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i) (0, 2)

ii) (2, 0)

iii) (4, 0)

iv) (\(\sqrt{2}\), 4\(\sqrt{2}\))

v) (1, 1)

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.

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