Equation with one variable unknow how to solve easily

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

favorite












I have



$$left|beginarrayccc
x & -2 & 3x-6 \
2x & phantom-0 & 2-x \
-x & phantom-5 & x-2
endarrayright| = 0$$



How can I solve this with a fast way? I thought I can do
$$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
and I will continue with the other two, but I don't know if I am right.






linear-algebra differential-equations







share|cite|improve this question



























    up vote
    4
    down vote

    favorite












    I have



    $$left|beginarrayccc
    x & -2 & 3x-6 \
    2x & phantom-0 & 2-x \
    -x & phantom-5 & x-2
    endarrayright| = 0$$



    How can I solve this with a fast way? I thought I can do
    $$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
    and I will continue with the other two, but I don't know if I am right.






    linear-algebra differential-equations







    share|cite|improve this question

























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      I have



      $$left|beginarrayccc
      x & -2 & 3x-6 \
      2x & phantom-0 & 2-x \
      -x & phantom-5 & x-2
      endarrayright| = 0$$



      How can I solve this with a fast way? I thought I can do
      $$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
      and I will continue with the other two, but I don't know if I am right.






      linear-algebra differential-equations







      share|cite|improve this question















      I have



      $$left|beginarrayccc
      x & -2 & 3x-6 \
      2x & phantom-0 & 2-x \
      -x & phantom-5 & x-2
      endarrayright| = 0$$



      How can I solve this with a fast way? I thought I can do
      $$x-2+3x-6=0 implies 4x=-8 implies x=-2$$
      and I will continue with the other two, but I don't know if I am right.






      linear-algebra differential-equations




      linear-algebra differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Blue

      44.2k868141




      44.2k868141










      asked 1 hour ago









      m.s

      514




      514




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          This is wrong:
          $$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$



          This is the right way to do it
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| = 0$$



          Apply $R3 rightarrow R3+R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          Apply $R2rightarrow R2-2R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          0 & phantom-4 & 14-7x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          $$xleft( 4(4x-8)-3(14-7x)right)=0$$
          $$37x(x-2)=0 implies x=0,2$$






          share|cite|improve this answer






















          • yes sorry.But is the right way this?
            – m.s
            1 hour ago










          • I can't understand what you are asking can you edit your question using mathjax
            – Deepesh Meena
            1 hour ago






          • 1




            imgur.com/a/qCgZxLj . this is what i ask
            – m.s
            1 hour ago











          • no , i cant get it ,where you going it.If you could continue i would be greatful
            – m.s
            1 hour ago










          • see the updated naswer
            – Deepesh Meena
            1 hour ago

















          up vote
          3
          down vote













          Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:



          $$0 = left|beginarrayccc
          phantom-x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| =
          x(x-2) left|beginarrayccc
          phantom-1 & -2 & phantom-3 \
          phantom-2& phantom-0 & -1 \
          -1& phantom-5 & phantom-1
          endarrayright| =
          37x(x-2) quadimpliesquad x=0,2$$






          share|cite|improve this answer




















          • can you explain please how you find 37x(x-2)?
            – m.s
            43 mins ago










          • The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
            – Mathematician 42
            41 mins ago










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2919854%2fequation-with-one-variable-unknow-how-to-solve-easily%23new-answer', 'question_page');

          );

          Post as a guest

















          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          This is wrong:
          $$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$



          This is the right way to do it
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| = 0$$



          Apply $R3 rightarrow R3+R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          Apply $R2rightarrow R2-2R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          0 & phantom-4 & 14-7x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          $$xleft( 4(4x-8)-3(14-7x)right)=0$$
          $$37x(x-2)=0 implies x=0,2$$






          share|cite|improve this answer






















          • yes sorry.But is the right way this?
            – m.s
            1 hour ago










          • I can't understand what you are asking can you edit your question using mathjax
            – Deepesh Meena
            1 hour ago






          • 1




            imgur.com/a/qCgZxLj . this is what i ask
            – m.s
            1 hour ago











          • no , i cant get it ,where you going it.If you could continue i would be greatful
            – m.s
            1 hour ago










          • see the updated naswer
            – Deepesh Meena
            1 hour ago














          up vote
          1
          down vote



          accepted










          This is wrong:
          $$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$



          This is the right way to do it
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| = 0$$



          Apply $R3 rightarrow R3+R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          Apply $R2rightarrow R2-2R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          0 & phantom-4 & 14-7x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          $$xleft( 4(4x-8)-3(14-7x)right)=0$$
          $$37x(x-2)=0 implies x=0,2$$






          share|cite|improve this answer






















          • yes sorry.But is the right way this?
            – m.s
            1 hour ago










          • I can't understand what you are asking can you edit your question using mathjax
            – Deepesh Meena
            1 hour ago






          • 1




            imgur.com/a/qCgZxLj . this is what i ask
            – m.s
            1 hour ago











          • no , i cant get it ,where you going it.If you could continue i would be greatful
            – m.s
            1 hour ago










          • see the updated naswer
            – Deepesh Meena
            1 hour ago












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          This is wrong:
          $$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$



          This is the right way to do it
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| = 0$$



          Apply $R3 rightarrow R3+R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          Apply $R2rightarrow R2-2R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          0 & phantom-4 & 14-7x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          $$xleft( 4(4x-8)-3(14-7x)right)=0$$
          $$37x(x-2)=0 implies x=0,2$$






          share|cite|improve this answer














          This is wrong:
          $$x-2+3x-6=0 implies 4x=-8 implies x=-2 $$



          This is the right way to do it
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| = 0$$



