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Solve the simultaneous equations: $3 x-y+z=5$; $2 x+2 y+3 z=4$  and $x+3 y-z=11$
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$3 x-y+z=5 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$
$2 x+2 y+3 z=4 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$
From, $(1)+(3): 4 x+2 y=16 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$
$3(3)+(2): 5 x+11 y=37 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$
Equations (4) and (5) are solved as simultaneous equations in two variables:
By elimination method and eliminating $x$ first:
$5(4)-4(5) ; \quad 34 y=68$, that is, $y=2$
Substituting for $\mathrm{y}$ in equation (4) gives:
$4 x+2(2)=16$
$4 \mathrm{x}=12$, i.e. $\mathrm{x}=3$
$\therefore \therefore \mathrm{x}=2, \mathrm{y}=3$
by Bronze Status (5,182 points)

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