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Suppose two absolute value equations have an odd number of distinct solutions. One of the equations is equal to $a$. What is $a$?
system of equations with an odd number of distinct solutions, with absolute value signProving that two systems of linear equations are equivalent if they have the same solutionsHow to see a plane is tangent to a sphere from their equationsThe equivalence of homogenous systems of linear equations in two unknowns that have the same solutionsAre approximate least square intersections unique?Determining the number of solutions to a system of equationssystem of two eqautions in three unknowns: finding the number of solutionsNumber of solutions of $(sqrt3sin x+cos x)^sqrtsqrt3sin2x-cos2x+2=4$How is the “two intersecting lines” picture of a system of two linear equations related to the “sum of two vectors” picture?What is the meaning of the difference between the equations of two non-intersecting circles represent?What's the general equation of a 3D right cone? And the intersection of the 3d cone with a line?
$begingroup$
This question has been asked here, but its results are inconclusive.
Suppose that $a$ is a number such that the system of equations
$$|2x| − y = 5$$
$$x − |2y + 2| = a$$
has an odd number of distinct solutions. What is the product of all
possible values of $a$?
I start with a little simplifying first.
The first equation gives $$y=pm2x-5$$
The second equation gives $$y=fracpm(x-a)-22$$
After getting these equations, I don't see any way to go further.
Therefore, I then try another approach using logic.
The first equation is a "v" shifted down $5$(hence the "-5") units and vertically stretched by $2$(hence the "2x") giving it a $22.5^circ$ angle with the $y$-axis.
As for the second one, I am not that sure how to graph. How do I graph the second equation?
The maximum number of odd intersections, that is, solutions to the two equations is obviously $3$.
Using similar logic, there is then cases where there are 1 intersections.
There are three cases with an odd number of solutions.
Case 1: 3 solutions; The second equation's "v" has to have 1 side of the "v" intersect with the first equation two times, and the second line of the "v" touch the vertex of the first equation.
Case 2: 3 solutions; The second equation has its vertex on a side of the "v" of the first equation, and has the lines of the "v" intersecting with the other line of the "v" of the first equation.
Case 3: 1 solution; The vertex of the second equation touches the side of the "v" of the first equation.
Are there any cases I have missed?
I can see that if I know that line of symmetry $y=c$(where $c$ is a constant) of the second equation, I can set equations for the two equations to find $a$ for all three cases I have found. How do I do that? Is there any way to find the line of symmetry for the second equation using its equation with a number missing?
Thanks! Your help is appreciated!
Max0815
linear-algebra algebra-precalculus analytic-geometry graphing-functions
asked Mar 27 at 0:57
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
This question has been asked here, but its results are inconclusive.
Suppose that $a$ is a number such that the system of equations
$$|2x| − y = 5$$
$$x − |2y + 2| = a$$
has an odd number of distinct solutions. What is the product of all
possible values of $a$?
I start with a little simplifying first.
The first equation gives $$y=pm2x-5$$
The second equation gives $$y=fracpm(x-a)-22$$
After getting these equations, I don't see any way to go further.
Therefore, I then try another approach using logic.
The first equation is a "v" shifted down $5$(hence the "-5") units and vertically stretched by $2$(hence the "2x") giving it a $22.5^circ$ angle with the $y$-axis.
As for the second one, I am not that sure how to graph. How do I graph the second equation?
The maximum number of odd intersections, that is, solutions to the two equations is obviously $3$.
Using similar logic, there is then cases where there are 1 intersections.
There are three cases with an odd number of solutions.
Case 1: 3 solutions; The second equation's "v" has to have 1 side of the "v" intersect with the first equation two times, and the second line of the "v" touch the vertex of the first equation.
Case 2: 3 solutions; The second equation has its vertex on a side of the "v" of the first equation, and has the lines of the "v" intersecting with the other line of the "v" of the first equation.
Case 3: 1 solution; The vertex of the second equation touches the side of the "v" of the first equation.
Are there any cases I have missed?
I can see that if I know that line of symmetry $y=c$(where $c$ is a constant) of the second equation, I can set equations for the two equations to find $a$ for all three cases I have found. How do I do that? Is there any way to find the line of symmetry for the second equation using its equation with a number missing?
