Solving independent linear equations











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begin{align}
&{-}2y+2z-1=0 tag{4} \[4px]
&{-}2x+4y-2z-2=0 tag{5} \[4px]
&phantom{-2}x-y+3/2=0 tag{6}
end{align}



Equation (6) is the sum of (4) and (5). There are only two independent equations.
Putting $z=0$ in (5) and (6) and solving for x and y, we have
begin{align}
x&=-2 \[4px]
y&=-1/2
end{align}





  • equation (6) is the sum of (4) and (5): OK, I see it


  • There are only two independent equations: I didn't get; what does this sentence mean?


  • Putting $z=0$ in (5) and (6): why putting z=0 in equation (5) and (6)?



Please help






linear-algebra







share|cite|improve this question




























    up vote
    2
    down vote

    favorite













    begin{align}
    &{-}2y+2z-1=0 tag{4} \[4px]
    &{-}2x+4y-2z-2=0 tag{5} \[4px]
    &phantom{-2}x-y+3/2=0 tag{6}
    end{align}



    Equation (6) is the sum of (4) and (5). There are only two independent equations.
    Putting $z=0$ in (5) and (6) and solving for x and y, we have
    begin{align}
    x&=-2 \[4px]
    y&=-1/2
    end{align}





    • equation (6) is the sum of (4) and (5): OK, I see it


    • There are only two independent equations: I didn't get; what does this sentence mean?


    • Putting $z=0$ in (5) and (6): why putting z=0 in equation (5) and (6)?



    Please help






    linear-algebra







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      begin{align}
      &{-}2y+2z-1=0 tag{4} \[4px]
      &{-}2x+4y-2z-2=0 tag{5} \[4px]
      &phantom{-2}x-y+3/2=0 tag{6}
      end{align}



      Equation (6) is the sum of (4) and (5). There are only two independent equations.
      Putting $z=0$ in (5) and (6) and solving for x and y, we have
      begin{align}
      x&=-2 \[4px]
      y&=-1/2
      end{align}





      • equation (6) is the sum of (4) and (5): OK, I see it


      • There are only two independent equations: I didn't get; what does this sentence mean?


      • Putting $z=0$ in (5) and (6): why putting z=0 in equation (5) and (6)?



      Please help






      linear-algebra







      share|cite|improve this question
















      begin{align}
      &{-}2y+2z-1=0 tag{4} \[4px]
      &{-}2x+4y-2z-2=0 tag{5} \[4px]
      &phantom{-2}x-y+3/2=0 tag{6}
      end{align}



      Equation (6) is the sum of (4) and (5). There are only two independent equations.
      Putting $z=0$ in (5) and (6) and solving for x and y, we have
      begin{align}
      x&=-2 \[4px]
      y&=-1/2
      end{align}





      • equation (6) is the sum of (4) and (5): OK, I see it


      • There are only two independent equations: I didn't get; what does this sentence mean?


      • Putting $z=0$ in (5) and (6): why putting z=0 in equation (5) and (6)?



      Please help






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 5 hours ago









      egreg

      176k1384198




      176k1384198










      asked 6 hours ago









      Akash

      686




      686






















          2 Answers
          2






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          up vote
          5
          down vote













          Any value given to $x$, $y$ and $z$ that satisfy equations (4) and (5) will also satisfy equation (6). So the last equation gives no new information.



          The fact that $z$ disappears if we sum (4) and (5) means that it is “free”: we can assign it any value and determine suitable values for $x$ and $y$ that satisfy the equations.



          Using $z=0$ will give one solution for the system, but it has infinitely many solutions, one for each value given to $z$.



          For instance, if we use $z=1$, we get $x=-1$ and $y=1/2$.



          The general solution is
          begin{cases}
          x=-2+z \[4px]
          y=-dfrac{1}{2}+z
          end{cases}






          share|cite|improve this answer





















          • Thank you, now my doubts are clear
            – Akash
            4 hours ago




















          up vote
          3
          down vote













          (This is too long for a comment so I post it as a solution).



