Why am I getting wrong solution to the system $2x+6y-3z=10$, $5x+2y-1z=12$? Suppose we have this system of equations:
$$\begin{align}
2x+6y-3z&=10\tag{1} \\
5x+2y-1z&=12\tag{2}
\end{align}$$
Why does solving as follows give an incorrect solution?
$$\begin{align}
2x+6y-3z-10&=5x+2y-1z-12 \\
-3x+4y-2z&=-2\tag{3}
\end{align}$$
According to this equation, $(0,0,1)$ is a solution. However, when I plug this into the two starting equations, the result is inconsistent.
What is wrong with the logic I used in finding the solution?
By the way, I was only looking for a single solution of the two starting equations, not a general solution. I know how to use matrices to solve the system, but I am still curious as to what is wrong with the approach I demonstrated above.
 A: Kavi Rama Murthy is right: your process creates a new plane that includes the set of solutions (a line) but also includes many other points (such as (0,0,1)) that do not solve the equation. For reference see the image below that has your two equations graphed in red and green, your new equation in grey, and the point (0,0,1) in blue.

A: If you really want just a single solution, set just one variable to zero and solve the resulting system. With different coefficients on each variable, you can be sure the line that is the general solution will pass through all three planes $x=0, y=0,z=0$. So you can set $z$ to zero to get:
$$
\begin{align}
2x+6y&= 10 \\
5x+2y &= 12
\end{align}
$$
From which you can quickly find the solution $(x,y,z) = (2,1,0)$. (You can find the more general solution by now setting $z=1$ and re-solving, but I three just leaving one variable free and solving the system that way is going to be faster.)
Setting two variables to zero can't work because there's no insurance that the line of the solution crosses any of the three axes.
A: Let's make this as simple as possible. Suppose you have the equations
$$x=2$$
$$y=1$$
Subtracting them gives
$$x-y=1$$
According to this equation, $(175,174)$ is a solution! But we know the only solution to the original equations is $(2,1)$.
A: 
Why does solving as follows give an incorrect solution?

Because in the process of your solving the system, you are not making an "equivalent" transformation of the system, which may introduce new solutions that are not solutions to the original system.
In the language of matrices, when solving $Ax=b$, you are not supposed to change the rank of the coefficient matrix $A$.
A: \begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align}
Matrix form: $AX=b$
$A=\begin{pmatrix} 2 & 6&-3\\5&2&-1\end{pmatrix}$
$X=\begin{pmatrix}x\\y\\z\end{pmatrix}
$
$b=\begin{pmatrix} 10\\12\end{pmatrix}$
Solve : $[A|b]=\begin{pmatrix} 2 & 6&-3&|10\\5&2&-1&|12\end{pmatrix}$
Then you get the solution :
$X=\{c\begin{pmatrix} 0\\1\\2\end{pmatrix}+\begin{pmatrix} 2\\1\\0\end{pmatrix}:c\in\Bbb{R} \}$

Now back to the original problem :
$-3x+4y-2z=-2$
represent a plane inside the $3$ dimensional space.But the solution of the original system lies on a line.
Also the line lies on the plane $-3x+4y-2z=-2$.
Hence any solution of the original system are also the solution of $-3x+4y-2z=-2$.

\begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align} $$\overset{(1')=(1)-(2)}\downarrow$$
\begin{align}
-3x+4y-2z&=-2 \tag {1'} \\
5x+2y-z&=12 \tag {2'}
\end{align}

Now solve the system $(1') , (2')$ and you will get the original solutions as both systems are equivalent system.This is the way how elimination works!
A: We try to visualize geometry of three planes and their intersections. The views could help avoiding going astray even in so many simple equations during a first learning.
The two plane equations
$$\begin{align}
P_1=2x+6y-3z&=10\tag{1} \\
P_2=5x+2y-1z&=12\tag{2}
\end{align}$$
along with another equation and plane
$$ P_3= ax+by+cz= d \tag 3 $$
result in the solution of three simultaneous linear equations as a single trihedral vertex point $P_{123} $ representing intersection of three concurrent intersecting straight lines.
Each plane extend into 3-space with direction cosines calculable from coefficients of $(x,y,z)$ in the plane equation. Intersection of two planes is a straight line and intersection of three planes is a solution point in 3-space.
If there is inconsistency of two equations (checked by slope or vanishing determinant) we have a a triangular prism instead of a triangular pyramid.
Linear combination $ P=\lambda P_1 +(1- \lambda) P_2  $ represents any plane through common axis of intersection $ (P_1 , P_2) $.