Thanks! Your help is appreciated!
Max0815
linear-algebra algebra-precalculus analytic-geometry graphing-functions
asked Mar 27 at 0:57
Max0815Max0815
81418
$endgroup$
$begingroup$
I think the three cases cover it - you can use the fact that the vertex of the second equation has co-ordinates $(a, -2)$
$endgroup$
– WW1
Mar 27 at 1:14
$begingroup$
@WW1 how did you get that?
$endgroup$
– Max0815
Mar 27 at 1:23
1
$begingroup$
sorry it's actually at $(a,-1)$ - the vertex occurs where the thing in the absolute value signs is zero $|2y+2|=0$ so $x=a$
$endgroup$
– WW1
Mar 27 at 2:27
$begingroup$
@WW1 Ah. I see what you mean, and the line of symmetry would by y=-1. Thank you.
$endgroup$
– Max0815
Mar 27 at 4:41
add a comment |
$begingroup$
This question has been asked here, but its results are inconclusive.
Suppose that $a$ is a number such that the system of equations
$$|2x| − y = 5$$
$$x − |2y + 2| = a$$
has an odd number of distinct solutions. What is the product of all
possible values of $a$?
I start with a little simplifying first.
The first equation gives $$y=pm2x-5$$
The second equation gives $$y=fracpm(x-a)-22$$
After getting these equations, I don't see any way to go further.
Therefore, I then try another approach using logic.
The first equation is a "v" shifted down $5$(hence the "-5") units and vertically stretched by $2$(hence the "2x") giving it a $22.5^circ$ angle with the $y$-axis.
As for the second one, I am not that sure how to graph. How do I graph the second equation?
The maximum number of odd intersections, that is, solutions to the two equations is obviously $3$.
Using similar logic, there is then cases where there are 1 intersections.
There are three cases with an odd number of solutions.
Case 1: 3 solutions; The second equation's "v" has to have 1 side of the "v" intersect with the first equation two times, and the second line of the "v" touch the vertex of the first equation.
Case 2: 3 solutions; The second equation has its vertex on a side of the "v" of the first equation, and has the lines of the "v" intersecting with the other line of the "v" of the first equation.
Case 3: 1 solution; The vertex of the second equation touches the side of the "v" of the first equation.
Are there any cases I have missed?
I can see that if I know that line of symmetry $y=c$(where $c$ is a constant) of the second equation, I can set equations for the two equations to find $a$ for all three cases I have found. How do I do that? Is there any way to find the line of symmetry for the second equation using its equation with a number missing?
Thanks! Your help is appreciated!
Max0815
linear-algebra algebra-precalculus analytic-geometry graphing-functions
asked Mar 27 at 0:57
Max0815Max0815
81418
$endgroup$
This question has been asked here, but its results are inconclusive.
Suppose that $a$ is a number such that the system of equations
$$|2x| − y = 5$$
$$x − |2y + 2| = a$$
has an odd number of distinct solutions. What is the product of all
possible values of $a$?
I start with a little simplifying first.
The first equation gives $$y=pm2x-5$$
The second equation gives $$y=fracpm(x-a)-22$$
After getting these equations, I don't see any way to go further.
Therefore, I then try another approach using logic.
The first equation is a "v" shifted down $5$(hence the "-5") units and vertically stretched by $2$(hence the "2x") giving it a $22.5^circ$ angle with the $y$-axis.
As for the second one, I am not that sure how to graph. How do I graph the second equation?
The maximum number of odd intersections, that is, solutions to the two equations is obviously $3$.
Using similar logic, there is then cases where there are 1 intersections.
There are three cases with an odd number of solutions.
Case 1: 3 solutions; The second equation's "v" has to have 1 side of the "v" intersect with the first equation two times, and the second line of the "v" touch the vertex of the first equation.
Case 2: 3 solutions; The second equation has its vertex on a side of the "v" of the first equation, and has the lines of the "v" intersecting with the other line of the "v" of the first equation.
Case 3: 1 solution; The vertex of the second equation touches the side of the "v" of the first equation.
