What is the algebraic proof that $|x|+|x-2|=2$ is an inequality
$begingroup$
$|x|+|x-2|=2$ is an inequality because the solution is $xle2$ (if you plug in $2$ or any value lower than $2$ for $x$, you will get an answer of $2$). What is the algebraic proof (with steps)? How can I write this out using algebra to get a value of $x≤2$?
algebra-precalculus inequality
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
$endgroup$
add a comment |
$begingroup$
$|x|+|x-2|=2$ is an inequality because the solution is $xle2$ (if you plug in $2$ or any value lower than $2$ for $x$, you will get an answer of $2$). What is the algebraic proof (with steps)? How can I write this out using algebra to get a value of $x≤2$?
algebra-precalculus inequality
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
$endgroup$
1
$begingroup$
Hint: For $x leq 2$, $|x-2|=2-x$.
$endgroup$
– greelious
Jan 15 at 21:47
$begingroup$
This is true if and only if $0 leq x leq 2$. In this case, there is a simple explanation writing it $|x-0|+|x-2| $ i.e., the sum of distances from $x$ to $0$ and from $x$ to $2$.
$endgroup$
– Jean Marie
Jan 15 at 21:48
3
$begingroup$
Your statement is incorrect. The statement is an equation, not an inequality. It may be equivalent to an inequality, but it is not itself an inequality.
$endgroup$
– MPW
Jan 15 at 21:49
$begingroup$
MPW, I am particularly asking for the algebraic proof that would enable the equation to be an inequality, as it does have a specific range so it can be proven (algebraically) to or be an inequality. Remember, ≤ is a combination of < and =. I can clearly see that the statement is an equation (thanks for pointing that out, though), but I want to know how it could be an inequality, and the algebraic proof. And it's a question, by the way, not a statement.
$endgroup$
– ZeeKay Mdludlu
Jan 16 at 2:29
add a comment |
$begingroup$
$|x|+|x-2|=2$ is an inequality because the solution is $xle2$ (if you plug in $2$ or any value lower than $2$ for $x$, you will get an answer of $2$). What is the algebraic proof (with steps)? How can I write this out using algebra to get a value of $x≤2$?
algebra-precalculus inequality
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
$endgroup$
$|x|+|x-2|=2$ is an inequality because the solution is $xle2$ (if you plug in $2$ or any value lower than $2$ for $x$, you will get an answer of $2$). What is the algebraic proof (with steps)? How can I write this out using algebra to get a value of $x≤2$?
algebra-precalculus inequality
algebra-precalculus inequality
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
edited Jan 15 at 23:04
J. W. Tanner
5,1651520
5,1651520
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
asked Jan 15 at 21:44
ZeeKay MdludluZeeKay Mdludlu
1
1
1
$begingroup$
Hint: For $x leq 2$, $|x-2|=2-x$.
$endgroup$
– greelious
Jan 15 at 21:47
$begingroup$
This is true if and only if $0 leq x leq 2$. In this case, there is a simple explanation writing it $|x-0|+|x-2| $ i.e., the sum of distances from $x$ to $0$ and from $x$ to $2$.
$endgroup$
– Jean Marie
Jan 15 at 21:48
3
$begingroup$
Your statement is incorrect. The statement is an equation, not an inequality. It may be equivalent to an inequality, but it is not itself an inequality.
$endgroup$
– MPW
Jan 15 at 21:49
$begingroup$
MPW, I am particularly asking for the algebraic proof that would enable the equation to be an inequality, as it does have a specific range so it can be proven (algebraically) to or be an inequality. Remember, ≤ is a combination of < and =. I can clearly see that the statement is an equation (thanks for pointing that out, though), but I want to know how it could be an inequality, and the algebraic proof. And it's a question, by the way, not a statement.
$endgroup$
– ZeeKay Mdludlu
Jan 16 at 2:29
add a comment |
1
$begingroup$
Hint: For $x leq 2$, $|x-2|=2-x$.
