Solving a $3\times3$ system of linear equations.

Question

I have the following as a given:

$2x + y = 22$

$2z + x = 28$

$2y + z = 38$

Solve for $x, y$ and $z$

P.S. I tried solving it by substitution and elimination but neither is working.

Thanks

Answer 1

$$9y= (2x+y)-2(2z+x) +4(2y+z) $$ $$=22-2\cdot 28+4\cdot 38$$ $$\implies y=...$$

Answer 2

if we add up all the equations, we get $x+y+z=\frac{88}{3}$ $$y=\frac{88}{3}-x-y $$ use first equation $x=22-\frac{88}{3}-z $ let's Express x through the previous equality $y=\frac{88}{3}-22+\frac{88}{3}+z-z$ $\Rightarrow $ $y=\frac{110}{3}$ use first equation of system $$2x=22-y=\frac{-22}{3} $$ And finally find $z $ $$2z=28+\frac{22}{3} $$ and $z=\frac{53}{3} $ Solution $(\frac{-22}{3},\frac{110}{3},\frac{53}{3}) $