What does it mean to have an absolute value equal an absolute value?
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I have no problem reading absolute value equations such as $|x -2| = 2$.
I know this means that the distance of some real number is $2$ away from the origin. Because the origin splits the number line into a negative side and positive side then the numbers inside the absolute value symbol will be $2$ and $-2$, since those are the only two numbers $2$ units away from the origin. Then, it's just a matter of finding the values of $x$ which will give $2$ and $-2$ inside the absolute value.
Therefore, $|x - 2| = 2$ which is
$x - 2 = 2$
or
$x - 2 = -2$
And the solutions are $0, 4$
But when I see $|3x - 1| = |x + 5|$, I have no idea know what this means. I know how to solve it, but I don't know how this relates to the distance from the origin or how to interpret this on a number line. My initial interpretation is to say, "the absolute value of some unknown number is the absolute value of some unknown number," but that doesn't tell me the distance from $0$.
My Algebra textbook gave the following definition:
If $|u| = |v|$, then $u = v$ or $u = -v$.
But I can't really tell why this is the case.
absolute-value
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I have no problem reading absolute value equations such as $|x -2| = 2$.
I know this means that the distance of some real number is $2$ away from the origin. Because the origin splits the number line into a negative side and positive side then the numbers inside the absolute value symbol will be $2$ and $-2$, since those are the only two numbers $2$ units away from the origin. Then, it's just a matter of finding the values of $x$ which will give $2$ and $-2$ inside the absolute value.
Therefore, $|x - 2| = 2$ which is
$x - 2 = 2$
or
$x - 2 = -2$
And the solutions are $0, 4$
But when I see $|3x - 1| = |x + 5|$, I have no idea know what this means. I know how to solve it, but I don't know how this relates to the distance from the origin or how to interpret this on a number line. My initial interpretation is to say, "the absolute value of some unknown number is the absolute value of some unknown number," but that doesn't tell me the distance from $0$.
My Algebra textbook gave the following definition:
If $|u| = |v|$, then $u = v$ or $u = -v$.
But I can't really tell why this is the case.
absolute-value
2
It might help to plot the equation $y = |3x -1| = 3|x-1/3|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots.
â Andy Walls
Sep 10 at 23:25
Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two.
â Bernard Massé
Sep 10 at 23:43
1
The algebraic $definition$ of $|x|$ is that $|x|=x$ if $xgeq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $x,-x.$ So if $|x|=|y|$ then one member of $x,-x $ is equal to one member of $y,-y,$ which implies $(x=ylor x=-ylor -x=ylor -x=-y)$, which is equivalent to $(x=y lor x=- y)$. Conversely , if $x=pm y,$ then the non-negative member of $x,-x,$ which is $|x|, $ must equal the non-negative member of $y,-y,$ which is $|y|.$
â DanielWainfleet
Sep 11 at 0:10
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I have no problem reading absolute value equations such as $|x -2| = 2$.
I know this means that the distance of some real number is $2$ away from the origin. Because the origin splits the number line into a negative side and positive side then the numbers inside the absolute value symbol will be $2$ and $-2$, since those are the only two numbers $2$ units away from the origin. Then, it's just a matter of finding the values of $x$ which will give $2$ and $-2$ inside the absolute value.
Therefore, $|x - 2| = 2$ which is
$x - 2 = 2$
or
$x - 2 = -2$
And the solutions are $0, 4$
But when I see $|3x - 1| = |x + 5|$, I have no idea know what this means. I know how to solve it, but I don't know how this relates to the distance from the origin or how to interpret this on a number line. My initial interpretation is to say, "the absolute value of some unknown number is the absolute value of some unknown number," but that doesn't tell me the distance from $0$.
My Algebra textbook gave the following definition:
If $|u| = |v|$, then $u = v$ or $u = -v$.
But I can't really tell why this is the case.
absolute-value
I have no problem reading absolute value equations such as $|x -2| = 2$.
I know this means that the distance of some real number is $2$ away from the origin. Because the origin splits the number line into a negative side and positive side then the numbers inside the absolute value symbol will be $2$ and $-2$, since those are the only two numbers $2$ units away from the origin. Then, it's just a matter of finding the values of $x$ which will give $2$ and $-2$ inside the absolute value.