$endgroup$
– greelious
Jan 15 at 21:47
$begingroup$
This is true if and only if $0 leq x leq 2$. In this case, there is a simple explanation writing it $|x-0|+|x-2| $ i.e., the sum of distances from $x$ to $0$ and from $x$ to $2$.
$endgroup$
– Jean Marie
Jan 15 at 21:48
3
$begingroup$
Your statement is incorrect. The statement is an equation, not an inequality. It may be equivalent to an inequality, but it is not itself an inequality.
$endgroup$
– MPW
Jan 15 at 21:49
$begingroup$
MPW, I am particularly asking for the algebraic proof that would enable the equation to be an inequality, as it does have a specific range so it can be proven (algebraically) to or be an inequality. Remember, ≤ is a combination of < and =. I can clearly see that the statement is an equation (thanks for pointing that out, though), but I want to know how it could be an inequality, and the algebraic proof. And it's a question, by the way, not a statement.
$endgroup$
– ZeeKay Mdludlu
Jan 16 at 2:29
1
1
$begingroup$
Hint: For $x leq 2$, $|x-2|=2-x$.
$endgroup$
– greelious
Jan 15 at 21:47
$begingroup$
Hint: For $x leq 2$, $|x-2|=2-x$.
$endgroup$
– greelious
Jan 15 at 21:47
$begingroup$
This is true if and only if $0 leq x leq 2$. In this case, there is a simple explanation writing it $|x-0|+|x-2| $ i.e., the sum of distances from $x$ to $0$ and from $x$ to $2$.
$endgroup$
– Jean Marie
Jan 15 at 21:48
$begingroup$
This is true if and only if $0 leq x leq 2$. In this case, there is a simple explanation writing it $|x-0|+|x-2| $ i.e., the sum of distances from $x$ to $0$ and from $x$ to $2$.
$endgroup$
– Jean Marie
Jan 15 at 21:48
3
3
$begingroup$
Your statement is incorrect. The statement is an equation, not an inequality. It may be equivalent to an inequality, but it is not itself an inequality.
$endgroup$
– MPW
Jan 15 at 21:49
$begingroup$
Your statement is incorrect. The statement is an equation, not an inequality. It may be equivalent to an inequality, but it is not itself an inequality.
$endgroup$
– MPW
Jan 15 at 21:49
$begingroup$
MPW, I am particularly asking for the algebraic proof that would enable the equation to be an inequality, as it does have a specific range so it can be proven (algebraically) to or be an inequality. Remember, ≤ is a combination of < and =. I can clearly see that the statement is an equation (thanks for pointing that out, though), but I want to know how it could be an inequality, and the algebraic proof. And it's a question, by the way, not a statement.
$endgroup$
– ZeeKay Mdludlu
Jan 16 at 2:29
$begingroup$
MPW, I am particularly asking for the algebraic proof that would enable the equation to be an inequality, as it does have a specific range so it can be proven (algebraically) to or be an inequality. Remember, ≤ is a combination of < and =. I can clearly see that the statement is an equation (thanks for pointing that out, though), but I want to know how it could be an inequality, and the algebraic proof. And it's a question, by the way, not a statement.
$endgroup$
– ZeeKay Mdludlu
Jan 16 at 2:29
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You have three cases:
$xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.
$0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.
$2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.
In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.
answered Jan 15 at 21:53
IuliaIulia
1,025415
$endgroup$
add a comment |
$begingroup$
By the triangle inequality:
$$2=|x|+|2-x|geq|x+2-x|=2.$$
The equality occurs for $xgeq0$ and $2-xgeq0$ or $0leq xleq2$, which is the answer.
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
$y=x-1$;
$|y+1|+|y-1|=2;$
$|2y| =|y+1+y-1| le $
$|y+1| +|y-1| =2$;
Hence
$|y| le 1$, i.e. $-1 le y le 1$;
With $y=x-1$:
$0 le x le 2$.
