Cannot solve linear system of equations

Question

It would be nice if somebody could find my mistake for the following linear system of equations:

$$ \left\{\begin{array}{rcrcrcr} -2x & - & 4y & - & z & = & -21 \\ -3x & + & y & + & 2z & = & -14 \\ x & - & 2y & - & z & = & a \end{array}\right. $$

It is solvable for all $a \in \Bbb R$.

I know that $y= 3- a$ , however when I try to solve this system I get a different result. The first thing I did was to subtract the first equation from the third so that I can eliminate $z$. The third equation now would be $$3x+2y=a+21.$$

After that I wanted to eradicate the $z$ in the second equation so I multiplied the first equation by two and added it to the second equation. My second equation now is: $$-7x-7y=-56.$$

I then went on to eliminate the $y$ in the third equation so I multiplied the second equation by 3 and the third equation by 7 and added the second equation to the third. The third equation would now be: $$-7y=7a-33$$ which is equivalent to $$y=-a+\frac{33}{7}$$ which is unequal to the actual solution. Where is my mistake?

Answer 1

Adding 3 times the second and 7 times the third equation produces $-7y = 7a - 168 + 147 = 7a - 21$.

Answer 2

From what you have already done,

$3x+2y=a+21$
$-7x-7y=-56$ or $x+y=8$ or $3x+3y=24$.

Subtracting 1 from 2, $y = 3 - a$