Solve $x + y = 1$ and $3 x + 2 y = 6$

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Solution

$x + y = 1 \Rightarrow y = 1 - x$
Put $y = 1 - x$ in $3 x + 2 y = 6$
$3 x + 2 (1 - x) = 6$
$\Rightarrow 3 x + 2 - 2 x = 6$
$\Rightarrow x = 6 - 2 = 4$
$∴ x = 4$
Put $x = 4$ in $y = 1 - x = 1 - 4 = - 3$
$∴ x = 4$ , $y = - 3$

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