An arbitrary point like the one you gave  $(0,0,1) $ or  $(6,1,7)$ lying on a co-intersection plane $P$ is not a solution. It is nowhere on any of three concurrent intersection lines nor on the solution apex point if a third line and its equation is considered for a full point solution.
Equation / plane $P_3$ is not plotted so as to avoid cluttering in their images.
We use Cramer's Rule with matrices etc. for a simultaneous concurrent lines common vertex solution.
Planes (1) and (2) have their sum and difference  internal and external bisector  planes as
$$ P_1= -3x+4y-2z =-2,\quad P_2= 7x+8y-4z=22\;\tag{4}$$
Among intersection planes $ P= P_1 \cap P_2,\; $ by $\lambda$ variation we can  choose the internal bisector plane $P_1+P_2$ or external bisector plane $P_1-P_2 $ depending on the quadrant in which they lie . They are plotted here using Mathematica
A point either on  $P1$ or $P2$ need not in general lie  on lines passing through intersection of planes which is the real check of solution, so does not qualify as solution.
A: You started with two equations and you correctly stated that if they are equal, then it is necessary that your third equation is true.
We have $$2x+6y-3z=10$$ end $$5x+2y-1z=12$$
From this we correctly conclude that $$-3x+4y-2z=-2$$ must be true.
But that does not mean that every point that satisfies this last equation will satisfy the first two. In other words we get that if a point $(a,b,c)$ is incident with the plane described by the first equation and the point is incident with the plane described by the second equation then that point must be incident with the plane described by the third equation. Logically, this has the form $$A\land B\implies C$$. Your example shows that the converse, $$C\implies A\land B$$ is not necessarily true.
In fact, we know that the intersection of two planes that are neither equal nor parallel must be a line. What the third equation shows is that the line of intersection must be incident with that plane. That does not mean that any point on that plane is on the line of intersection.
A: Your solution strategy is not wrong, but you go astray before reaching your target.
Or in terms of a  quotation by Yogi Berra:
"You've got to be very careful if you don't know where you're going, because you might not get there."
Given the system of linear equations $(1)\:\&\:(2)\,$ you are deducing from it the equation
$$-3x+4y-2z \:=\: -2 \tag{3) = (1)$\,-\,$(2}$$
which may substitute either equation $(2)$ or $(1)$ to arrive at the equivalent linear systems
$$\begin{align}
2x+6y-3z & \:=\: 10\tag{1} \\
-3x+4y-2z & \:=\: -2\tag{3}
\end{align}$$
or
$$\begin{align}
-3x+4y-2z & \:=\: -2\tag{3} \\
5x+2y-1z& \:=\: 12\tag{2}
\end{align}$$
both having the same solution set as the initial system.
Then you lose the track when dropping one equation and restricting attention to $(3)$ only. Ignoring relevant constraints results in general in enlarging the set of solutions, and the additional solutions do not satisfy the ignored constraints.

Notice that, when looking for a single solution of the two starting equation, you may eliminate $y$ and $z$ at one stroke in building the linear combination $\,3\cdot(2)-(1)$ which yields the equation $13x= 26$.
Thus, every single solution must satisfy $x=2\,$.
Insertion into $(1)$, or $(2),$ it doesn't matter which one, gives the further condition $2(y-1) = z\,.$
A: If $2x+6y-3z=10$ and $5x+2y-1z=12$, then $2x+6y-3z-10=5x+2y-1z-12$.
But the converse is not true, i.e., if $2x+6y-3z-10=5x+2y-1z-12$, it is not necessary that $2x+6y-3z=10$ and $5x+2y-1z=12$.
The correct way to solve an equation (or system of equations) is to find  statements that are equivalent to the original equation.
The word equivalent means that A is true if and only if B is true.
Example: Solve the system of equation
\begin{gather}
    2x+1=3 \\
    x+y=3
\end{gather}
You should verify that the following statements are all equalivent to each other.

