I am trying to solve this linear equation but I am getting wrong solution.
$$4x-3y+z=-1 \tag1$$ $$x+5y-z=12 \tag2$$ $$2y+z=10\tag3$$
I tried substituting $z = 10-2y$ and $x = 12+2/5y$ the answer I did not get I tried solving the both the equation putting the value to first equation.
After the substitution of $z$,
$$4x-5y=-11,\\x+7y=22.$$
Then substituting $x=22-7y$,
$$-33y=-99.$$