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Solve the following system of equations.

$21 x + 47 y = 110$

$47 x + 21 y = 162, x, y ≠ 0$

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14, 1

− 13, 1

13, 1

13, − 1

answer is C.

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Detailed Solution

Given system of equations are,

21 x + 47 y = 110......... 1 47 x + 21 y = 162 … … 2

Add eq (1) and (2) and simplify.

21 x + 47 y = 110

47 x + 21 y = 162

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68 x + 68 y = 272

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Taking 68 as common from LHS, we get,

68 1 x + 1 y = 272 1 x + 1 y = 27268 1 x + 1 y = 4 … … . (3)

Subtract eq (2) from (1) and simplify.

21 x + 47 y = 110 47 x + 21 y = 162

(-) (-) (-)
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− 26 x + 26 y = − 52

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Taking 26 as common from LHS, we get,

26 − 1 x + 1 y = − 52 − 1 x + 1 y = − 5226 − 1 x + 1 y = − 2 … … (4)

Add eq (3) and (4).

1 x + 1 y = 4 − 1 x + 1 y = − 2 − − − − − − − − − 2 y = 2 − − − − − − − − − 2 y = 2 y = 22 y = 1

Substitute the value of y = 1 in eq (3) and find the value of x.

1 x + 1 y = 4 1 x + 11 = 4 1 x + 1 = 4 1 x = 4 − 1 1 x = 3 3 x = 1 x = 13

Hence, the solution for the given equations is

13, 1

.
The correct option is (3).

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