What is the algebraic proof that $|x|+|x-2|=2$ is an inequality












-3












$begingroup$


$|x|+|x-2|=2$ 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







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$endgroup$








  • 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
















-3












$begingroup$


$|x|+|x-2|=2$ 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







share|cite|improve this question











$endgroup$








  • 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














-3












-3








-3





$begingroup$


$|x|+|x-2|=2$ 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







share|cite|improve this question











$endgroup$




$|x|+|x-2|=2$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 23:04









J. W. Tanner

5,1651520




5,1651520










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














  • 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










5 Answers
5






active

oldest

votes


















3












$begingroup$

You have three cases:





  1. $xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.


  2. $0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.


  3. $2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.


In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $y=x-1$;



      $|y+1|+|y-1|=2;$



      $|2y| =|y+1+y-1| le $



      $|y+1| +|y-1| =2$;



      Hence



      $|y| le 1$, i.e. $-1 le y le 1$;



      With $y=x-1$:



      $0 le x le 2$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        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 $$0leq x leq 2.$$






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          Alternatively, you can square both sides:
          $$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
          |x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$






          share|cite|improve this answer









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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            You have three cases:





            1. $xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.


            2. $0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.


            3. $2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.


            In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              You have three cases:





              1. $xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.


              2. $0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.


              3. $2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.


              In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                You have three cases:





                1. $xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.


                2. $0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.


                3. $2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.


                In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.






                share|cite|improve this answer









                $endgroup$



                You have three cases:





                1. $xleq 0$ The equality becomes $-x+2-x=2 Rightarrow 0=2x Rightarrow x=0$.


                2. $0leq xleq 2$ The equality becomes $x+2-x=2 Rightarrow 2=2$ which is true $forall xin[0,2]$.


                3. $2leq x$ The equality becomes $x+x-2=2Rightarrow 2x=4Rightarrow x=2$.


                In conclusion $xin[0,2]$ is the set of solutions: $0leq xleq 2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 21:53









                IuliaIulia

                1,025415




                1,025415























                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$


















                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$
















                        0












                        0








                        0





                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$



                        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.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 15 at 21:51









                        Michael RozenbergMichael Rozenberg

                        111k1897201




                        111k1897201























                            0












                            $begingroup$

                            $y=x-1$;



                            $|y+1|+|y-1|=2;$



                            $|2y| =|y+1+y-1| le $



                            $|y+1| +|y-1| =2$;



                            Hence



                            $|y| le 1$, i.e. $-1 le y le 1$;



                            With $y=x-1$:



                            $0 le x le 2$.






                            share|cite|improve this answer









                            $endgroup$


















                              0












                              $begingroup$

                              $y=x-1$;



                              $|y+1|+|y-1|=2;$



                              $|2y| =|y+1+y-1| le $



                              $|y+1| +|y-1| =2$;



                              Hence



                              $|y| le 1$, i.e. $-1 le y le 1$;



                              With $y=x-1$:



                              $0 le x le 2$.






                              share|cite|improve this answer









                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                $y=x-1$;



                                $|y+1|+|y-1|=2;$



                                $|2y| =|y+1+y-1| le $



                                $|y+1| +|y-1| =2$;



                                Hence



                                $|y| le 1$, i.e. $-1 le y le 1$;



                                With $y=x-1$:



                                $0 le x le 2$.






                                share|cite|improve this answer









                                $endgroup$



                                $y=x-1$;



                                $|y+1|+|y-1|=2;$



                                $|2y| =|y+1+y-1| le $



                                $|y+1| +|y-1| =2$;



                                Hence



                                $|y| le 1$, i.e. $-1 le y le 1$;



                                With $y=x-1$:



                                $0 le x le 2$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 15 at 22:36









                                Peter SzilasPeter Szilas

                                12k2822




                                12k2822























                                    0












                                    $begingroup$

                                    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 $$0leq x leq 2.$$






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      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 $$0leq x leq 2.$$






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        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 $$0leq x leq 2.$$






                                        share|cite|improve this answer









                                        $endgroup$



                                        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 $$0leq x leq 2.$$







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jan 16 at 10:52









                                        user376343user376343

                                        3,9934829




                                        3,9934829























                                            0












                                            $begingroup$

                                            Alternatively, you can square both sides:
                                            $$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
                                            |x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Alternatively, you can square both sides:
                                              $$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
                                              |x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Alternatively, you can square both sides:
                                                $$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
                                                |x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$






                                                share|cite|improve this answer









                                                $endgroup$



                                                Alternatively, you can square both sides:
                                                $$|x|+|x-2|=2 Rightarrow x^2+2|x^2-2x|+(x-2)^2=4 Rightarrow \
                                                |x^2-2x|=-(x^2-2x) Rightarrow x^2-2xle 0 Rightarrow xin[0,2].$$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jan 16 at 13:08









                                                farruhotafarruhota

                                                22.5k2942




                                                22.5k2942






























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                                                    Bressuire

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                                                    Cabo Verde

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                                                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                                                    Gyllenstierna

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