Are there any cases I have missed?
I can see that if I know that line of symmetry $y=c$(where $c$ is a constant) of the second equation, I can set equations for the two equations to find $a$ for all three cases I have found. How do I do that? Is there any way to find the line of symmetry for the second equation using its equation with a number missing?
Thanks! Your help is appreciated!
Max0815
linear-algebra algebra-precalculus analytic-geometry graphing-functions
linear-algebra algebra-precalculus analytic-geometry graphing-functions
asked Mar 27 at 0:57
Max0815Max0815
81418
asked Mar 27 at 0:57
Max0815Max0815
81418
asked Mar 27 at 0:57
Max0815Max0815
81418
asked Mar 27 at 0:57
Max0815Max0815
81418
asked Mar 27 at 0:57
Max0815Max0815
81418
81418
$begingroup$
I think the three cases cover it - you can use the fact that the vertex of the second equation has co-ordinates $(a, -2)$
$endgroup$
– WW1
Mar 27 at 1:14
$begingroup$
@WW1 how did you get that?
$endgroup$
– Max0815
Mar 27 at 1:23
1
$begingroup$
sorry it's actually at $(a,-1)$ - the vertex occurs where the thing in the absolute value signs is zero $|2y+2|=0$ so $x=a$
$endgroup$
– WW1
Mar 27 at 2:27
$begingroup$
@WW1 Ah. I see what you mean, and the line of symmetry would by y=-1. Thank you.
$endgroup$
– Max0815
Mar 27 at 4:41
add a comment |
$begingroup$
I think the three cases cover it - you can use the fact that the vertex of the second equation has co-ordinates $(a, -2)$
$endgroup$
– WW1
Mar 27 at 1:14
$begingroup$
@WW1 how did you get that?
$endgroup$
– Max0815
Mar 27 at 1:23
1
$begingroup$
sorry it's actually at $(a,-1)$ - the vertex occurs where the thing in the absolute value signs is zero $|2y+2|=0$ so $x=a$
$endgroup$
– WW1
Mar 27 at 2:27
$begingroup$
@WW1 Ah. I see what you mean, and the line of symmetry would by y=-1. Thank you.
$endgroup$
– Max0815
Mar 27 at 4:41
$begingroup$
I think the three cases cover it - you can use the fact that the vertex of the second equation has co-ordinates $(a, -2)$
$endgroup$
– WW1
Mar 27 at 1:14
$begingroup$
I think the three cases cover it - you can use the fact that the vertex of the second equation has co-ordinates $(a, -2)$
$endgroup$
– WW1
Mar 27 at 1:14
$begingroup$
@WW1 how did you get that?
$endgroup$
– Max0815
Mar 27 at 1:23
$begingroup$
@WW1 how did you get that?
$endgroup$
– Max0815
Mar 27 at 1:23
1
1
$begingroup$
sorry it's actually at $(a,-1)$ - the vertex occurs where the thing in the absolute value signs is zero $|2y+2|=0$ so $x=a$
$endgroup$
– WW1
Mar 27 at 2:27
$begingroup$
sorry it's actually at $(a,-1)$ - the vertex occurs where the thing in the absolute value signs is zero $|2y+2|=0$ so $x=a$
$endgroup$
– WW1
Mar 27 at 2:27
$begingroup$
@WW1 Ah. I see what you mean, and the line of symmetry would by y=-1. Thank you.
$endgroup$
– Max0815
Mar 27 at 4:41
$begingroup$
@WW1 Ah. I see what you mean, and the line of symmetry would by y=-1. Thank you.
$endgroup$
– Max0815
Mar 27 at 4:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note: Thank you to @WW1 for hinting. He gets an upvote! :)
There are 2 absolute values. An absolute value means two solutions, so in this problem, there are $2+2=4$ cases. However, there are three cases that have odd solutions, given in the diagram:
The first absolute value is a "v" with a vertical stretch of $2$ and a vertex at $(0, -5)$. We note that the vertex occurs whenever the stuff in the absolute value signs is equal to $0$. So for the second equation, $|2y+2|=0$ and we conclude that $x=a$.
