### Solve the following pair of linear equations by using cross multiplication. \begin{aligned} &u x - v y = w \\ &v x - u y = 1 - w \end{aligned}

Answer:

$x = \dfrac { v } { (v - u ) ( v + u ) } - \dfrac { w } { v - u } \text{ and } y = \dfrac { u } { (v - u ) ( v + u ) } - \dfrac { w } { u - v }$

Step by Step Explanation:
1. The given system of equations can be written as \begin{aligned} &u x - v y = w {\space} {\space} &&{\implies} {\space}{\space} u x - v y - w = 0 \\ &v x - u y = 1 - w {\space} {\space} &&{\implies} {\space}{\space} v x - u y - ( 1 - w) = 0 \end{aligned}
2. By cross-multiplication, we have \begin{aligned} & $$\dfrac { x } { \begin{matrix} - v \\ - u \end{matrix} { \rlap {\nearrow} {\searrow}} \begin{matrix} -w \\ -(1 - w) \end{matrix} } = \dfrac { -y } { \begin{matrix} u \\ v \end{matrix} \rlap {\nearrow} {\searrow} \begin{matrix} -w \\ -(1 - w)\end{matrix} } = \dfrac { 1 } { \begin{matrix} u \\ v\end{matrix} \rlap {\nearrow} {\searrow} \begin{matrix} - v\\ - u \end{matrix} }$$ \\ {\implies} & \dfrac { x } { v(1 - w) - u w } = \dfrac { -y } { -u(1 - w) + vw} = \dfrac { 1 } { - u^2 + v^2 } \\ {\implies} & \dfrac { x } { v - vw - u w } = \dfrac { -y } { -u + uw + vw } = \dfrac { 1 } { - u^2 + v^2 } \\ {\implies} & \dfrac { x } { v - w(v + u ) } = \dfrac { -y } { -u + w( u + v ) } = \dfrac { 1 } { (v - u ) ( v + u ) } \\ {\implies} & x = \dfrac { v - w(v + u ) } { (v - u ) ( v + u ) } \text{ and } y = \dfrac { u - w( u + v ) } { (v - u ) ( v + u ) } \\ {\implies}& x = \dfrac { v } { (v - u ) ( v + u ) } - \dfrac { w } { v - u } \space \space \text{ and } \space \space y = \dfrac { u } { (v - u ) ( v + u ) } - \dfrac { w } { u - v } \end{aligned}
3. Hence, the value of $x$ is $\dfrac { v } { (v - u ) ( v + u ) } - \dfrac { w } { v - u }$ and the value of $y$ is $\dfrac { u } { (v - u ) ( v + u ) } - \dfrac { w } { u - v } .$

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