Mixing
How to do $lvert x - 1 rvert < lvert x-3 rvert$
$begingroup$
Solve for $x$ such that $$lvert x - 1 rvert < lvert x-3 rvert$$
I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work?
algebra-precalculus inequality absolute-value
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
asked Sep 30 '18 at 18:13
user2793618user2793618
997
$endgroup$
add a comment |
$begingroup$
Solve for $x$ such that $$lvert x - 1 rvert < lvert x-3 rvert$$
I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work?
algebra-precalculus inequality absolute-value
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
asked Sep 30 '18 at 18:13
user2793618user2793618
997
$endgroup$
$begingroup$
$|A| < B$ means $-B < A < B$. In your case, $x neq 3$, so we can write it as $left|frac{x-1}{x-3}right| < 1$. Now take it from here.
$endgroup$
– Anurag A
Sep 30 '18 at 18:16
$begingroup$
I'd break it into regions where one or the other signs change. Thus consider $x<1,1<x<3, x>3$.
$endgroup$
– lulu
Sep 30 '18 at 18:16
add a comment |
$begingroup$
Solve for $x$ such that $$lvert x - 1 rvert < lvert x-3 rvert$$
I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work?
algebra-precalculus inequality absolute-value
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
asked Sep 30 '18 at 18:13
user2793618user2793618
997
$endgroup$
Solve for $x$ such that $$lvert x - 1 rvert < lvert x-3 rvert$$
I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work?
algebra-precalculus inequality absolute-value
algebra-precalculus inequality absolute-value
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
asked Sep 30 '18 at 18:13
user2793618user2793618
997
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
asked Sep 30 '18 at 18:13
user2793618user2793618
997
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
edited Jan 23 at 16:10
Martin Sleziak
44.9k10119273
44.9k10119273
asked Sep 30 '18 at 18:13
user2793618user2793618
997
asked Sep 30 '18 at 18:13
user2793618user2793618
997
asked Sep 30 '18 at 18:13
user2793618user2793618
997
997
$begingroup$
$|A| < B$ means $-B < A < B$. In your case, $x neq 3$, so we can write it as $left|frac{x-1}{x-3}right| < 1$. Now take it from here.
$endgroup$
– Anurag A
Sep 30 '18 at 18:16
$begingroup$
I'd break it into regions where one or the other signs change. Thus consider $x<1,1<x<3, x>3$.
$endgroup$
– lulu
Sep 30 '18 at 18:16
add a comment |
$begingroup$
$|A| < B$ means $-B < A < B$. In your case, $x neq 3$, so we can write it as $left|frac{x-1}{x-3}right| < 1$. Now take it from here.
$endgroup$
– Anurag A
Sep 30 '18 at 18:16
$begingroup$
I'd break it into regions where one or the other signs change. Thus consider $x<1,1<x<3, x>3$.
$endgroup$
– lulu
Sep 30 '18 at 18:16
$begingroup$
$|A| < B$ means $-B < A < B$. In your case, $x neq 3$, so we can write it as $left|frac{x-1}{x-3}right| < 1$. Now take it from here.
$endgroup$
– Anurag A
Sep 30 '18 at 18:16
$begingroup$
$|A| < B$ means $-B < A < B$. In your case, $x neq 3$, so we can write it as $left|frac{x-1}{x-3}right| < 1$. Now take it from here.
$endgroup$
– Anurag A
Sep 30 '18 at 18:16
$begingroup$
I'd break it into regions where one or the other signs change. Thus consider $x<1,1<x<3, x>3$.
$endgroup$
– lulu
Sep 30 '18 at 18:16
$begingroup$
I'd break it into regions where one or the other signs change. Thus consider $x<1,1<x<3, x>3$.
$endgroup$
– lulu
Sep 30 '18 at 18:16
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
hint: Square both sides ! and solve $(x-1)^2 < (x-3)^2$ . Can you continue ?
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
$endgroup$
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
add a comment |
$begingroup$
The intuition here is that $x$ is closer to $1$ than to $3$, and this is clearly whenever $x$ is less than $2$.
This intuition can also help to see what is going on in other situations - for example in two or three dimensions when similar expressions are used.
You need to take care, though, that the signs are right. $|x-a|lt|x-b|$ means that $x$ is closer to $a$ than $b$. If it were $|x+a|$ instead you'd be looking at the distance from $-a$. This is one of the easier places to trip up in a hurry when you are learning.
