Question
Solve $\frac{1}{3}(x+y-5)=y-z=2 x-11=9-(x+2 z)$
✓
Answer
$\begin{aligned} & \frac{1}{3}(x+y-5)=y-z \\ & x+y-5=3 y-3 z \\ & x+y-3 y+3 z=5 \\ & x-2 y+3 z=5 \ldots(1) \\ & y-z-2 x-11 \\ & -2 x+y-z=-11 \\ & 2 x-y+z=11 \ldots(2) \\ & 2 x-11=9-(x+2 z) \\ & 2 x-11=9-x-2 z \\ & 2 x+x+2 z=9+11 \\ & 3 x+2 z=20 \ldots(3)\end{aligned}$
$\begin{aligned} & (1) \times 1 \Rightarrow x-2 y+3 z=5 \\ & (2) \times 2 \Rightarrow \underline{4 x-2 y+2 z=22} \\ & \text { (-) }( \pm) \_(-) \_(-) \\ & (1)-(4) \Rightarrow-3 x+0+z=-17 \\ & \end{aligned}$
$
3 x-z=17
$
Subtracting (3) and (5) we get
$
\begin{aligned}
& (3) \Rightarrow 3 x+2 z=20 \\
& (5) \Rightarrow 3 x-z=+17 \\
& \ \ \ \ \ \ \ \ (-) \ (+)\ \ \ \ \ (-) \\
& \hline 3 z=3
\end{aligned}
$
$
z=\frac{3}{3}=1
$
Substitute the value of $z=1$ in (3)
$
\begin{aligned}
& 3 x+2(1)=20 \\
& 3 x=20-2 \\
& 3 x=18 \\
& x=\frac{18}{3} \\
& =6
\end{aligned}
$
Substitute the value of x = 6, z = 1 in (2)
2(6) – y + 1 = 11
12 – y + 1 = 11
13 – y = 11
– y = 11 – 13
– y = – 2
y = 2
∴ The value of x = 6, y = 2 and z = 1
$\begin{aligned} & (1) \times 1 \Rightarrow x-2 y+3 z=5 \\ & (2) \times 2 \Rightarrow \underline{4 x-2 y+2 z=22} \\ & \text { (-) }( \pm) \_(-) \_(-) \\ & (1)-(4) \Rightarrow-3 x+0+z=-17 \\ & \end{aligned}$
$
3 x-z=17
$
Subtracting (3) and (5) we get
$
\begin{aligned}
& (3) \Rightarrow 3 x+2 z=20 \\
& (5) \Rightarrow 3 x-z=+17 \\
& \ \ \ \ \ \ \ \ (-) \ (+)\ \ \ \ \ (-) \\
& \hline 3 z=3
\end{aligned}
$
$
z=\frac{3}{3}=1
$
Substitute the value of $z=1$ in (3)
$
\begin{aligned}
& 3 x+2(1)=20 \\
& 3 x=20-2 \\
& 3 x=18 \\
& x=\frac{18}{3} \\
& =6
\end{aligned}
$
Substitute the value of x = 6, z = 1 in (2)
2(6) – y + 1 = 11
12 – y + 1 = 11
13 – y = 11
– y = 11 – 13
– y = – 2
y = 2
∴ The value of x = 6, y = 2 and z = 1
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