*

*$2x+1=3$ and $x+y=3$.

*$x=1$ and $x+y=3$.

*$x=1$ and $1+y=3$.

*$x=1$ and $y=2$.

Since they are all equalivent to each other, $2x+1=3$ and $x+y=3$ if and only if $x=1$ and $y=2$, so we can see $x=1$ and $y=2$ are the only solution to the original equation.
:)
A: In (1) you demand that $$A=2x+6y-3z-10$$ is $0$. In (2) you demand that $$B=5x+2y-z-12$$ is $0$. In (3) you just demand that these $A$ and $B$ are equal. If they are both $0$ they will of course be equal, but the other implication does not hold.
A: First let's see the equations;
$$(2x+6y-3z = 10) \text{ - 1 equation}$$
$$(5x+2y-1z = 12) \text{ - 2 equation}$$
We can multiply the whole second equation by $3$ and we get the following equation;
$$3(5x+2y-1z = 12)$$
$$15x+6y-3z = 36$$
Now mark this equation as the second. Then by subtracting the first from the second(through the elimination method) we get,
$$+ 15x+6y-3z = 36$$
$$-(2x+6y-3z = 10)$$
We get
$$13x=26$$
$$x=2$$
Now we got the value of one variable.
By plugging the value in the equations we can get a system of linear equations in two variables then we can find the rest of the value. By doing this we get the following solutions. The two equations will form a coincident system of lines.$$$$
$(2x+6y-3z = 10)$ putting x = 2 we get $$2y-z=6$$
$(5x+2y-z = 12)$ putting x = 2 we get $$2y-z=6$$
Hence the solution to this system of equation is as follows;
$$x = 2$$
$$y \in [- \infty, \infty]$$
$$z \in [- \infty, \infty]$$
But $y$ and $z$ are dependant on each other through the equation $2y-z=6$. So if you put any value of $y$ or $z$ you can find the respective variable.
$$$$
For example the solutions to this equation can be (2,0,-6) or of the form (2,  6+z/2, z) or (2,y,2y-6).$$$$
Hope you found this answer useful.
Thank you
A: Just don't complicate things with matrices and other stuff,
@TonyK 's answer gives you the idea of the problem, But the cause is
You have missed this symbol $\iff$, that is, you have missed the word phrase "If and Only If"
Lets say ,
$$
a=c$$$$
b=c
$$

*

*"then $a=b$" Is WRONG,but this is what we usually think

*"$a=b \iff a=c,b=c$" Is Right,
Notice that   $a=b \iff a=c,b=c$ is different from $a=c,b=c \iff a=b$,which is even wrong
The first statement is reversible i.e.  $$
\text{"if }a=c,b=c,\text{ then } a=b \text{" can be reversed as}$$$$
\text{"if } a=b,\text{ then }a=c,b=c\text{"(reversed format)} 
$$
The statement is literally right, but its reversed format makes it wrong
But the second statement is said as "if and ONLY if", which makes it irreversible, and right
what I mean here is


*

*Values that satisfy your first and second equation($2x+6y-3z=10,5x+2y-1z=12$),
will satisfy third equation($-3x+4y-2z=-2$)

*But NOT THE REVERSE, Values that satisfy third equation($-3x+4y-2z=-2$) need not satisfy your first and second equation($2x+6y-3z=10,5x+2y-1z=12$)
So never mind to substitute $(0,0,1)$, as it satisfies third equation, because it NEED NOT satisfy the first and second equation
but you may substitute the value that satisfies first and second equation, into third equation, which will definitely work

Work $\iff$ you gain profit ; )
A: You derived $(3)$ from $(1)$ and $(2)$, so any solution satisfying the $(1)$ and $(2)$ must satisfy $(3)$. The converse of this statement i.e. any solution satisfying $(3)$ must satisfy $(1)$ and $(2)$ need not to be true. However, its contrapositive i.e. the values not satisfying $(3)$ must not satisfy $(1)$ and $(2)$ is always true; you can verify it.