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
Geometric approach
Absolute value $|a-b|$ denotes the distance between the points on $x-$axis, representing $a;$ and $;b.$
In the present case we have $$|x|+|x-2|=|x-0|+|x-2|$$ which has to give $2.$ In distance wording, the sum of distances from $x$ to $0$ and from the same $x$ to $2$ gives $2.$
The points out the interval $;[0,2];$ are too far from $0$ or from $2$, thus none of them satisfies. But each point lying in $[0,2]$ satisfies. The given equation $;|x|+|x-2|=2;$ can be rewritten $$0leq x leq 2.$$
answered Jan 16 at 10:52
user376343user376343
3,9934829
$endgroup$
add a comment |
$begingroup$
Alternatively, you can square both sides:
$$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
|x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have three cases:
$xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.
$0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.
$2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.
In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.
answered Jan 15 at 21:53
IuliaIulia
1,025415
$endgroup$
add a comment |
$begingroup$
You have three cases:
$xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.
$0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.
$2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.
In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.
answered Jan 15 at 21:53
IuliaIulia
1,025415
$endgroup$
add a comment |
$begingroup$
You have three cases:
$xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.
$0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.
$2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.
In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.
answered Jan 15 at 21:53
IuliaIulia
1,025415
$endgroup$
You have three cases:
$xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.
$0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.
$2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.
In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.
answered Jan 15 at 21:53
IuliaIulia
1,025415
answered Jan 15 at 21:53
IuliaIulia
1,025415
answered Jan 15 at 21:53
IuliaIulia
1,025415
answered Jan 15 at 21:53
IuliaIulia
1,025415
1,025415
add a comment |
add a comment |
$begingroup$
By the triangle inequality:
$$2=|x|+|2-x|geq|x+2-x|=2.$$
The equality occurs for $xgeq0$ and $2-xgeq0$ or $0leq xleq2$, which is the answer.
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
By the triangle inequality:
$$2=|x|+|2-x|geq|x+2-x|=2.$$
The equality occurs for $xgeq0$ and $2-xgeq0$ or $0leq xleq2$, which is the answer.
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
By the triangle inequality:
$$2=|x|+|2-x|geq|x+2-x|=2.$$
The equality occurs for $xgeq0$ and $2-xgeq0$ or $0leq xleq2$, which is the answer.
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
By the triangle inequality:
$$2=|x|+|2-x|geq|x+2-x|=2.$$
The equality occurs for $xgeq0$ and $2-xgeq0$ or $0leq xleq2$, which is the answer.
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 15 at 21:51
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
$begingroup$
$y=x-1$;
$|y+1|+|y-1|=2;$
$|2y| =|y+1+y-1| le $
$|y+1| +|y-1| =2$;
Hence
$|y| le 1$, i.e. $-1 le y le 1$;
With $y=x-1$:
$0 le x le 2$.
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
$y=x-1$;
$|y+1|+|y-1|=2;$
$|2y| =|y+1+y-1| le $
$|y+1| +|y-1| =2$;
Hence
$|y| le 1$, i.e. $-1 le y le 1$;
With $y=x-1$:
$0 le x le 2$.
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
$y=x-1$;
$|y+1|+|y-1|=2;$
$|2y| =|y+1+y-1| le $
$|y+1| +|y-1| =2$;
Hence
$|y| le 1$, i.e. $-1 le y le 1$;
With $y=x-1$:
$0 le x le 2$.
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
$endgroup$
$y=x-1$;
$|y+1|+|y-1|=2;$
$|2y| =|y+1+y-1| le $
$|y+1| +|y-1| =2$;
Hence
$|y| le 1$, i.e. $-1 le y le 1$;
With $y=x-1$:
$0 le x le 2$.