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
$endgroup$
add a comment |
$begingroup$
The question is asking for you to find those values of $x$ that make $|x-1| < |x-3|$ true. For example, $x=0$ makes it work. To make things easier, we can rewrite our inequality in the equivalent inequality
$$ 0 leq |x-3| - |x-1| $$
And now this translates to find value of $x$ for which $f(x) = |x-3|-|x-1|$ is positive. can you find them? Here is a graph:
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
add a comment |
$begingroup$
Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $ge 0$ and $x-3$ can be $<$ or $ge 0$.
Or in other words $x$ can be $<$ or $ge 1$ or $x$ can be $< $ or $ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x ge 3$ both) or redundant (if $xge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases. $x < 1; 1 le x < 3; xge 3$.
Case 1: $x < 1$. Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-ximplies 1 < 3$ ... which is always true. This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2: $1 le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x implies 2x < 4implies x < 2$. This means if $1 le x < 3$ then we know further than $x < 2$ and that $2le x < 3$ is impossible.
Case 3: $x ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3implies -1 < -3$. That is simply impossible. So we know that $x ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$. 2) $x$ could be any value so that $le x < 2$ but $x$ can not be $2 le x < 3$. And 3) $x ge 3$ is impossible.
Combining those three results we get .... $x < 2$. And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.
If $x ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$ and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 < 3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$sqrt {k^2} = |k|$ and if $0 le a$ and $0 le b$ then $a^2 < b^2 iff a < b$.
So $sqrt {(x-1)^2} = |x-1| < |x - 3| = sqrt{(x-2)^2} iff $
$(x-1)^2 < (x-3)^2 iff$
$x^2 -2x + 1 < x^2 - 6x + 9 iff$
$4x < 8 iff$
$x < 2$.
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
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add a comment |
$begingroup$
$y:=x-1$;
Let
A) $y not =0$.
$|y| lt |y-2|$;
$1lt |1-2/y|;$
1) $1lt 1-2/y$
$2/y lt 0$, or
$y <0$.
2)$1 lt 2/y -1$,
$1lt 1/y.$
$0<y <1.$
B) Now check for $y=0$.
Combining
1) $y <0$ and 2) $0<y<1$, with
B) $y=0$:
we have $y <1$, or $x <2$.
(Recall $x= y+1$.)
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
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– AfronPie
Sep 30 '18 at 19:18
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
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– Peter Szilas
Sep 30 '18 at 20:11
add a comment |
$begingroup$
Hint:
Let $x=a+ib$ where $a,b$ are real
$|x-1|=sqrt{(a-1)^2+b^2}>0$
Now if $p>q, $ $$p^2-q ^2=(p-q)(p+q)>0$$ for $p+q>0$
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
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$begingroup$
hint: Square both sides ! and solve $(x-1)^2 < (x-3)^2$ . Can you continue ?
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
$endgroup$
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
add a comment |
$begingroup$
hint: Square both sides ! and solve $(x-1)^2 < (x-3)^2$ . Can you continue ?
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
$endgroup$
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
add a comment |
$begingroup$
hint: Square both sides ! and solve $(x-1)^2 < (x-3)^2$ . Can you continue ?
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
$endgroup$
hint: Square both sides ! and solve $(x-1)^2 < (x-3)^2$ . Can you continue ?
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
answered Sep 30 '18 at 18:16
DeepSeaDeepSea
71.3k54488
71.3k54488
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
add a comment |
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
Why do you square both sides?
$endgroup$
– user2793618
Sep 30 '18 at 18:19
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
$begingroup$
@user2793618 Both sides are non-negative. Squaring gets rid of the absolute value signs and gives an algebraic inequality which may be easier to handle.
$endgroup$
– Mark Bennet
Sep 30 '18 at 18:24
add a comment |
$begingroup$
The intuition here is that $x$ is closer to $1$ than to $3$, and this is clearly whenever $x$ is less than $2$.
This intuition can also help to see what is going on in other situations - for example in two or three dimensions when similar expressions are used.
You need to take care, though, that the signs are right. $|x-a|lt|x-b|$ means that $x$ is closer to $a$ than $b$. If it were $|x+a|$ instead you'd be looking at the distance from $-a$. This is one of the easier places to trip up in a hurry when you are learning.
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
$endgroup$
add a comment |
$begingroup$
The intuition here is that $x$ is closer to $1$ than to $3$, and this is clearly whenever $x$ is less than $2$.
This intuition can also help to see what is going on in other situations - for example in two or three dimensions when similar expressions are used.
