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ax/b - by/a = a+b ax …

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ax/b - by/a = a+b ax - 2ab = by Solve the equations
  • 1 answers

Sia ? 6 years, 6 months ago

The given equations are
{tex}\frac { a x } { b } - \frac { b y } { a } - ( a + b ) = 0{/tex}
{tex}ax - by - 2ab = 0{/tex}
By cross multiplication, we have
{tex}\frac { x } { \left( - \frac { b } { a } \right) \times ( - 2 a b ) - ( - b ) \times ( - ( a + b ) ) } ={/tex}{tex}\frac { y } { - ( a + b ) \times a - ( - 2 a b ) \times \frac { a } { b } }{/tex}{tex}= \frac { 1 } { \frac { \mathrm { a } } { \mathrm { b } } \times ( - \mathrm { b } ) - \mathrm { a } \times \left( - \frac { \mathrm { b } } { \mathrm { a } } \right) } {/tex}
{tex}\Rightarrow \frac { x } { 2 b ^ { 2 } - b ( a + b ) } = \frac { y } { - a ( a + b ) + 2 a ^ { 2 } } = \frac { 1 } { - a + b }{/tex}
or, {tex}\frac { x } { 2 b ^ { 2 } - a b - b ^ { 2 } } = \frac { y } { - a ^ { 2 } - a b + 2 a ^ { 2 } } = \frac { 1 } { - a + b }{/tex}
{tex}\Rightarrow \frac { x } { b ^ { 2 } - a b } = \frac { y } { a ^ { 2 } - a b } = \frac { 1 } { - ( a - b ) }{/tex}
{tex}\Rightarrow \frac { x } { - b ( a - b ) } = \frac { y } { a ( a - b ) } = \frac { 1 } { - ( a - b ) }{/tex}
{tex}\therefore \frac { x } { - b ( a - b ) } = \frac { 1 } { - ( a - b ) }{/tex} and {tex}\frac { y } { a ( a - b ) } = \frac { 1 } { - ( a - b ) }{/tex}
{tex}\therefore x = \frac { - b ( a - b ) } { - ( a - b ) } \text { and } y = \frac { a ( a - b ) } { - ( a - b ) }{/tex}
{tex} \Rightarrow{/tex} {tex}x = b,\ and\ y = -a{/tex}
{tex}\therefore{/tex} the solution is {tex}x = b, y = -a{/tex}

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