x or y? y or z? x or z? x, y, z?

1st equation

$x + y + z = 4y + 3$ [Shifting $y$ from L.H.S. to R.H.S.]

$x + z = 4y + 3 - y$ [Shifting $z$ from L.H.S to R.H.S. and subtracting $y$ from $4y$ ]

$x = 3y + 3 - z$

3rd equation

$z - y - x = y$ [Putting $3y + 3 - z$ in place of $x$ ]

$z - y - (3y + 3 - z) = y$ [Removing brackets]

$z - y - 3y - 3 + z = y$ [Solving L.H.S.]

$2z - 4y - 3 = y$ [Shifting $-4y$ from L.H.S to R.H.S.]

$2z - 3 = y + 4y$

$2z - 3 = 5y$ [Shifting 5 from R.H.S to L.H.S.]

$\frac{2z - 3}{5}$ $= y$

2nd equation

$y + z - 7 = 3y$ [Putting $\frac{2z - 3}{5}$ in place of $y$ ]

$\frac{2z - 3}{5}$ $+ z - 7 = 3 ($ $\frac{2z - 3}{5}$ $)$ [Multiplying in R.H.S. and adding in L.H.S.]

$\frac{2z - 3 + 5z - 35}{5}$ $=$ $\frac{(3*2z)-(3*3)}{5}$

$\frac{7z - 38}{5}$ $=$ $\frac{6z - 9}{5}$ [Removing 5 in denominator from both sides]

$7z - 38 = 6z - 9$ [Shifting $-38$ and $6z$ ]

$7z - 6z = -9 + 38$ [Solving L.H.S and R.H.S.]

$z = 29$

Again 1st equation

$x + y + z = 4y + 3$ [Putting $\frac{2z - 3}{5}$ in place of $y$ ]

$x +$ $\frac{2z - 3}{5}$ $+ z = 4 ($ $\frac{2z - 3}{5}$ $)+ 3$ [Putting 29 in place of z]

$x +$ $\frac{2(29) - 3}{5}$ $+ 29 =$ $\frac{(4*2*29)-(4*3)}{5}$ $+ 3$ [Solving it]

$x +$ $\frac{58 - 3}{5}$ $+ 29 =$ $\frac{232-12}{5}$ $+ 3$

$x +$ $\frac{55}{5}$ $+ 29 =$ $\frac{220}{5}$ $+ 3$

$x + 11 + 29 = 44 +3$

$x + 40 = 47$ [Shifting 40 from L.H.S. to R.H.S.]

$x = 47 - 40$

$x = 7$

Again 1st equation

$x + y + z = 4y + 3$ [Putting 7 in place of $x$ and 29 in place of $z$ ]

$7 + y + 29 = 4y + 3$ [Solving it]

$y + 36 = 4y + 3$ [Shifting $y$ from L.H.S. to R.H.S. and 3 from R.H.S. to L.H.S.]

$36 - 3 = 4y - y$

$33 = 3y$ [Shifting 3 from R.H.S. to L.H.S.]

$\frac{33}{3}$ $= y$

$y = 11$

Thus, $x = 7; y = 11; z = 29$

So, the value of $2x + 3y + z$ is

= $2(7) + 3(11) + 29$

= $14 + 33 + 29$

= $76$ = $Ans$

${\begin{cases}x + y + z = 4y + 3 \ ...(1) \\ y + z - 7 = 3y \ ...(2) \\ z - y - x = y \ ...(3)\end{cases}}$

$(1)+(3)$

$\implies 2z=5y+3 \ ...(4)$

Putting $(4)$ in $(2)$ gives:

$\implies y=11$

Putting $y$ in $(4)$ gives:

$\implies z=29$

Putting $y$ and $z$ in $(3)$ gives:

$\implies x=7$

$\implies 2x+3y+z=(2×7)+(3×11)+29=\boxed{76}$ .

$\begin{cases} x + y + z = 4y + 3 & \implies x - 3y + z = 3 & ...(1) \\ y + z - 7 = 3y & \implies - 2y + z = 7 & ...(2) \\ z - y - x = y & \implies x + 2y - z = 0 & ...(3) \end{cases}$

$\begin{aligned} (2)+(3): \ \ x & = 7 & ...\color{#3D99F6}{(4)} \end{aligned}$

$\begin{aligned} (1)+(3): \ \ 2x - y & = 3 \\ \implies y & = 2x - 3 \\ & = 2 \times 7_{\color{#3D99F6}{(4)}} - 3 \\ \implies y & = 11 & ...\color{#D61F06}{(5)} \end{aligned}$

$\begin{aligned} (1)+(2)+(3): \ \ 2x - 3y + z & = 10 \\ 2x - 3y + z + 6y & = 10 + 6y \\ 2x + 3y + z & = 10 + 6 \times 11_{\color{#D61F06}{(5)}} \\ \implies 2x + 3y + z & = \boxed{76} \end{aligned}$