          In order to solve any equation with multiple variables, a good way to approach is to think how much "information" each equation gives.



          We see the first equation, so it gives us one information. The second gives us a second information, because it cannot be derived from previous information.



          But looking at the third equation (or number 6), we see that you can get this result by adding the two previous equations together. Therefore, it does not give us any new information. Now we have three unknowns and two equations that give us information about them. These two equations are said to be independent because they give us information about the variables. The third equation is a linear combination of the other two, so it's NOT independent of the two others.



          Because we have two equations and three unknowns, we have an infinite amount of solutions. Apparently, in the example, we are interested in obtaining at least one of them. Due to the linearity of the equations, we we are free to set any value to any of the variables, and we will obtain the other two. In this case, the author has decided to set $z=0$ in order to get a solution.






          share|cite|improve this answer





















          • Thank you i got it
            – Akash
            4 hours ago











          Your Answer





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          2 Answers
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          active

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          2 Answers
          2






          active

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          active

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          up vote
          5
          down vote













          Any value given to $x$, $y$ and $z$ that satisfy equations (4) and (5) will also satisfy equation (6). So the last equation gives no new information.



          The fact that $z$ disappears if we sum (4) and (5) means that it is “free”: we can assign it any value and determine suitable values for $x$ and $y$ that satisfy the equations.



          Using $z=0$ will give one solution for the system, but it has infinitely many solutions, one for each value given to $z$.



          For instance, if we use $z=1$, we get $x=-1$ and $y=1/2$.



          The general solution is
          begin{cases}
          x=-2+z \[4px]
          y=-dfrac{1}{2}+z
          end{cases}






          share|cite|improve this answer





















          • Thank you, now my doubts are clear
            – Akash
            4 hours ago

















          up vote
          5
          down vote













          Any value given to $x$, $y$ and $z$ that satisfy equations (4) and (5) will also satisfy equation (6). So the last equation gives no new information.



          The fact that $z$ disappears if we sum (4) and (5) means that it is “free”: we can assign it any value and determine suitable values for $x$ and $y$ that satisfy the equations.



          Using $z=0$ will give one solution for the system, but it has infinitely many solutions, one for each value given to $z$.



          For instance, if we use $z=1$, we get $x=-1$ and $y=1/2$.



          The general solution is
          begin{cases}
          x=-2+z \[4px]
          y=-dfrac{1}{2}+z
          end{cases}






          share|cite|improve this answer





















          • Thank you, now my doubts are clear
            – Akash
            4 hours ago















          up vote
          5
          down vote










          up vote
          5
          down vote









          Any value given to $x$, $y$ and $z$ that satisfy equations (4) and (5) will also satisfy equation (6). So the last equation gives no new information.



          The fact that $z$ disappears if we sum (4) and (5) means that it is “free”: we can assign it any value and determine suitable values for $x$ and $y$ that satisfy the equations.



          Using $z=0$ will give one solution for the system, but it has infinitely many solutions, one for each value given to $z$.



          For instance, if we use $z=1$, we get $x=-1$ and $y=1/2$.



          The general solution is
          begin{cases}
          x=-2+z \[4px]
          y=-dfrac{1}{2}+z
          end{cases}






          share|cite|improve this answer












          Any value given to $x$, $y$ and $z$ that satisfy equations (4) and (5) will also satisfy equation (6). So the last equation gives no new information.



          The fact that $z$ disappears if we sum (4) and (5) means that it is “free”: we can assign it any value and determine suitable values for $x$ and $y$ that satisfy the equations.



          Using $z=0$ will give one solution for the system, but it has infinitely many solutions, one for each value given to $z$.



          For instance, if we use $z=1$, we get $x=-1$ and $y=1/2$.



          The general solution is
          begin{cases}
          x=-2+z \[4px]
          y=-dfrac{1}{2}+z
          end{cases}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          egreg

          176k1384198




          176k1384198












          • Thank you, now my doubts are clear
            – Akash
            4 hours ago




















          • Thank you, now my doubts are clear
            – Akash
            4 hours ago


















          Thank you, now my doubts are clear
          – Akash
          4 hours ago






          Thank you, now my doubts are clear
          – Akash
          4 hours ago












          up vote
          3
          down vote













          (This is too long for a comment so I post it as a solution).