Therefore, $|x - 2| = 2$ which is
$x - 2 = 2$
or
$x - 2 = -2$
And the solutions are $0, 4$
But when I see $|3x - 1| = |x + 5|$, I have no idea know what this means. I know how to solve it, but I don't know how this relates to the distance from the origin or how to interpret this on a number line. My initial interpretation is to say, "the absolute value of some unknown number is the absolute value of some unknown number," but that doesn't tell me the distance from $0$.
My Algebra textbook gave the following definition:
If $|u| = |v|$, then $u = v$ or $u = -v$.
But I can't really tell why this is the case.
absolute-value
absolute-value
edited Sep 11 at 6:44
Carmeister
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2,3291720
asked Sep 10 at 23:15
Slecker
1358
1358
2
It might help to plot the equation $y = |3x -1| = 3|x-1/3|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots.
â Andy Walls
Sep 10 at 23:25
Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two.
â Bernard Massé
Sep 10 at 23:43
1
The algebraic $definition$ of $|x|$ is that $|x|=x$ if $xgeq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $x,-x.$ So if $|x|=|y|$ then one member of $x,-x $ is equal to one member of $y,-y,$ which implies $(x=ylor x=-ylor -x=ylor -x=-y)$, which is equivalent to $(x=y lor x=- y)$. Conversely , if $x=pm y,$ then the non-negative member of $x,-x,$ which is $|x|, $ must equal the non-negative member of $y,-y,$ which is $|y|.$
â DanielWainfleet
Sep 11 at 0:10
add a comment |Â
2
It might help to plot the equation $y = |3x -1| = 3|x-1/3|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots.
â Andy Walls
Sep 10 at 23:25
Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two.
â Bernard Massé
Sep 10 at 23:43
1
The algebraic $definition$ of $|x|$ is that $|x|=x$ if $xgeq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $x,-x.$ So if $|x|=|y|$ then one member of $x,-x $ is equal to one member of $y,-y,$ which implies $(x=ylor x=-ylor -x=ylor -x=-y)$, which is equivalent to $(x=y lor x=- y)$. Conversely , if $x=pm y,$ then the non-negative member of $x,-x,$ which is $|x|, $ must equal the non-negative member of $y,-y,$ which is $|y|.$
â DanielWainfleet
Sep 11 at 0:10
2
2
It might help to plot the equation $y = |3x -1| = 3|x-1/3|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots.
â Andy Walls
Sep 10 at 23:25
It might help to plot the equation $y = |3x -1| = 3|x-1/3|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots.
â Andy Walls
Sep 10 at 23:25
Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two.
â Bernard Massé
Sep 10 at 23:43
Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two.
â Bernard Massé
Sep 10 at 23:43
1
1
The algebraic $definition$ of $|x|$ is that $|x|=x$ if $xgeq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $x,-x.$ So if $|x|=|y|$ then one member of $x,-x $ is equal to one member of $y,-y,$ which implies $(x=ylor x=-ylor -x=ylor -x=-y)$, which is equivalent to $(x=y lor x=- y)$. Conversely , if $x=pm y,$ then the non-negative member of $x,-x,$ which is $|x|, $ must equal the non-negative member of $y,-y,$ which is $|y|.$
â DanielWainfleet
Sep 11 at 0:10
The algebraic $definition$ of $|x|$ is that $|x|=x$ if $xgeq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $x,-x.$ So if $|x|=|y|$ then one member of $x,-x $ is equal to one member of $y,-y,$ which implies $(x=ylor x=-ylor -x=ylor -x=-y)$, which is equivalent to $(x=y lor x=- y)$. Conversely , if $x=pm y,$ then the non-negative member of $x,-x,$ which is $|x|, $ must equal the non-negative member of $y,-y,$ which is $|y|.$
â DanielWainfleet
Sep 11 at 0:10
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5 Answers
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Your interpretation is good.
Any value $v$ is at distance $|v|$ from origin. Sometimes we are given the distance and are asked to find original value. When something ($in mathbbR$) is at distance $|w|$ from origin, it has a value either $w$ or $-w$.
As you said, $|x-2|=2$ means that $(x-2)$ is at distance $2$ from origin.
The same goes for the example that confuses you;
$$|3x-1|=|x+5|$$
means that $(3x-1)$ is at distance $|x+5|$ from origin.
And what can we conclude from this?
That the value of $(3x-1)$ is either $(x+5)$ or $-(x+5)$, and that is what your textbook says using $u$ and $v$.
[Also, you can flip it and say that $(x+5)$ is at distance $|3x-1|$ from origin, and those are the redundant cases you see mentioned in other answers.]
And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin.
