Question

2(ax - by) +(a + 4b) = 0
2(bx + ay) + (b - 4a) = 0
solve for x and y
please its very important
give the correct answer
i will mark the answer brainliest ❤️❤️❤️​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

Solve for x and y :-

2(ax - by) +(a + 4b) = 0

2(bx + ay) + (b - 4a) = 0

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$\huge{AηsωeR} ✍$

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☆ Consider

$\bf \:  ⟼ 2(ax - by) +(a + 4b) = 0$

can be rewritten as

$\sf \:  ⟼ \: 2ax - 2by = - a - \: 4b \: \sf \:  ⟼ \: (1)$

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☆ Now consider,

$\bf \:  ⟼ \: 2(bx + ay) + (b - 4a) = 0$

can be rewritten as,

$\sf \:  ⟼ \: 2bx + 2ay = 4a - b \: \sf \:  ⟼ \: (2)$

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Now, multiply equation (1) by a and (2) by b, we get

$\sf \:  ⟼ \: {2a}^{2} x - 2aby = - {a}^{2} - 4ab \: \sf \:  ⟼ \: (3)$

$\sf \:  ⟼ \: {2b}^{2} x + 2aby = 4ab - {b}^{2} \: \sf \:  ⟼ \: (4)$

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☆ Now, on adding equation (3) and (4), we get

$\sf \:  ⟼ \: 2x( {a}^{2} + {b}^{2} ) = - {a}^{2} - {b}^{2}$

$\sf \:  ⟼ \: 2x \cancel{( {a}^{2} + {b}^{2} )} = - \cancel{( {a}^{2} + {b}^{2} )}$

$\bf\implies \: \: x = - \dfrac{1}{2}$

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On substituting the value of x in equation (1), we get

$\sf \:  ⟼ \: - \cancel{2}a (\dfrac{1}{ \cancel{2}}) - 2by = - a - \: 4b \: \sf \:$

$\sf \:  ⟼ \: - a - 2by = - a \: - 4b$

$\sf \:  ⟼ \: - 2by = - \: 4b \: \sf \:$

$\bf\implies \: \: y = 2$

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$\large{\boxed{\boxed{\bf{Hence, \: x \: = \: - \: \dfrac{1}{2} \: and \: y \: = \: 2}}}}$

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