You need to take care, though, that the signs are right. $|x-a|lt|x-b|$ means that $x$ is closer to $a$ than $b$. If it were $|x+a|$ instead you'd be looking at the distance from $-a$. This is one of the easier places to trip up in a hurry when you are learning.
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
$endgroup$
add a comment |
$begingroup$
The intuition here is that $x$ is closer to $1$ than to $3$, and this is clearly whenever $x$ is less than $2$.
This intuition can also help to see what is going on in other situations - for example in two or three dimensions when similar expressions are used.
You need to take care, though, that the signs are right. $|x-a|lt|x-b|$ means that $x$ is closer to $a$ than $b$. If it were $|x+a|$ instead you'd be looking at the distance from $-a$. This is one of the easier places to trip up in a hurry when you are learning.
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
$endgroup$
The intuition here is that $x$ is closer to $1$ than to $3$, and this is clearly whenever $x$ is less than $2$.
This intuition can also help to see what is going on in other situations - for example in two or three dimensions when similar expressions are used.
You need to take care, though, that the signs are right. $|x-a|lt|x-b|$ means that $x$ is closer to $a$ than $b$. If it were $|x+a|$ instead you'd be looking at the distance from $-a$. This is one of the easier places to trip up in a hurry when you are learning.
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
answered Sep 30 '18 at 19:24
Mark BennetMark Bennet
81.6k984181
81.6k984181
add a comment |
add a comment |
$begingroup$
The question is asking for you to find those values of $x$ that make $|x-1| < |x-3|$ true. For example, $x=0$ makes it work. To make things easier, we can rewrite our inequality in the equivalent inequality
$$ 0 leq |x-3| - |x-1| $$
And now this translates to find value of $x$ for which $f(x) = |x-3|-|x-1|$ is positive. can you find them? Here is a graph:
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
add a comment |
$begingroup$
The question is asking for you to find those values of $x$ that make $|x-1| < |x-3|$ true. For example, $x=0$ makes it work. To make things easier, we can rewrite our inequality in the equivalent inequality
$$ 0 leq |x-3| - |x-1| $$
And now this translates to find value of $x$ for which $f(x) = |x-3|-|x-1|$ is positive. can you find them? Here is a graph:
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
add a comment |
$begingroup$
The question is asking for you to find those values of $x$ that make $|x-1| < |x-3|$ true. For example, $x=0$ makes it work. To make things easier, we can rewrite our inequality in the equivalent inequality
$$ 0 leq |x-3| - |x-1| $$
And now this translates to find value of $x$ for which $f(x) = |x-3|-|x-1|$ is positive. can you find them? Here is a graph:
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
$endgroup$
The question is asking for you to find those values of $x$ that make $|x-1| < |x-3|$ true. For example, $x=0$ makes it work. To make things easier, we can rewrite our inequality in the equivalent inequality
$$ 0 leq |x-3| - |x-1| $$
And now this translates to find value of $x$ for which $f(x) = |x-3|-|x-1|$ is positive. can you find them? Here is a graph:
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
answered Sep 30 '18 at 18:18
Jimmy SabaterJimmy Sabater
3,013325
3,013325
add a comment |
add a comment |
$begingroup$
Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $ge 0$ and $x-3$ can be $<$ or $ge 0$.
Or in other words $x$ can be $<$ or $ge 1$ or $x$ can be $< $ or $ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x ge 3$ both) or redundant (if $xge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases. $x < 1; 1 le x < 3; xge 3$.
Case 1: $x < 1$. Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-ximplies 1 < 3$ ... which is always true. This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2: $1 le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x implies 2x < 4implies x < 2$. This means if $1 le x < 3$ then we know further than $x < 2$ and that $2le x < 3$ is impossible.
Case 3: $x ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3implies -1 < -3$. That is simply impossible. So we know that $x ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$. 2) $x$ could be any value so that $le x < 2$ but $x$ can not be $2 le x < 3$. And 3) $x ge 3$ is impossible.
Combining those three results we get .... $x < 2$. And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.
If $x ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$ and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 < 3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$sqrt {k^2} = |k|$ and if $0 le a$ and $0 le b$ then $a^2 < b^2 iff a < b$.
So $sqrt {(x-1)^2} = |x-1| < |x - 3| = sqrt{(x-2)^2} iff $
$(x-1)^2 < (x-3)^2 iff$
$x^2 -2x + 1 < x^2 - 6x + 9 iff$
$4x < 8 iff$
$x < 2$.
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $ge 0$ and $x-3$ can be $<$ or $ge 0$.
