Solve for x and y given …
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Sia ? 6 years, 5 months ago
Taking {tex} \frac { 1 } { x } = u{/tex} and {tex} \frac { 1 } { y } = v.{/tex}The given system of equations become
{tex} 2 u + \frac { 2 } { 3 } v = \frac { 1 } { 6 }{/tex}
Therefore, {tex} 12u+4v=1{/tex}............(i)
and, {tex}3u+2v=0{/tex}..........(ii)
Multiplying (ii) by 2 and subtracting from (i), we get
{tex} 6 u = 1 \Rightarrow u = \frac { 1 } { 6 }{/tex}
Putting {tex} u = \frac { 1 } { 6 }{/tex}in (i), we get
{tex} 2 + 4 v = 1 \Rightarrow v = - \frac { 1 } { 4 }{/tex}
Hence, {tex} x = \frac { 1 } { u } = 6{/tex} and {tex} y = \frac { 1 } { v } = - 4{/tex}
So. the solution of the given system of equations is {tex}x=6,y=-4{/tex}
Putting x = 6, y = -4 in {tex}y=ax-4{/tex}, we get
{tex}-4=6a-4{/tex}
{tex} \Rightarrow a=0{/tex}
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