0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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: $$x+3y+z=10$$ $$x+3y+z+0=10$$ $$x+3y+z+(x-2y-z+6)=10$$ $$2x+y=4$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.