Solve for by the method of elimination $x$ and $y$ :
$x + y = 3$ and $3 x - 2 y - 4 = 0$ .

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Solution

x + y = 3 __ (1)
3x - 2y = 4 _ (2)
__
3 (x + y = 3)
1 (3x - 2y = 4)
_
3x + 3y = 9
3x - 2y = 4
(-) (+) (-)
______
5y = 5 $∴$ y = 5/5 = 1
substituting y = 1 in equation (1) we get
x + y = 3 $\Rightarrow$ x = 3 - y = 3 - 1 = 2
$∴$ x = 2
Hence x = 2, y = 1

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