          In order to solve any equation with multiple variables, a good way to approach is to think how much "information" each equation gives.



          We see the first equation, so it gives us one information. The second gives us a second information, because it cannot be derived from previous information.



          But looking at the third equation (or number 6), we see that you can get this result by adding the two previous equations together. Therefore, it does not give us any new information. Now we have three unknowns and two equations that give us information about them. These two equations are said to be independent because they give us information about the variables. The third equation is a linear combination of the other two, so it's NOT independent of the two others.



          Because we have two equations and three unknowns, we have an infinite amount of solutions. Apparently, in the example, we are interested in obtaining at least one of them. Due to the linearity of the equations, we we are free to set any value to any of the variables, and we will obtain the other two. In this case, the author has decided to set $z=0$ in order to get a solution.






          share|cite|improve this answer





















          • Thank you i got it
            – Akash
            4 hours ago















          up vote
          3
          down vote













          (This is too long for a comment so I post it as a solution).



          In order to solve any equation with multiple variables, a good way to approach is to think how much "information" each equation gives.



          We see the first equation, so it gives us one information. The second gives us a second information, because it cannot be derived from previous information.



          But looking at the third equation (or number 6), we see that you can get this result by adding the two previous equations together. Therefore, it does not give us any new information. Now we have three unknowns and two equations that give us information about them. These two equations are said to be independent because they give us information about the variables. The third equation is a linear combination of the other two, so it's NOT independent of the two others.



          Because we have two equations and three unknowns, we have an infinite amount of solutions. Apparently, in the example, we are interested in obtaining at least one of them. Due to the linearity of the equations, we we are free to set any value to any of the variables, and we will obtain the other two. In this case, the author has decided to set $z=0$ in order to get a solution.






          share|cite|improve this answer





















          • Thank you i got it
            – Akash
            4 hours ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          (This is too long for a comment so I post it as a solution).



          In order to solve any equation with multiple variables, a good way to approach is to think how much "information" each equation gives.



          We see the first equation, so it gives us one information. The second gives us a second information, because it cannot be derived from previous information.



          But looking at the third equation (or number 6), we see that you can get this result by adding the two previous equations together. Therefore, it does not give us any new information. Now we have three unknowns and two equations that give us information about them. These two equations are said to be independent because they give us information about the variables. The third equation is a linear combination of the other two, so it's NOT independent of the two others.



          Because we have two equations and three unknowns, we have an infinite amount of solutions. Apparently, in the example, we are interested in obtaining at least one of them. Due to the linearity of the equations, we we are free to set any value to any of the variables, and we will obtain the other two. In this case, the author has decided to set $z=0$ in order to get a solution.






          share|cite|improve this answer












          (This is too long for a comment so I post it as a solution).



          In order to solve any equation with multiple variables, a good way to approach is to think how much "information" each equation gives.



          We see the first equation, so it gives us one information. The second gives us a second information, because it cannot be derived from previous information.



          But looking at the third equation (or number 6), we see that you can get this result by adding the two previous equations together. Therefore, it does not give us any new information. Now we have three unknowns and two equations that give us information about them. These two equations are said to be independent because they give us information about the variables. The third equation is a linear combination of the other two, so it's NOT independent of the two others.



          Because we have two equations and three unknowns, we have an infinite amount of solutions. Apparently, in the example, we are interested in obtaining at least one of them. Due to the linearity of the equations, we we are free to set any value to any of the variables, and we will obtain the other two. In this case, the author has decided to set $z=0$ in order to get a solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          Matti P.

          1,711413




          1,711413












          • Thank you i got it
            – Akash
            4 hours ago


















          • Thank you i got it
            – Akash
            4 hours ago
















          Thank you i got it
          – Akash
          4 hours ago




          Thank you i got it
          – Akash
          4 hours ago


















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