So $$|u|=|v|$$ means that $u$ and $v$ are equaly far from origin. How far specificaly? Exactly $|u|$ (or $|v|$, because they are equal).
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
1
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
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Maybe trying to answer the following question can help you..
In real line, when do two points $u$ and $v$ have the same absolute value (the same distance away from the origin)?
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If youâÂÂre having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as $3t - 1$ and the other one as $t + 5$ (this would imply that at the moment of time $t=0$, presumably when you started observing the situation, one was at $-1$ and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).
Now youâÂÂre asked the question, at what time one was as far away from the origin as the other one? (Maybe youâÂÂre a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) ThatâÂÂs what your equation $$|3t - 1| = |t + 5|$$ says. Intuitively, because the first objectâÂÂs velocity is greater but at $t=0$ itâÂÂs to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation $3t - 1 = -(t + 5)$ corresponds to the former case (coordinates are $-4$ and $4$ respectively), and $3t-1 = t+5$ to the latter (both objects at $8$).
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
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if $|u| = |v|$ there are actually four possibilities, but they are two redundant pairs.
The four possibilities are.
A) $u ge 0; vge 0$ and therefore $u = |u|; v=|v|=|u| = u$ and RESULT 1) $u = v$.
B) $u < 0; v < 0$ and therefore $u = -|u|; v = -|v|=-|u| = u$ and RESULT 1) $u = v$. (that's redudant.)
C) $uge 0; v < 0$ and therefore $u = |u|; v = -|v|=-|u|=-u$ and RESULT 2) $u = -v$
D) $u < 0; v ge 0$ and therefore $u = -|u|; v = |v| = |u| = -u$ and RESULT 2) $u = -v$. (that's reducant).
Now if $|3x -1| = |x+5|$ we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.
Let's do it to see what happens and see if we can learn from it:
A) $3x - 1 ge 0$ and $x + 5 ge 0$ and therefore $3x-1 = x + 5$.
In other words $3x ge 1$ and $x ge -5$ and $2x = 6$
In other words $x ge frac 13$ and $x ge -5$ and $x = 3$.
Or in other words $x = 3$.
B) $3x -1 < 0$ and $x + 5 < 0$ and therefore $3x -1 =x+5$.
Or $x < frac 13$ and $x < -5$ and $x =3$.
That's a contradiction. Notice there was no reason to consider the two case whether $3x -1$ and $x + 5$ are greater or less than $0$. That was just a waste of time. It would have been just as well to only consider that $3x - 1= x+5implies x = 3$. That's all we had to do.
C) $3x - 1< 0$ and $x + 5 ge 0$ and therefore $3x -1 = -x -5$
Or $x < frac 13$ and $x ge -5$ and $4x = -4implies x = -1$. So $x =-1$ is possible.
D) $3x -1 ge 0$ and $x + 5 < 0$ and therefore $3x-1 = -x -5$
Or $x ge frac 13$ and $x < -5$ and $4x = -4implies x = -1$. That's a contradiction.
But again we didn't have to do both C) and D). That was redundant.
It's enough to know that if $|3x -1| = |x+5|$ then either $3x - 1 = x+5$ (and we don't care if they are both positive or both negative-- we can figure that out later) or $3x - 1 = -x -5$ (ditto).
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If $|u| = |v|$, then $u = v$ or $u = -v$.
An easy way to understand why this is true is to square both sides of the equation. The statement $|u|=|v|$ is the same statement as $u^2=v^2$ (since $sqrtx^2=|x|$ for any $x$). Rearrange this to $u^2-v^2=0$, then factor the LHS to obtain
$(u-v)(u+v)=0$.
Conclude that the original statement $|u|=|v|$ is equivalent to the statement $u-v=0$ or $u+v=0$.
You can similarly solve $|3x-1|=|x+5|$ by squaring both sides.
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5 Answers
5
active
oldest
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Your interpretation is good.
Any value $v$ is at distance $|v|$ from origin. Sometimes we are given the distance and are asked to find original value. When something ($in mathbbR$) is at distance $|w|$ from origin, it has a value either $w$ or $-w$.
As you said, $|x-2|=2$ means that $(x-2)$ is at distance $2$ from origin.
The same goes for the example that confuses you;
$$|3x-1|=|x+5|$$
means that $(3x-1)$ is at distance $|x+5|$ from origin.
And what can we conclude from this?
That the value of $(3x-1)$ is either $(x+5)$ or $-(x+5)$, and that is what your textbook says using $u$ and $v$.