          Apply $R3 rightarrow R3+R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          Apply $R2rightarrow R2-2R1$
          $$left|beginarrayccc
          x & -2 & 3x-6 \
          0 & phantom-4 & 14-7x \
          0 & phantom-3 & 4x-8
          endarrayright| = 0$$



          $$xleft( 4(4x-8)-3(14-7x)right)=0$$
          $$37x(x-2)=0 implies x=0,2$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          Deepesh Meena

          3,7012825




          3,7012825











          • yes sorry.But is the right way this?
            – m.s
            1 hour ago










          • I can't understand what you are asking can you edit your question using mathjax
            – Deepesh Meena
            1 hour ago






          • 1




            imgur.com/a/qCgZxLj . this is what i ask
            – m.s
            1 hour ago











          • no , i cant get it ,where you going it.If you could continue i would be greatful
            – m.s
            1 hour ago










          • see the updated naswer
            – Deepesh Meena
            1 hour ago
















          • yes sorry.But is the right way this?
            – m.s
            1 hour ago










          • I can't understand what you are asking can you edit your question using mathjax
            – Deepesh Meena
            1 hour ago






          • 1




            imgur.com/a/qCgZxLj . this is what i ask
            – m.s
            1 hour ago











          • no , i cant get it ,where you going it.If you could continue i would be greatful
            – m.s
            1 hour ago










          • see the updated naswer
            – Deepesh Meena
            1 hour ago















          yes sorry.But is the right way this?
          – m.s
          1 hour ago




          yes sorry.But is the right way this?
          – m.s
          1 hour ago












          I can't understand what you are asking can you edit your question using mathjax
          – Deepesh Meena
          1 hour ago




          I can't understand what you are asking can you edit your question using mathjax
          – Deepesh Meena
          1 hour ago




          1




          1




          imgur.com/a/qCgZxLj . this is what i ask
          – m.s
          1 hour ago





          imgur.com/a/qCgZxLj . this is what i ask
          – m.s
          1 hour ago













          no , i cant get it ,where you going it.If you could continue i would be greatful
          – m.s
          1 hour ago




          no , i cant get it ,where you going it.If you could continue i would be greatful
          – m.s
          1 hour ago












          see the updated naswer
          – Deepesh Meena
          1 hour ago




          see the updated naswer
          – Deepesh Meena
          1 hour ago










          up vote
          3
          down vote













          Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:



          $$0 = left|beginarrayccc
          phantom-x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| =
          x(x-2) left|beginarrayccc
          phantom-1 & -2 & phantom-3 \
          phantom-2& phantom-0 & -1 \
          -1& phantom-5 & phantom-1
          endarrayright| =
          37x(x-2) quadimpliesquad x=0,2$$






          share|cite|improve this answer




















          • can you explain please how you find 37x(x-2)?
            – m.s
            43 mins ago










          • The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
            – Mathematician 42
            41 mins ago














          up vote
          3
          down vote













          Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:



          $$0 = left|beginarrayccc
          phantom-x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| =
          x(x-2) left|beginarrayccc
          phantom-1 & -2 & phantom-3 \
          phantom-2& phantom-0 & -1 \
          -1& phantom-5 & phantom-1
          endarrayright| =
          37x(x-2) quadimpliesquad x=0,2$$






          share|cite|improve this answer




















          • can you explain please how you find 37x(x-2)?
            – m.s
            43 mins ago










          • The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
            – Mathematician 42
            41 mins ago












          up vote
          3
          down vote










          up vote
          3
          down vote









          Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:



          $$0 = left|beginarrayccc
          phantom-x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| =
          x(x-2) left|beginarrayccc
          phantom-1 & -2 & phantom-3 \
          phantom-2& phantom-0 & -1 \
          -1& phantom-5 & phantom-1
          endarrayright| =
          37x(x-2) quadimpliesquad x=0,2$$






          share|cite|improve this answer












          Elements of the first column are all multiples of $x$; elements in the third column are all multiples of $x-2$. You can factor the determinant thusly:



          $$0 = left|beginarrayccc
          phantom-x & -2 & 3x-6 \
          2x & phantom-0 & 2-x \
          -x & phantom-5 & x-2
          endarrayright| =
          x(x-2) left|beginarrayccc
          phantom-1 & -2 & phantom-3 \
          phantom-2& phantom-0 & -1 \
          -1& phantom-5 & phantom-1
          endarrayright| =
          37x(x-2) quadimpliesquad x=0,2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Blue

          44.2k868141




          44.2k868141











          • can you explain please how you find 37x(x-2)?
            – m.s
            43 mins ago










          • The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
            – Mathematician 42
            41 mins ago
















          • can you explain please how you find 37x(x-2)?
            – m.s
            43 mins ago










          • The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
            – Mathematician 42
            41 mins ago















          can you explain please how you find 37x(x-2)?
          – m.s
          43 mins ago




          can you explain please how you find 37x(x-2)?
          – m.s
          43 mins ago












          The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
          – Mathematician 42
          41 mins ago




          The first sentence he wrote literally explains what he did. In my opinion this is by far the best and fastest way of doing it.
          – Mathematician 42
          41 mins ago

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2919854%2fequation-with-one-variable-unknow-how-to-solve-easily%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What does second last employer means? [closed]

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
          Read more

          List of Gilmore Girls characters

          Image
          The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
          Read more

          Confectionery

          Image
          Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of ...
          Read more