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
answered Jan 15 at 22:36
Peter SzilasPeter Szilas
12k2822
12k2822
add a comment |
add a comment |
$begingroup$
Geometric approach
Absolute value $|a-b|$ denotes the distance between the points on $x-$axis, representing $a;$ and $;b.$
In the present case we have $$|x|+|x-2|=|x-0|+|x-2|$$ which has to give $2.$ In distance wording, the sum of distances from $x$ to $0$ and from the same $x$ to $2$ gives $2.$
The points out the interval $;[0,2];$ are too far from $0$ or from $2$, thus none of them satisfies. But each point lying in $[0,2]$ satisfies. The given equation $;|x|+|x-2|=2;$ can be rewritten $$0leq x leq 2.$$
answered Jan 16 at 10:52
user376343user376343
3,9934829
$endgroup$
add a comment |
$begingroup$
Geometric approach
Absolute value $|a-b|$ denotes the distance between the points on $x-$axis, representing $a;$ and $;b.$
In the present case we have $$|x|+|x-2|=|x-0|+|x-2|$$ which has to give $2.$ In distance wording, the sum of distances from $x$ to $0$ and from the same $x$ to $2$ gives $2.$
The points out the interval $;[0,2];$ are too far from $0$ or from $2$, thus none of them satisfies. But each point lying in $[0,2]$ satisfies. The given equation $;|x|+|x-2|=2;$ can be rewritten $$0leq x leq 2.$$
answered Jan 16 at 10:52
user376343user376343
3,9934829
$endgroup$
add a comment |
$begingroup$
Geometric approach
Absolute value $|a-b|$ denotes the distance between the points on $x-$axis, representing $a;$ and $;b.$
In the present case we have $$|x|+|x-2|=|x-0|+|x-2|$$ which has to give $2.$ In distance wording, the sum of distances from $x$ to $0$ and from the same $x$ to $2$ gives $2.$
The points out the interval $;[0,2];$ are too far from $0$ or from $2$, thus none of them satisfies. But each point lying in $[0,2]$ satisfies. The given equation $;|x|+|x-2|=2;$ can be rewritten $$0leq x leq 2.$$
answered Jan 16 at 10:52
user376343user376343
3,9934829
$endgroup$
Geometric approach
Absolute value $|a-b|$ denotes the distance between the points on $x-$axis, representing $a;$ and $;b.$
In the present case we have $$|x|+|x-2|=|x-0|+|x-2|$$ which has to give $2.$ In distance wording, the sum of distances from $x$ to $0$ and from the same $x$ to $2$ gives $2.$
The points out the interval $;[0,2];$ are too far from $0$ or from $2$, thus none of them satisfies. But each point lying in $[0,2]$ satisfies. The given equation $;|x|+|x-2|=2;$ can be rewritten $$0leq x leq 2.$$
answered Jan 16 at 10:52
user376343user376343
3,9934829
answered Jan 16 at 10:52
user376343user376343
3,9934829
answered Jan 16 at 10:52
user376343user376343
3,9934829
answered Jan 16 at 10:52
user376343user376343
3,9934829
3,9934829
add a comment |
add a comment |
$begingroup$
Alternatively, you can square both sides:
$$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
|x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
Alternatively, you can square both sides:
$$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
|x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
Alternatively, you can square both sides:
$$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
|x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
$endgroup$
Alternatively, you can square both sides:
$$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
|x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
answered Jan 16 at 13:08
farruhotafarruhota
22.5k2942
22.5k2942
add a comment |
add a comment |
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$begingroup$
Hint: For $x leq 2$, $|x-2|=2-x$.
$endgroup$
– greelious
Jan 15 at 21:47
$begingroup$
This is true if and only if $0 leq x leq 2$. In this case, there is a simple explanation writing it $|x-0|+|x-2| $ i.e., the sum of distances from $x$ to $0$ and from $x$ to $2$.
$endgroup$
– Jean Marie
Jan 15 at 21:48
3
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Your statement is incorrect. The statement is an equation, not an inequality. It may be equivalent to an inequality, but it is not itself an inequality.
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– MPW
Jan 15 at 21:49
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MPW, I am particularly asking for the algebraic proof that would enable the equation to be an inequality, as it does have a specific range so it can be proven (algebraically) to or be an inequality. Remember, ≤ is a combination of < and =. I can clearly see that the statement is an equation (thanks for pointing that out, though), but I want to know how it could be an inequality, and the algebraic proof. And it's a question, by the way, not a statement.
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– ZeeKay Mdludlu
Jan 16 at 2:29