[Also, you can flip it and say that $(x+5)$ is at distance $|3x-1|$ from origin, and those are the redundant cases you see mentioned in other answers.]
And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin.
So $$|u|=|v|$$ means that $u$ and $v$ are equaly far from origin. How far specificaly? Exactly $|u|$ (or $|v|$, because they are equal).
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
1
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
add a comment |Â
up vote
4
down vote
accepted
Your interpretation is good.
Any value $v$ is at distance $|v|$ from origin. Sometimes we are given the distance and are asked to find original value. When something ($in mathbbR$) is at distance $|w|$ from origin, it has a value either $w$ or $-w$.
As you said, $|x-2|=2$ means that $(x-2)$ is at distance $2$ from origin.
The same goes for the example that confuses you;
$$|3x-1|=|x+5|$$
means that $(3x-1)$ is at distance $|x+5|$ from origin.
And what can we conclude from this?
That the value of $(3x-1)$ is either $(x+5)$ or $-(x+5)$, and that is what your textbook says using $u$ and $v$.
[Also, you can flip it and say that $(x+5)$ is at distance $|3x-1|$ from origin, and those are the redundant cases you see mentioned in other answers.]
And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin.
So $$|u|=|v|$$ means that $u$ and $v$ are equaly far from origin. How far specificaly? Exactly $|u|$ (or $|v|$, because they are equal).
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
1
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Your interpretation is good.
Any value $v$ is at distance $|v|$ from origin. Sometimes we are given the distance and are asked to find original value. When something ($in mathbbR$) is at distance $|w|$ from origin, it has a value either $w$ or $-w$.
As you said, $|x-2|=2$ means that $(x-2)$ is at distance $2$ from origin.
The same goes for the example that confuses you;
$$|3x-1|=|x+5|$$
means that $(3x-1)$ is at distance $|x+5|$ from origin.
And what can we conclude from this?
That the value of $(3x-1)$ is either $(x+5)$ or $-(x+5)$, and that is what your textbook says using $u$ and $v$.
[Also, you can flip it and say that $(x+5)$ is at distance $|3x-1|$ from origin, and those are the redundant cases you see mentioned in other answers.]
And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin.
So $$|u|=|v|$$ means that $u$ and $v$ are equaly far from origin. How far specificaly? Exactly $|u|$ (or $|v|$, because they are equal).
Your interpretation is good.
Any value $v$ is at distance $|v|$ from origin. Sometimes we are given the distance and are asked to find original value. When something ($in mathbbR$) is at distance $|w|$ from origin, it has a value either $w$ or $-w$.
As you said, $|x-2|=2$ means that $(x-2)$ is at distance $2$ from origin.
The same goes for the example that confuses you;
$$|3x-1|=|x+5|$$
means that $(3x-1)$ is at distance $|x+5|$ from origin.
And what can we conclude from this?
That the value of $(3x-1)$ is either $(x+5)$ or $-(x+5)$, and that is what your textbook says using $u$ and $v$.
[Also, you can flip it and say that $(x+5)$ is at distance $|3x-1|$ from origin, and those are the redundant cases you see mentioned in other answers.]
And to directly refer to your title question; Having two absolute values equal means, in terms of distance from origin, that they are both equaly far from origin.
So $$|u|=|v|$$ means that $u$ and $v$ are equaly far from origin. How far specificaly? Exactly $|u|$ (or $|v|$, because they are equal).
edited Sep 11 at 0:56
answered Sep 10 at 23:45
Sandro LovniÃÂki
22615
22615
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
1
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
add a comment |Â
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
1
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
Thanks for your answer, it's cleared up most of my confusion, just one more stumbling block. The solution set, 0, 4, for |x - 2| = 2 makes sense to me since, when plugged back in, they give |2| = 2 and |-2| = 2 which are the absolute values that are two away from the origin. The solution set for |3x - 1| = |x + 5| is -1, 3. When I plug these values back into the equation I get |8| = |8| and |-4| = |4|. I don't get how there could be two different absolute values. Does this mean that the points -4, 4, and 8 are the absolute values, or the points filled on the number line?
â Slecker
Sep 11 at 1:48
1
1
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Slecker As opposed to the simpler equation of the form $|u| = 2$ that asks you âÂÂwhich values of $x$ cause $u$ to be 2 away from the originâÂÂ, you have an equation $|u|=|v|$ and the question is âÂÂwhich values of $x$ cause $u$ to be as far away from the origin as $v$ isâÂÂ. $x=-1$ causes $u=-4$ and $v=4$ which fits (both are 4 away from the origin, on different sides), and $x=3$ results in $u=v=8$ (both are obviously 8 away from the origin on the same side, on the same exact point).
