I want to know why exactly does the solution of the linear equations doesn't change upon adding and subtracting them when you get a new equation?

For example:

When you solve these two equations below you get a new one:

$x-2y-z = -6$
$x+3y+z = 10$

$2x+y = 4$

So, why the solution of this would still be the same? How are we sure?

  • $\begingroup$ It's worth noting that all solutions to the system of two equations are solutions to the last equation you wrote, but it's not the other way round. $\endgroup$ – I want to make games Sep 28 '15 at 18:15
  • $\begingroup$ Can you explain "why" using my example? $\endgroup$ – user963241 Sep 28 '15 at 18:26
  • $\begingroup$ If the first two equations are true, than the third one is true too, because it basically says: if I add A+B I get A+B (where A and B are the things at both left and right side of equations respectively 1 and 2). But for example, (x=1, y=2, z=1000) is a solution to the last equation, but not a solution to the system 1) and 2). Do you understand? $\endgroup$ – I want to make games Sep 28 '15 at 19:34
  • $\begingroup$ No, it is a solution to all the equations. In my case for example the solution to all the equations is $(x=1, y=2, y=3)$ It means that all the equations satisfy these values. So, if you plug these into equation (1) you get $[1-2(2)-3] = -6$. and if the same way you try to plug these into (2) and (3) the equations satisfy. Is that what you said? $\endgroup$ – user963241 Sep 28 '15 at 19:59
  • $\begingroup$ Let me rephrase: 1) x-2y-z=-6 || 2) x+3y+z=10 || 3) 2x+y=4 || If values ($x_0, y_0, z_0$) are a solution to both 1) and 2) they HAVE to be a solution to 3). But if values ($x_1, y_1, z_1$) are a solution to equation 3) they don't have to satisfy 1) and 2). For example, if you take x=1, y=2, z=1000, the equation 3) is satisfied (it simply doesn't say anything about z). But the equations 1) and 2) are not satisfied. I mentioned it's worth noting because I had a feeling you might not be aware of this. It's not a hint for solving this system, it's just some general knowledge I wanted to share. $\endgroup$ – I want to make games Sep 28 '15 at 20:10

The trick is that adding zero to an equation does not change the equation.

If you re-arrange the first equation: $$x-2y-z=-6$$ $$x-2y-z+6=0$$

Now that the first equation equals zero, you can add it to the second equation and not change anything in the second equation: $$x+3y+z=10$$ $$x+3y+z+0=10$$ $$x+3y+z+(x-2y-z+6)=10$$ $$2x+y=4$$

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Call x−2y−z A, −6 B, x−2y−z C, −6 D. So you have A=B and C=D. If I tell you that A+C = B+D, would you find it strange?

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  • $\begingroup$ I think it makes sense. $\endgroup$ – user963241 Sep 28 '15 at 18:26

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