\(\displaystyle \begin{array}{rlrl}2 x+5 y & =\frac{15}{\lambda} & 5 x+2 y=\frac{20}{\lambda} \\\\ x & =\frac{10}{3 \lambda} & y =\frac{5}{3 \lambda}\end{array} \)
\(1.) \) \(2 x+5 y =\frac{15}{\lambda}\)
\(2.) \) \(5 x+2 y =\frac{20}{\lambda}\)
\(5* 1.):\) \(10 x+25 y =\frac{75}{\lambda}\)
\(2* 2.):\) \(10 x+4 y =\frac{40}{\lambda}\)
\( 21y=\frac{35}{\lambda} \) → \( y=\frac{5}{3\lambda} \)
\(1.) \) \(2 x+5 *\frac{5}{3\lambda} =\frac{15}{\lambda}\)
\(1.) \) \(2 x+\frac{25}{3\lambda} =\frac{45}{3\lambda}\)
\(1.) \) \( x =\frac{10}{3\lambda}\)