â Roman Odaisky
Sep 11 at 2:48
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
@Roman Aha! Thank you very much for that last bit of insight: âÂÂ|u|=|v| and the question is 'which values of x cause u to be as far away from the origin as v is.'" It helps me tremendously fit the last conceptual puzzle piece in my head!
â Slecker
Sep 11 at 3:04
add a comment |Â
up vote
3
down vote
Maybe trying to answer the following question can help you..
In real line, when do two points $u$ and $v$ have the same absolute value (the same distance away from the origin)?
add a comment |Â
up vote
3
down vote
Maybe trying to answer the following question can help you..
In real line, when do two points $u$ and $v$ have the same absolute value (the same distance away from the origin)?
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Maybe trying to answer the following question can help you..
In real line, when do two points $u$ and $v$ have the same absolute value (the same distance away from the origin)?
Maybe trying to answer the following question can help you..
In real line, when do two points $u$ and $v$ have the same absolute value (the same distance away from the origin)?
answered Sep 10 at 23:40
rldias
2,5731422
2,5731422
add a comment |Â
add a comment |Â
up vote
1
down vote
If youâÂÂre having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as $3t - 1$ and the other one as $t + 5$ (this would imply that at the moment of time $t=0$, presumably when you started observing the situation, one was at $-1$ and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).
Now youâÂÂre asked the question, at what time one was as far away from the origin as the other one? (Maybe youâÂÂre a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) ThatâÂÂs what your equation $$|3t - 1| = |t + 5|$$ says. Intuitively, because the first objectâÂÂs velocity is greater but at $t=0$ itâÂÂs to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation $3t - 1 = -(t + 5)$ corresponds to the former case (coordinates are $-4$ and $4$ respectively), and $3t-1 = t+5$ to the latter (both objects at $8$).
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
add a comment |Â
up vote
1
down vote
If youâÂÂre having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as $3t - 1$ and the other one as $t + 5$ (this would imply that at the moment of time $t=0$, presumably when you started observing the situation, one was at $-1$ and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).
Now youâÂÂre asked the question, at what time one was as far away from the origin as the other one? (Maybe youâÂÂre a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) ThatâÂÂs what your equation $$|3t - 1| = |t + 5|$$ says. Intuitively, because the first objectâÂÂs velocity is greater but at $t=0$ itâÂÂs to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation $3t - 1 = -(t + 5)$ corresponds to the former case (coordinates are $-4$ and $4$ respectively), and $3t-1 = t+5$ to the latter (both objects at $8$).
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If youâÂÂre having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as $3t - 1$ and the other one as $t + 5$ (this would imply that at the moment of time $t=0$, presumably when you started observing the situation, one was at $-1$ and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).
Now youâÂÂre asked the question, at what time one was as far away from the origin as the other one? (Maybe youâÂÂre a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) ThatâÂÂs what your equation $$|3t - 1| = |t + 5|$$ says. Intuitively, because the first objectâÂÂs velocity is greater but at $t=0$ itâÂÂs to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation $3t - 1 = -(t + 5)$ corresponds to the former case (coordinates are $-4$ and $4$ respectively), and $3t-1 = t+5$ to the latter (both objects at $8$).
If youâÂÂre having trouble imagining what the equation might represent, consider two objects moving along the number line. Let the coordinate of one of those change with time as $3t - 1$ and the other one as $t + 5$ (this would imply that at the moment of time $t=0$, presumably when you started observing the situation, one was at $-1$ and was moving at a constant speed of 3 units of distance per unit of time in the positive direction, while the second one was at the point 5 and its speed was 1).
Now youâÂÂre asked the question, at what time one was as far away from the origin as the other one? (Maybe youâÂÂre a spy trying to determine the timestamp of some event, but all you know is that the signal from two ships or planes or whatever was equally strong at that point.) ThatâÂÂs what your equation $$|3t - 1| = |t + 5|$$ says. Intuitively, because the first objectâÂÂs velocity is greater but at $t=0$ itâÂÂs to the left of the second object, this would happen two times: once in the past, when the origin was exactly between the objects, and once in the future, when one object overtakes the other. The equation $3t - 1 = -(t + 5)$ corresponds to the former case (coordinates are $-4$ and $4$ respectively), and $3t-1 = t+5$ to the latter (both objects at $8$).
answered Sep 11 at 2:36
Roman Odaisky
1915
1915
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
add a comment |Â
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
That's actually a good way of conceptualizing it, and I never could think of a way to apply absolute values to real life, so your illustration with the spy is quite interesting and helpful.
