiplying by say $(x^3+1)/x>1$ or something. When you multiply both sides by x, why don't you need to flip the sign? Because there's the possibility x is negative! Why don't you need to construct to equations, one where the sign is the same, and one where the sign is flipped.
As an extension, let's say you were multiplying both sides of $(x^3+1)/-x>1$ by -x. Do you need to change the sign too?
Thank you in advance.
algebra-precalculus inequality
asked Aug 2 at 13:48
Ethan Chan
591322
When solving inequality equations, we have to change the sign of the inequality when we divide/multiply by a negative number. And I get why. Because your making one side positive and one side negative.
But let's say you were multiplying by say $(x^3+1)/x>1$ or something. When you multiply both sides by x, why don't you need to flip the sign? Because there's the possibility x is negative! Why don't you need to construct to equations, one where the sign is the same, and one where the sign is flipped.
As an extension, let's say you were multiplying both sides of $(x^3+1)/-x>1$ by -x. Do you need to change the sign too?
Thank you in advance.
algebra-precalculus inequality
asked Aug 2 at 13:48
Ethan Chan
591322
asked Aug 2 at 13:48
Ethan Chan
591322
asked Aug 2 at 13:48
Ethan Chan
591322
asked Aug 2 at 13:48
Ethan Chan
591322
591322
1
When multiplying by an expression whose sign may vary (according to some variable, $x$ for instance), then one must separate the inequality into two different inequalities, one for each case, and solve each of them separately.
– Suzet
Aug 2 at 13:50
@Suzet So does one need to flip the sign in the 2 situations ($(x^3+1)/x>1$ etc) above? If so, why?
– Ethan Chan
Aug 2 at 13:52
See Inequalities.
– Mauro ALLEGRANZA
Aug 2 at 13:53
"And I get why. Because your making one side positive and one side negative." - not sure I understand what you are saying here, but it does not sound correct to me.
– NickD
Aug 2 at 15:40
add a comment |Â
1
When multiplying by an expression whose sign may vary (according to some variable, $x$ for instance), then one must separate the inequality into two different inequalities, one for each case, and solve each of them separately.
– Suzet
Aug 2 at 13:50
@Suzet So does one need to flip the sign in the 2 situations ($(x^3+1)/x>1$ etc) above? If so, why?
– Ethan Chan
Aug 2 at 13:52
See Inequalities.
– Mauro ALLEGRANZA
Aug 2 at 13:53
"And I get why. Because your making one side positive and one side negative." - not sure I understand what you are saying here, but it does not sound correct to me.
– NickD
Aug 2 at 15:40
1
1
When multiplying by an expression whose sign may vary (according to some variable, $x$ for instance), then one must separate the inequality into two different inequalities, one for each case, and solve each of them separately.
– Suzet
Aug 2 at 13:50
When multiplying by an expression whose sign may vary (according to some variable, $x$ for instance), then one must separate the inequality into two different inequalities, one for each case, and solve each of them separately.
– Suzet
Aug 2 at 13:50
@Suzet So does one need to flip the sign in the 2 situations ($(x^3+1)/x>1$ etc) above? If so, why?
– Ethan Chan
Aug 2 at 13:52
@Suzet So does one need to flip the sign in the 2 situations ($(x^3+1)/x>1$ etc) above? If so, why?
– Ethan Chan
Aug 2 at 13:52
See Inequalities.
– Mauro ALLEGRANZA
Aug 2 at 13:53
See Inequalities.
– Mauro ALLEGRANZA
Aug 2 at 13:53
"And I get why. Because your making one side positive and one side negative." - not sure I understand what you are saying here, but it does not sound correct to me.
– NickD
Aug 2 at 15:40
"And I get why. Because your making one side positive and one side negative." - not sure I understand what you are saying here, but it does not sound correct to me.
– NickD
Aug 2 at 15:40
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note that there is not exception and the following
$$fracx^3+1x>1 iff x^3+1>x$$
holds for $x>0$ while for $x<0$ we have
$$fracx^3+1x>1 iff x^3+1<x$$
To avoid this kind of manipulation and distinction in two cases, we can proceed, as an alternative, as follow
$$fracx^3+1x>1 iff fracx^3+1x-1>0 iff fracx^3+1-xx>0$$
and consider separetely the sign of numerator and denominator.
answered Aug 2 at 13:52
gimusi
63.8k73480
add a comment |Â
up vote
0
down vote
When you multiply the sides by $x$, you don't need to flip the sign if you assume that $x>0$. If you assume that $x<0$, then you do have to flip the sign.