Furthermore, setting $a=0$ and graphing, we can conclude from the data that the second equation's line of symmetry is $y=-1$. We can then conclude from this that the vertex of the second equation is $(a, -1)$, and the vertex shifts with the change of $a$.
According the the diagram, we can see there are three cases. Let's start with the first case(the furthest one on the left):
We see that the vertex of the first equation is the solution to both. Plugging in the $x$ and $y$, we can solve for the $a$: I set $0=|2(-5)+2|+a$ so $a=-8$.
For the second case, we can see that there is a solution in the form of $(x, -1)$. Plugging this in the first equation gives $x=-2$. Plugging this into the second equation gives $a=-2$.
The third case has a solution at $(x, -1)$ again. Realizing that this is an absolute value equation, $x$ can also equal $2$. So we have the pair $(2, -1)$ which gives $a=2$ when I plug this into the second equation.
As a result, we have three solutions for $a$: $-8, -2$, and $2$.
Multiplying these together gives $(-8)cdot(-2)cdot2=boxed32$
answered Mar 29 at 22:26
Max0815Max0815
81418
$endgroup$
add a comment |
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$begingroup$
Note: Thank you to @WW1 for hinting. He gets an upvote! :)
There are 2 absolute values. An absolute value means two solutions, so in this problem, there are $2+2=4$ cases. However, there are three cases that have odd solutions, given in the diagram:
The first absolute value is a "v" with a vertical stretch of $2$ and a vertex at $(0, -5)$. We note that the vertex occurs whenever the stuff in the absolute value signs is equal to $0$. So for the second equation, $|2y+2|=0$ and we conclude that $x=a$.
Furthermore, setting $a=0$ and graphing, we can conclude from the data that the second equation's line of symmetry is $y=-1$. We can then conclude from this that the vertex of the second equation is $(a, -1)$, and the vertex shifts with the change of $a$.
According the the diagram, we can see there are three cases. Let's start with the first case(the furthest one on the left):
We see that the vertex of the first equation is the solution to both. Plugging in the $x$ and $y$, we can solve for the $a$: I set $0=|2(-5)+2|+a$ so $a=-8$.
For the second case, we can see that there is a solution in the form of $(x, -1)$. Plugging this in the first equation gives $x=-2$. Plugging this into the second equation gives $a=-2$.
The third case has a solution at $(x, -1)$ again. Realizing that this is an absolute value equation, $x$ can also equal $2$. So we have the pair $(2, -1)$ which gives $a=2$ when I plug this into the second equation.
As a result, we have three solutions for $a$: $-8, -2$, and $2$.
Multiplying these together gives $(-8)cdot(-2)cdot2=boxed32$
answered Mar 29 at 22:26
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
Note: Thank you to @WW1 for hinting. He gets an upvote! :)
There are 2 absolute values. An absolute value means two solutions, so in this problem, there are $2+2=4$ cases. However, there are three cases that have odd solutions, given in the diagram:
The first absolute value is a "v" with a vertical stretch of $2$ and a vertex at $(0, -5)$. We note that the vertex occurs whenever the stuff in the absolute value signs is equal to $0$. So for the second equation, $|2y+2|=0$ and we conclude that $x=a$.
Furthermore, setting $a=0$ and graphing, we can conclude from the data that the second equation's line of symmetry is $y=-1$. We can then conclude from this that the vertex of the second equation is $(a, -1)$, and the vertex shifts with the change of $a$.
According the the diagram, we can see there are three cases. Let's start with the first case(the furthest one on the left):
We see that the vertex of the first equation is the solution to both. Plugging in the $x$ and $y$, we can solve for the $a$: I set $0=|2(-5)+2|+a$ so $a=-8$.
For the second case, we can see that there is a solution in the form of $(x, -1)$. Plugging this in the first equation gives $x=-2$. Plugging this into the second equation gives $a=-2$.
The third case has a solution at $(x, -1)$ again. Realizing that this is an absolute value equation, $x$ can also equal $2$. So we have the pair $(2, -1)$ which gives $a=2$ when I plug this into the second equation.
As a result, we have three solutions for $a$: $-8, -2$, and $2$.