â Slecker
Sep 11 at 3:19
add a comment |Â
up vote
0
down vote
if $|u| = |v|$ there are actually four possibilities, but they are two redundant pairs.
The four possibilities are.
A) $u ge 0; vge 0$ and therefore $u = |u|; v=|v|=|u| = u$ and RESULT 1) $u = v$.
B) $u < 0; v < 0$ and therefore $u = -|u|; v = -|v|=-|u| = u$ and RESULT 1) $u = v$. (that's redudant.)
C) $uge 0; v < 0$ and therefore $u = |u|; v = -|v|=-|u|=-u$ and RESULT 2) $u = -v$
D) $u < 0; v ge 0$ and therefore $u = -|u|; v = |v| = |u| = -u$ and RESULT 2) $u = -v$. (that's reducant).
Now if $|3x -1| = |x+5|$ we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.
Let's do it to see what happens and see if we can learn from it:
A) $3x - 1 ge 0$ and $x + 5 ge 0$ and therefore $3x-1 = x + 5$.
In other words $3x ge 1$ and $x ge -5$ and $2x = 6$
In other words $x ge frac 13$ and $x ge -5$ and $x = 3$.
Or in other words $x = 3$.
B) $3x -1 < 0$ and $x + 5 < 0$ and therefore $3x -1 =x+5$.
Or $x < frac 13$ and $x < -5$ and $x =3$.
That's a contradiction. Notice there was no reason to consider the two case whether $3x -1$ and $x + 5$ are greater or less than $0$. That was just a waste of time. It would have been just as well to only consider that $3x - 1= x+5implies x = 3$. That's all we had to do.
C) $3x - 1< 0$ and $x + 5 ge 0$ and therefore $3x -1 = -x -5$
Or $x < frac 13$ and $x ge -5$ and $4x = -4implies x = -1$. So $x =-1$ is possible.
D) $3x -1 ge 0$ and $x + 5 < 0$ and therefore $3x-1 = -x -5$
Or $x ge frac 13$ and $x < -5$ and $4x = -4implies x = -1$. That's a contradiction.
But again we didn't have to do both C) and D). That was redundant.
It's enough to know that if $|3x -1| = |x+5|$ then either $3x - 1 = x+5$ (and we don't care if they are both positive or both negative-- we can figure that out later) or $3x - 1 = -x -5$ (ditto).
add a comment |Â
up vote
0
down vote
if $|u| = |v|$ there are actually four possibilities, but they are two redundant pairs.
The four possibilities are.
A) $u ge 0; vge 0$ and therefore $u = |u|; v=|v|=|u| = u$ and RESULT 1) $u = v$.
B) $u < 0; v < 0$ and therefore $u = -|u|; v = -|v|=-|u| = u$ and RESULT 1) $u = v$. (that's redudant.)
C) $uge 0; v < 0$ and therefore $u = |u|; v = -|v|=-|u|=-u$ and RESULT 2) $u = -v$
D) $u < 0; v ge 0$ and therefore $u = -|u|; v = |v| = |u| = -u$ and RESULT 2) $u = -v$. (that's reducant).
Now if $|3x -1| = |x+5|$ we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.
Let's do it to see what happens and see if we can learn from it:
A) $3x - 1 ge 0$ and $x + 5 ge 0$ and therefore $3x-1 = x + 5$.
In other words $3x ge 1$ and $x ge -5$ and $2x = 6$
In other words $x ge frac 13$ and $x ge -5$ and $x = 3$.
Or in other words $x = 3$.
B) $3x -1 < 0$ and $x + 5 < 0$ and therefore $3x -1 =x+5$.
Or $x < frac 13$ and $x < -5$ and $x =3$.
That's a contradiction. Notice there was no reason to consider the two case whether $3x -1$ and $x + 5$ are greater or less than $0$. That was just a waste of time. It would have been just as well to only consider that $3x - 1= x+5implies x = 3$. That's all we had to do.
C) $3x - 1< 0$ and $x + 5 ge 0$ and therefore $3x -1 = -x -5$
Or $x < frac 13$ and $x ge -5$ and $4x = -4implies x = -1$. So $x =-1$ is possible.