Or in other words $x$ can be $<$ or $ge 1$ or $x$ can be $< $ or $ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x ge 3$ both) or redundant (if $xge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases. $x < 1; 1 le x < 3; xge 3$.
Case 1: $x < 1$. Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-ximplies 1 < 3$ ... which is always true. This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2: $1 le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x implies 2x < 4implies x < 2$. This means if $1 le x < 3$ then we know further than $x < 2$ and that $2le x < 3$ is impossible.
Case 3: $x ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3implies -1 < -3$. That is simply impossible. So we know that $x ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$. 2) $x$ could be any value so that $le x < 2$ but $x$ can not be $2 le x < 3$. And 3) $x ge 3$ is impossible.
Combining those three results we get .... $x < 2$. And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.
If $x ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$ and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 < 3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$sqrt {k^2} = |k|$ and if $0 le a$ and $0 le b$ then $a^2 < b^2 iff a < b$.
So $sqrt {(x-1)^2} = |x-1| < |x - 3| = sqrt{(x-2)^2} iff $
$(x-1)^2 < (x-3)^2 iff$
$x^2 -2x + 1 < x^2 - 6x + 9 iff$
$4x < 8 iff$
$x < 2$.
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $ge 0$ and $x-3$ can be $<$ or $ge 0$.
Or in other words $x$ can be $<$ or $ge 1$ or $x$ can be $< $ or $ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x ge 3$ both) or redundant (if $xge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases. $x < 1; 1 le x < 3; xge 3$.
Case 1: $x < 1$. Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-ximplies 1 < 3$ ... which is always true. This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2: $1 le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x implies 2x < 4implies x < 2$. This means if $1 le x < 3$ then we know further than $x < 2$ and that $2le x < 3$ is impossible.
Case 3: $x ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3implies -1 < -3$. That is simply impossible. So we know that $x ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$. 2) $x$ could be any value so that $le x < 2$ but $x$ can not be $2 le x < 3$. And 3) $x ge 3$ is impossible.
Combining those three results we get .... $x < 2$. And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.
If $x ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$ and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 < 3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$sqrt {k^2} = |k|$ and if $0 le a$ and $0 le b$ then $a^2 < b^2 iff a < b$.
So $sqrt {(x-1)^2} = |x-1| < |x - 3| = sqrt{(x-2)^2} iff $
$(x-1)^2 < (x-3)^2 iff$
$x^2 -2x + 1 < x^2 - 6x + 9 iff$
$4x < 8 iff$
$x < 2$.
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
$endgroup$
Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $ge 0$ and $x-3$ can be $<$ or $ge 0$.
Or in other words $x$ can be $<$ or $ge 1$ or $x$ can be $< $ or $ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x ge 3$ both) or redundant (if $xge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases. $x < 1; 1 le x < 3; xge 3$.
Case 1: $x < 1$. Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-ximplies 1 < 3$ ... which is always true. This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2: $1 le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x implies 2x < 4implies x < 2$. This means if $1 le x < 3$ then we know further than $x < 2$ and that $2le x < 3$ is impossible.
Case 3: $x ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3implies -1 < -3$. That is simply impossible. So we know that $x ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$. 2) $x$ could be any value so that $le x < 2$ but $x$ can not be $2 le x < 3$. And 3) $x ge 3$ is impossible.
Combining those three results we get .... $x < 2$. And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.
If $x ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$ and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 < 3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$sqrt {k^2} = |k|$ and if $0 le a$ and $0 le b$ then $a^2 < b^2 iff a < b$.
So $sqrt {(x-1)^2} = |x-1| < |x - 3| = sqrt{(x-2)^2} iff $
$(x-1)^2 < (x-3)^2 iff$
$x^2 -2x + 1 < x^2 - 6x + 9 iff$
$4x < 8 iff$
$x < 2$.
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
edited Sep 30 '18 at 18:53
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
answered Sep 30 '18 at 18:35
fleabloodfleablood
72.3k22687
72.3k22687
add a comment |
add a comment |
$begingroup$
$y:=x-1$;
Let
A) $y not =0$.
$|y| lt |y-2|$;
$1lt |1-2/y|;$
1) $1lt 1-2/y$
$2/y lt 0$, or
$y <0$.
2)$1 lt 2/y -1$,
$1lt 1/y.$
$0<y <1.$
B) Now check for $y=0$.
Combining
1) $y <0$ and 2) $0<y<1$, with
B) $y=0$:
we have $y <1$, or $x <2$.
(Recall $x= y+1$.)
