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Answer :
If (x, y) are the solutions of the equation x – 2y = 4 then, they satisfies that equation.
i) Let's put x = 0 and y = 2,
we get, L.H.S. = (0) - 2(2) = -4
so, since, L.H.S. \(\ne\) R.H.S.
(0, 2) are not the solutions of given equation.
ii) Let's put x = 2 and y = 0,
we get, L.H.S. = (2) - 2(0) = 2
so, since, L.H.S. \(\ne\) R.H.S.
(2, 0) are not the solutions of given equation.
iii) Let's put x = 4 and y = 0,
we get, L.H.S. = (4) - 2(0) = 4
so, since, L.H.S. = R.H.S.
(4, 0) are the solutions of given equation.
iv) Let's put x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\),
we get, L.H.S. = (\(\sqrt{2}\)) - 2(4\(\sqrt{2}\)) = -7\(\sqrt{2}\)
so, since, L.H.S. \(\ne\) R.H.S.
(\(\sqrt{2}\), 4\(\sqrt{2}\)) are not the solutions of given equation.
v) Let's put x = 1 and y = 1,
we get, L.H.S. = (1) - 2(1) = -1
so, since, L.H.S. \(\ne\) R.H.S.
(1, 1) are not the solutions of given equation.