D) $3x -1 ge 0$ and $x + 5 < 0$ and therefore $3x-1 = -x -5$
Or $x ge frac 13$ and $x < -5$ and $4x = -4implies x = -1$. That's a contradiction.
But again we didn't have to do both C) and D). That was redundant.
It's enough to know that if $|3x -1| = |x+5|$ then either $3x - 1 = x+5$ (and we don't care if they are both positive or both negative-- we can figure that out later) or $3x - 1 = -x -5$ (ditto).
add a comment |Â
up vote
0
down vote
up vote
0
down vote
if $|u| = |v|$ there are actually four possibilities, but they are two redundant pairs.
The four possibilities are.
A) $u ge 0; vge 0$ and therefore $u = |u|; v=|v|=|u| = u$ and RESULT 1) $u = v$.
B) $u < 0; v < 0$ and therefore $u = -|u|; v = -|v|=-|u| = u$ and RESULT 1) $u = v$. (that's redudant.)
C) $uge 0; v < 0$ and therefore $u = |u|; v = -|v|=-|u|=-u$ and RESULT 2) $u = -v$
D) $u < 0; v ge 0$ and therefore $u = -|u|; v = |v| = |u| = -u$ and RESULT 2) $u = -v$. (that's reducant).
Now if $|3x -1| = |x+5|$ we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.
Let's do it to see what happens and see if we can learn from it:
A) $3x - 1 ge 0$ and $x + 5 ge 0$ and therefore $3x-1 = x + 5$.
In other words $3x ge 1$ and $x ge -5$ and $2x = 6$
In other words $x ge frac 13$ and $x ge -5$ and $x = 3$.
Or in other words $x = 3$.
B) $3x -1 < 0$ and $x + 5 < 0$ and therefore $3x -1 =x+5$.
Or $x < frac 13$ and $x < -5$ and $x =3$.
That's a contradiction. Notice there was no reason to consider the two case whether $3x -1$ and $x + 5$ are greater or less than $0$. That was just a waste of time. It would have been just as well to only consider that $3x - 1= x+5implies x = 3$. That's all we had to do.
C) $3x - 1< 0$ and $x + 5 ge 0$ and therefore $3x -1 = -x -5$
Or $x < frac 13$ and $x ge -5$ and $4x = -4implies x = -1$. So $x =-1$ is possible.
D) $3x -1 ge 0$ and $x + 5 < 0$ and therefore $3x-1 = -x -5$
Or $x ge frac 13$ and $x < -5$ and $4x = -4implies x = -1$. That's a contradiction.
But again we didn't have to do both C) and D). That was redundant.
It's enough to know that if $|3x -1| = |x+5|$ then either $3x - 1 = x+5$ (and we don't care if they are both positive or both negative-- we can figure that out later) or $3x - 1 = -x -5$ (ditto).
if $|u| = |v|$ there are actually four possibilities, but they are two redundant pairs.
The four possibilities are.
A) $u ge 0; vge 0$ and therefore $u = |u|; v=|v|=|u| = u$ and RESULT 1) $u = v$.
B) $u < 0; v < 0$ and therefore $u = -|u|; v = -|v|=-|u| = u$ and RESULT 1) $u = v$. (that's redudant.)
C) $uge 0; v < 0$ and therefore $u = |u|; v = -|v|=-|u|=-u$ and RESULT 2) $u = -v$
D) $u < 0; v ge 0$ and therefore $u = -|u|; v = |v| = |u| = -u$ and RESULT 2) $u = -v$. (that's reducant).
Now if $|3x -1| = |x+5|$ we could solve by doing all four cases but that is unnecessary as we will get redundant and contradictory results.
Let's do it to see what happens and see if we can learn from it:
A) $3x - 1 ge 0$ and $x + 5 ge 0$ and therefore $3x-1 = x + 5$.
In other words $3x ge 1$ and $x ge -5$ and $2x = 6$
In other words $x ge frac 13$ and $x ge -5$ and $x = 3$.
Or in other words $x = 3$.
B) $3x -1 < 0$ and $x + 5 < 0$ and therefore $3x -1 =x+5$.
Or $x < frac 13$ and $x < -5$ and $x =3$.
That's a contradiction. Notice there was no reason to consider the two case whether $3x -1$ and $x + 5$ are greater or less than $0$. That was just a waste of time. It would have been just as well to only consider that $3x - 1= x+5implies x = 3$. That's all we had to do.