Multiplying these together gives $(-8)cdot(-2)cdot2=boxed32$
answered Mar 29 at 22:26
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
Note: Thank you to @WW1 for hinting. He gets an upvote! :)
There are 2 absolute values. An absolute value means two solutions, so in this problem, there are $2+2=4$ cases. However, there are three cases that have odd solutions, given in the diagram:
The first absolute value is a "v" with a vertical stretch of $2$ and a vertex at $(0, -5)$. We note that the vertex occurs whenever the stuff in the absolute value signs is equal to $0$. So for the second equation, $|2y+2|=0$ and we conclude that $x=a$.
Furthermore, setting $a=0$ and graphing, we can conclude from the data that the second equation's line of symmetry is $y=-1$. We can then conclude from this that the vertex of the second equation is $(a, -1)$, and the vertex shifts with the change of $a$.
According the the diagram, we can see there are three cases. Let's start with the first case(the furthest one on the left):
We see that the vertex of the first equation is the solution to both. Plugging in the $x$ and $y$, we can solve for the $a$: I set $0=|2(-5)+2|+a$ so $a=-8$.
For the second case, we can see that there is a solution in the form of $(x, -1)$. Plugging this in the first equation gives $x=-2$. Plugging this into the second equation gives $a=-2$.
The third case has a solution at $(x, -1)$ again. Realizing that this is an absolute value equation, $x$ can also equal $2$. So we have the pair $(2, -1)$ which gives $a=2$ when I plug this into the second equation.
As a result, we have three solutions for $a$: $-8, -2$, and $2$.
Multiplying these together gives $(-8)cdot(-2)cdot2=boxed32$
answered Mar 29 at 22:26
Max0815Max0815
81418
$endgroup$
Note: Thank you to @WW1 for hinting. He gets an upvote! :)
There are 2 absolute values. An absolute value means two solutions, so in this problem, there are $2+2=4$ cases. However, there are three cases that have odd solutions, given in the diagram:
The first absolute value is a "v" with a vertical stretch of $2$ and a vertex at $(0, -5)$. We note that the vertex occurs whenever the stuff in the absolute value signs is equal to $0$. So for the second equation, $|2y+2|=0$ and we conclude that $x=a$.
Furthermore, setting $a=0$ and graphing, we can conclude from the data that the second equation's line of symmetry is $y=-1$. We can then conclude from this that the vertex of the second equation is $(a, -1)$, and the vertex shifts with the change of $a$.
According the the diagram, we can see there are three cases. Let's start with the first case(the furthest one on the left):
We see that the vertex of the first equation is the solution to both. Plugging in the $x$ and $y$, we can solve for the $a$: I set $0=|2(-5)+2|+a$ so $a=-8$.
For the second case, we can see that there is a solution in the form of $(x, -1)$. Plugging this in the first equation gives $x=-2$. Plugging this into the second equation gives $a=-2$.
The third case has a solution at $(x, -1)$ again. Realizing that this is an absolute value equation, $x$ can also equal $2$. So we have the pair $(2, -1)$ which gives $a=2$ when I plug this into the second equation.
As a result, we have three solutions for $a$: $-8, -2$, and $2$.
Multiplying these together gives $(-8)cdot(-2)cdot2=boxed32$
answered Mar 29 at 22:26
Max0815Max0815
81418
answered Mar 29 at 22:26
Max0815Max0815
81418
answered Mar 29 at 22:26
Max0815Max0815
81418
answered Mar 29 at 22:26
Max0815Max0815
81418
81418
add a comment |
add a comment |
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I think the three cases cover it - you can use the fact that the vertex of the second equation has co-ordinates $(a, -2)$
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– WW1
Mar 27 at 1:14
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@WW1 how did you get that?
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– Max0815
Mar 27 at 1:23
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sorry it's actually at $(a,-1)$ - the vertex occurs where the thing in the absolute value signs is zero $|2y+2|=0$ so $x=a$
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– WW1
Mar 27 at 2:27
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@WW1 Ah. I see what you mean, and the line of symmetry would by y=-1. Thank you.
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– Max0815
Mar 27 at 4:41