The graphs of the equations $$3x + y - 5 = 0$$ and $$2x - y - 5 = 0$$ intersect at the point $$P(\alpha, \beta)$$. What is the value of $$(3\alpha + \beta)$$?
When graphs of the equations intersect at the point $$P(\alpha, \beta)$$ then,
$$3\alpha + \beta - 5 = 0$$ ---(1)
$$2\alpha - \beta - 5 = 0$$ ---(2),
On eq(1) +Â (2),
$$5\alpha - 10 = 0$$
$$\alpha = 2$$
From the eq(2),
$$3 \times 2+ \beta - 5 = 0$$
$$\beta = -1$$
Now,
$$(3\alpha + \beta)$$ = 3 $$\times$$ 2 - 1 = 6 - 1 = 5
$$\therefore$$ The correct answer is option D.
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