Mauritius
Solve the following pair of linear equations by the method of cross-multiplication.
@^
\begin{aligned}
a x - b y = a + b \\
b x + a y = b - a
\end{aligned}
@^
Answer:
^@ x = 1 \space \space \text{and} \space \space y = -1 ^@
- The given pair of equations is @^ \begin{aligned} a x - b y = a + b {\space}{\space}{\space}{\implies} a x - b y - (a + b) = 0\\ b x + a y = b - a {\space}{\space}{\space}{\implies} b x + a y - (b - a) = 0 \end{aligned} @^
- By cross-multiplication, we get @^ \begin{aligned} & \begin{equation} \dfrac { x } { \begin{matrix} - b \\ a\end{matrix} { \rlap {\nearrow} {\searrow}} \begin{matrix} - (a + b) \\ - (b - a) \end{matrix} } = \dfrac { -y } { \begin{matrix} a \\ b \end{matrix} \rlap {\nearrow} {\searrow} \begin{matrix} - (a + b) \\ - (b - a)\end{matrix} } = \dfrac { 1 } { \begin{matrix} a \\ b\end{matrix} \rlap {\nearrow} {\searrow} \begin{matrix} - b\\ a\end{matrix} } \end{equation} \\ &{\implies} \dfrac { x } { - b {\times}- (b - a) - a {\times} - (a + b)} = \dfrac { -y } {-(b - a) {\times} a + (a + b) {\times}b } = \dfrac { 1 } { a{\times}a - b {\times} (-b) }\\ &{\implies} \dfrac { x } { b (b - a) + a (a + b)} = \dfrac { -y } {- a (b - a)+ b(a + b) } = \dfrac { 1 } { ( a^2 + b^2) } \\ &{\implies} \dfrac { x } { b^2 - b a + a^2 + ab } = \dfrac { -y } { -ab + a^2 + ba + b^2 } = \dfrac { 1 } { (a^2 + b^2) } \\ &{\implies} \dfrac { x } { ( b^2 + a^2 ) } = \dfrac { -y } { ( b^2 + a^2 )} = \dfrac { 1 } { (a^2 + b^2) }\\ &{\implies} \dfrac { x } { ( b^2 + a^2 ) } = \dfrac { 1 } { (a^2 + b^2) } \space \space \text{and} \space \space \dfrac { -y } { ( b^2 + a^2 )} = \dfrac { 1 } { (a^2 + b^2) }\\ &{\implies} x = \dfrac { ( b^2 + a^2 ) } { (a^2 + b^2) } \space \space \text{and} \space \space y = -\dfrac{ ( b^2 + a^2 ) } { (a^2 + b^2) }\\ &{\implies} x = 1 \space \space \text{and} \space \space y = -1 \\ \end{aligned} @^
- Hence, the required solution is ^@ x = 1^@ and ^@y = -1^@.