C) $3x - 1< 0$ and $x + 5 ge 0$ and therefore $3x -1 = -x -5$
Or $x < frac 13$ and $x ge -5$ and $4x = -4implies x = -1$. So $x =-1$ is possible.
D) $3x -1 ge 0$ and $x + 5 < 0$ and therefore $3x-1 = -x -5$
Or $x ge frac 13$ and $x < -5$ and $4x = -4implies x = -1$. That's a contradiction.
But again we didn't have to do both C) and D). That was redundant.
It's enough to know that if $|3x -1| = |x+5|$ then either $3x - 1 = x+5$ (and we don't care if they are both positive or both negative-- we can figure that out later) or $3x - 1 = -x -5$ (ditto).
answered Sep 11 at 0:11
fleablood
62.4k22678
62.4k22678
add a comment |Â
add a comment |Â
up vote
0
down vote
If $|u| = |v|$, then $u = v$ or $u = -v$.
An easy way to understand why this is true is to square both sides of the equation. The statement $|u|=|v|$ is the same statement as $u^2=v^2$ (since $sqrtx^2=|x|$ for any $x$). Rearrange this to $u^2-v^2=0$, then factor the LHS to obtain
$(u-v)(u+v)=0$.
Conclude that the original statement $|u|=|v|$ is equivalent to the statement $u-v=0$ or $u+v=0$.
You can similarly solve $|3x-1|=|x+5|$ by squaring both sides.
add a comment |Â
up vote
0
down vote
If $|u| = |v|$, then $u = v$ or $u = -v$.
An easy way to understand why this is true is to square both sides of the equation. The statement $|u|=|v|$ is the same statement as $u^2=v^2$ (since $sqrtx^2=|x|$ for any $x$). Rearrange this to $u^2-v^2=0$, then factor the LHS to obtain
$(u-v)(u+v)=0$.
Conclude that the original statement $|u|=|v|$ is equivalent to the statement $u-v=0$ or $u+v=0$.
You can similarly solve $|3x-1|=|x+5|$ by squaring both sides.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $|u| = |v|$, then $u = v$ or $u = -v$.
An easy way to understand why this is true is to square both sides of the equation. The statement $|u|=|v|$ is the same statement as $u^2=v^2$ (since $sqrtx^2=|x|$ for any $x$). Rearrange this to $u^2-v^2=0$, then factor the LHS to obtain
$(u-v)(u+v)=0$.
Conclude that the original statement $|u|=|v|$ is equivalent to the statement $u-v=0$ or $u+v=0$.
You can similarly solve $|3x-1|=|x+5|$ by squaring both sides.
If $|u| = |v|$, then $u = v$ or $u = -v$.
An easy way to understand why this is true is to square both sides of the equation. The statement $|u|=|v|$ is the same statement as $u^2=v^2$ (since $sqrtx^2=|x|$ for any $x$). Rearrange this to $u^2-v^2=0$, then factor the LHS to obtain
$(u-v)(u+v)=0$.
Conclude that the original statement $|u|=|v|$ is equivalent to the statement $u-v=0$ or $u+v=0$.
You can similarly solve $|3x-1|=|x+5|$ by squaring both sides.
answered Sep 11 at 5:32
grand_chat
18.6k11122
18.6k11122
add a comment |Â
add a comment |Â
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2
It might help to plot the equation $y = |3x -1| = 3|x-1/3|$, and then plot the equation $y=|x+5|$ on the same graph. Your equation is valid at the intersection point(s) of those two plots.
â Andy Walls
Sep 10 at 23:25
Maybe this will help: $|u|=|v|$ is equivalent to $u=v$ or $u=-v$ or $-u=v$ or $-u=-v$. But of course, the last two of these are redundant being equivalent to the first two.
â Bernard Massé
Sep 10 at 23:43
1
The algebraic $definition$ of $|x|$ is that $|x|=x$ if $xgeq 0$ and $|x|=-x$ if $x<0$. In all cases, $|x|$ is the non-negative member of $x,-x.$ So if $|x|=|y|$ then one member of $x,-x $ is equal to one member of $y,-y,$ which implies $(x=ylor x=-ylor -x=ylor -x=-y)$, which is equivalent to $(x=y lor x=- y)$. Conversely , if $x=pm y,$ then the non-negative member of $x,-x,$ which is $|x|, $ must equal the non-negative member of $y,-y,$ which is $|y|.$
â DanielWainfleet
Sep 11 at 0:10