Matrices are MUCH more efficient for large systems, but, if the system is soluble at all, it can always be solved by the method of substitution.
System:
\(\displaystyle ax = k + x + y + z\)
\(\displaystyle by = k + x + y + z\)
\(\displaystyle cz = k + x + y + z\)
Solution (assuming all pronumerals are positive, a \(\displaystyle \ne \) 1, ab > a + b, and abc > ab + ac + bc)
\(\displaystyle ax = k + x + y + z \implies x(a - 1) = k + y + z \implies x = \dfrac{k + y + z}{a - 1}.\) Now substitute for x in the other two equations.
\(\displaystyle So\ by = k + x + y + z \implies by = k + \dfrac{k + y + z}{a - 1} + y + z \implies aby - by = ak - k + k + y + z + ay - y + az - z \implies y(ab - a - b) = a(k + z).\)
\(\displaystyle So\ cz = k + x + y + z \implies cz = k + \dfrac{k + y + z}{a - 1} + y + z \implies acz - cz = ak - k + k + y + z + ay - y + az - z \implies z(ac - a - c) = a(k + y).\)
Now we have two equations in two variables, y and z. So do it again.
\(\displaystyle y(ab - a - b) = a(k + z) \implies y = \dfrac{a(k + z)}{ab - a - b}.\) Now substitute for y in the other equation.
\(\displaystyle So\ z(ac - a - c) = a(k + y) \implies z(ac - a - c) = ak + a\left(\dfrac{a(k + z)}{ab - a - b}\right) \implies z(ab - a - b)(ac - a - c) = ak(ab - a - b) + a^2k + a^2z \implies\)
\(\displaystyle z(a^2bc - a^2b - abc - a^2c + a^2 + ac - abc + ab + bc) = a^2bk - a^2k - abk + a^2k + a^2z \implies\)
\(\displaystyle z[a^2(bc - b - c) - a(bc - b - c) - bc(a - 1)]= z[(a^2 - a)(bc - b - c) - bc(a - 1)] = (a - 1)z(abc - ab - ac - bc) = abk(a - 1) \implies\)
\(\displaystyle z = \dfrac{abk}{abc - ab - ac - bc} \implies\)
\(\displaystyle a(k + z) = \dfrac{ak(abc - ab - ac - bc) + a^2bk}{abc - ab - ac - bc} = \dfrac{ak(abc - ac - bc)}{abc - ab - ac - bc} = \dfrac{ack(ab - a - b)}{abc - ab - ac - bc} \implies\)
\(\displaystyle y = \dfrac{ack(ab - a - b)}{abc - ab - ac - bc} \div (ab - a - b) = \dfrac{ack}{abc - ab - ac - bc} \implies\)
\(\displaystyle k + y + z = k + \dfrac{abk + ack}{abc - ab - ac - bc} = \dfrac{abck - abk - ack - bck + abk + ack}{abc - ab - ac - bc} = \dfrac{bck(a - 1)}{abc - ab - ac - bc} \implies\)
\(\displaystyle x = \left(\dfrac{bck(a - 1)}{abc - ab -ac -bc}\right) \div (a - 1) = \dfrac{bck}{abc - ab - ac - bc}.\)
Now you should see the pattern if you decide to add more variables. I am not going to work this out for four variables, and I am certainly not going to prove the pattern for n variables. But you now have 2 ways to do this kind of problem, substitution and matrices, plus a pattern that makes intuitive sense. So good luck.