In general, when solving the inequality $$fracx^3+1x > 1$$ you must separate two cases:
- If $x>0$, then the inequality is equivalent to $x^3+1 > x$
- If $x<0$, then the inequality is equivalent to $x^3+1<x$
answered Aug 2 at 13:52
5xum
81.8k382146
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that there is not exception and the following
$$fracx^3+1x>1 iff x^3+1>x$$
holds for $x>0$ while for $x<0$ we have
$$fracx^3+1x>1 iff x^3+1<x$$
To avoid this kind of manipulation and distinction in two cases, we can proceed, as an alternative, as follow
$$fracx^3+1x>1 iff fracx^3+1x-1>0 iff fracx^3+1-xx>0$$
and consider separetely the sign of numerator and denominator.
answered Aug 2 at 13:52
gimusi
63.8k73480
add a comment |Â
up vote
1
down vote
accepted
Note that there is not exception and the following
$$fracx^3+1x>1 iff x^3+1>x$$
holds for $x>0$ while for $x<0$ we have
$$fracx^3+1x>1 iff x^3+1<x$$
To avoid this kind of manipulation and distinction in two cases, we can proceed, as an alternative, as follow
$$fracx^3+1x>1 iff fracx^3+1x-1>0 iff fracx^3+1-xx>0$$
and consider separetely the sign of numerator and denominator.
answered Aug 2 at 13:52
gimusi
63.8k73480
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that there is not exception and the following
$$fracx^3+1x>1 iff x^3+1>x$$
holds for $x>0$ while for $x<0$ we have
$$fracx^3+1x>1 iff x^3+1<x$$
To avoid this kind of manipulation and distinction in two cases, we can proceed, as an alternative, as follow
$$fracx^3+1x>1 iff fracx^3+1x-1>0 iff fracx^3+1-xx>0$$
and consider separetely the sign of numerator and denominator.
answered Aug 2 at 13:52
gimusi
63.8k73480
Note that there is not exception and the following
$$fracx^3+1x>1 iff x^3+1>x$$
holds for $x>0$ while for $x<0$ we have
$$fracx^3+1x>1 iff x^3+1<x$$
To avoid this kind of manipulation and distinction in two cases, we can proceed, as an alternative, as follow
$$fracx^3+1x>1 iff fracx^3+1x-1>0 iff fracx^3+1-xx>0$$
and consider separetely the sign of numerator and denominator.
answered Aug 2 at 13:52
gimusi
63.8k73480
answered Aug 2 at 13:52
gimusi
63.8k73480
answered Aug 2 at 13:52
gimusi
63.8k73480
answered Aug 2 at 13:52
gimusi
63.8k73480
63.8k73480
add a comment |Â
add a comment |Â
up vote
0
down vote
When you multiply the sides by $x$, you don't need to flip the sign if you assume that $x>0$. If you assume that $x<0$, then you do have to flip the sign.
In general, when solving the inequality $$fracx^3+1x > 1$$ you must separate two cases:
- If $x>0$, then the inequality is equivalent to $x^3+1 > x$
- If $x<0$, then the inequality is equivalent to $x^3+1<x$
answered Aug 2 at 13:52
5xum
81.8k382146
add a comment |Â
up vote
0
down vote
When you multiply the sides by $x$, you don't need to flip the sign if you assume that $x>0$. If you assume that $x<0$, then you do have to flip the sign.
In general, when solving the inequality $$fracx^3+1x > 1$$ you must separate two cases:
- If $x>0$, then the inequality is equivalent to $x^3+1 > x$
- If $x<0$, then the inequality is equivalent to $x^3+1<x$
answered Aug 2 at 13:52
5xum
81.8k382146
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When you multiply the sides by $x$, you don't need to flip the sign if you assume that $x>0$. If you assume that $x<0$, then you do have to flip the sign.
In general, when solving the inequality $$fracx^3+1x > 1$$ you must separate two cases:
- If $x>0$, then the inequality is equivalent to $x^3+1 > x$
- If $x<0$, then the inequality is equivalent to $x^3+1<x$
answered Aug 2 at 13:52
5xum
81.8k382146
When you multiply the sides by $x$, you don't need to flip the sign if you assume that $x>0$. If you assume that $x<0$, then you do have to flip the sign.
In general, when solving the inequality $$fracx^3+1x > 1$$ you must separate two cases:
- If $x>0$, then the inequality is equivalent to $x^3+1 > x$
- If $x<0$, then the inequality is equivalent to $x^3+1<x$
answered Aug 2 at 13:52
5xum
81.8k382146
answered Aug 2 at 13:52
5xum
81.8k382146
answered Aug 2 at 13:52
5xum
81.8k382146
answered Aug 2 at 13:52
5xum
81.8k382146
81.8k382146
add a comment |Â
add a comment |Â
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1
When multiplying by an expression whose sign may vary (according to some variable, $x$ for instance), then one must separate the inequality into two different inequalities, one for each case, and solve each of them separately.
– Suzet
Aug 2 at 13:50
@Suzet So does one need to flip the sign in the 2 situations ($(x^3+1)/x>1$ etc) above? If so, why?
– Ethan Chan
Aug 2 at 13:52
See Inequalities.
– Mauro ALLEGRANZA
Aug 2 at 13:53
"And I get why. Because your making one side positive and one side negative." - not sure I understand what you are saying here, but it does not sound correct to me.
– NickD
Aug 2 at 15:40