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
$endgroup$
– AfronPie
Sep 30 '18 at 19:18
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
$endgroup$
– Peter Szilas
Sep 30 '18 at 20:11
add a comment |
$begingroup$
$y:=x-1$;
Let
A) $y not =0$.
$|y| lt |y-2|$;
$1lt |1-2/y|;$
1) $1lt 1-2/y$
$2/y lt 0$, or
$y <0$.
2)$1 lt 2/y -1$,
$1lt 1/y.$
$0<y <1.$
B) Now check for $y=0$.
Combining
1) $y <0$ and 2) $0<y<1$, with
B) $y=0$:
we have $y <1$, or $x <2$.
(Recall $x= y+1$.)
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
$endgroup$
– AfronPie
Sep 30 '18 at 19:18
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
$endgroup$
– Peter Szilas
Sep 30 '18 at 20:11
add a comment |
$begingroup$
$y:=x-1$;
Let
A) $y not =0$.
$|y| lt |y-2|$;
$1lt |1-2/y|;$
1) $1lt 1-2/y$
$2/y lt 0$, or
$y <0$.
2)$1 lt 2/y -1$,
$1lt 1/y.$
$0<y <1.$
B) Now check for $y=0$.
Combining
1) $y <0$ and 2) $0<y<1$, with
B) $y=0$:
we have $y <1$, or $x <2$.
(Recall $x= y+1$.)
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
$y:=x-1$;
Let
A) $y not =0$.
$|y| lt |y-2|$;
$1lt |1-2/y|;$
1) $1lt 1-2/y$
$2/y lt 0$, or
$y <0$.
2)$1 lt 2/y -1$,
$1lt 1/y.$
$0<y <1.$
B) Now check for $y=0$.
Combining
1) $y <0$ and 2) $0<y<1$, with
B) $y=0$:
we have $y <1$, or $x <2$.
(Recall $x= y+1$.)
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
edited Oct 2 '18 at 21:00
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
answered Sep 30 '18 at 18:46
Peter SzilasPeter Szilas
11.5k2822
11.5k2822
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
$endgroup$
– AfronPie
Sep 30 '18 at 19:18
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
$endgroup$
– Peter Szilas
Sep 30 '18 at 20:11
add a comment |
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
$endgroup$
– AfronPie
Sep 30 '18 at 19:18
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
$endgroup$
– Peter Szilas
Sep 30 '18 at 20:11
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
$endgroup$
– AfronPie
Sep 30 '18 at 19:18
$begingroup$
You made an error. If you revert back to $x$, you get $1<x<2$, while the correct answer is simply $x<2$
$endgroup$
– AfronPie
Sep 30 '18 at 19:18
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
$endgroup$
– Peter Szilas
Sep 30 '18 at 20:11
$begingroup$
Aaron.You refer 2) which yields 1<x<2.But there is 1)which yields y<0 or x <1 , and include y=0 or x=1 ,then you get x <2.Your thoughts?
$endgroup$
– Peter Szilas
Sep 30 '18 at 20:11
add a comment |
$begingroup$
Hint:
Let $x=a+ib$ where $a,b$ are real
$|x-1|=sqrt{(a-1)^2+b^2}>0$
Now if $p>q, $ $$p^2-q ^2=(p-q)(p+q)>0$$ for $p+q>0$
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Let $x=a+ib$ where $a,b$ are real
$|x-1|=sqrt{(a-1)^2+b^2}>0$
Now if $p>q, $ $$p^2-q ^2=(p-q)(p+q)>0$$ for $p+q>0$
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Let $x=a+ib$ where $a,b$ are real
$|x-1|=sqrt{(a-1)^2+b^2}>0$
Now if $p>q, $ $$p^2-q ^2=(p-q)(p+q)>0$$ for $p+q>0$
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
Hint:
Let $x=a+ib$ where $a,b$ are real
$|x-1|=sqrt{(a-1)^2+b^2}>0$
Now if $p>q, $ $$p^2-q ^2=(p-q)(p+q)>0$$ for $p+q>0$
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
answered Sep 30 '18 at 18:23
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
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$begingroup$
$|A| < B$ means $-B < A < B$. In your case, $x neq 3$, so we can write it as $left|frac{x-1}{x-3}right| < 1$. Now take it from here.
$endgroup$
– Anurag A
Sep 30 '18 at 18:16
$begingroup$
I'd break it into regions where one or the other signs change. Thus consider $x<1,1<x<3, x>3$.
$endgroup$
– lulu
Sep 30 '18 at 18:16