Question

Ahola...!!!


Solve the following pair of linear equations by cross-multiplication method :-

2 (ax-by)+a+4b=0 ,
2 (bx+ay)+b-4a=0 .


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Answer 1

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Answer 2

SOLUTION

Given,

$= > 2(ax - by) + a + 4b = 0 \\ = > 2(bx + ay) + b - 4a = 0 \\ = > to \: find \: the \: solution \: of \: the \: system \: by \: the \: method \: of \: cross \: multiplication \\ = > 2(ax - by) + a + 4b = 0 \\ = > 2(bx + ay) + b - 4a = 0 \\ = > after \: rewriting \: the \: equations \: we \: have \\ = > 2ax - 2by + (a + 4b) = 0 \\ = > 2bx + 2ay + (b - 4a) = 0 \\ \\ = > by \: cross \: multiplication \: method \: we \: get \\ = > \frac{x}{(b - 4a)( - 2b) - (a + 4b)(2a)} = \frac{ - y}{(b - 4a)(2a) - (a + 4b)(2b)} = \frac{1}{4a {}^{2} + b {}^{2} } \\ \\ = > \frac{x}{( - 2b {}^{2} + 8ab) - (2a {}^{2} + 8ab)} = \frac{ - y}{(2ab - 8a {}^{2}) - (2ba + 8b {}^{2}) } = \frac{1}{4a {}^{2} + 4b {}^{2} } \\ \\ = > \frac{x}{ - 2(b {}^{2} + a {}^{2} )} = \frac{y}{8(a {}^{2} + {b}^{2} )} = \frac{1}{4a {}^{2} + 4b {}^{2} } \\ \\ = > \frac{x}{ - 2(b {}^{2} + a {}^{2} )} = \frac{1}{4a {}^{2} + b {}^{2} } = x = - \frac{1}{2} \\ = > for \: y \: consider \: the \: following \\ = > \frac{y}{8( {a}^{2} + {b}^{2}) } = \frac{1}{4a {}^{2} + 4b {}^{2} } = y = 2 \\ \\ = > hence \: the \: value \: of \: x \: = - \frac{1}{2} \: \: and \: \: y = 2$

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