[tex]\displaystiles\\\begin{cases}x\Big(2\sqrt{5}-3\Big)+y\sqrt{8}=3(1-x)-21\\y\sqrt{45}+y\Big(\sqrt{2}-2\Big)=2(1-y)-41\\\end{cases}\\\\\\\begin{cases}2\sqrt{5}x-3x+2\sqrt{2}y=3-3x-21\\3\sqrt{5}x+\sqrt{2}y-2y=2-2y-41\\\end{cases}\\\\\\\begin{cases}2\sqrt{5}x-3x+2\sqrt{2}y=-3x-18~~~\Big|+3x\\3\sqrt{5}x+\sqrt{2}y-2y=-2y-39~~~\Big|+2y\\\end{cases}[/tex]
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[tex]\displaystiles\\\begin{cases}2\sqrt{5}x+2\sqrt{2}y=-18\\3\sqrt{5}x+\sqrt{2}y=-39\\\end{cases}\\\\\\\text{Facem urmatoarele substitutii:}\\\\\boxed{\sqrt{5}x=a}\\\\\boxed{\sqrt{2}y=b}\\\\\\\begin{cases}2a+2b=-18~~~\Big|:(-2)\\3a+b=-39\\\end{cases}\\\\\\\begin{cases}-a-b=9\\3a+b=-39\\\end{cases}\\\\\\\text{Adunam ecuatiile.}\\\\2a=-30\\\\a=\dfrac{-30}{2}\\\\\boxed{a=-15}\\\\ 3a+b=-39\\\\3\cdot(-15)+b=-39\\\\-45+b=-39\\\\b=-39+45\\\\\boxed{b=6}[/tex]
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[tex]\displaystyle\\\text{Ne intoarcem la substitutii.}\\\\\boxed{\sqrt{5}x=a}~~~\text{si}~~~\boxed{a=-15}\\\\\implies~\sqrt{5}x=-15\implies~x=\frac{-15}{\sqrt{5}}=\frac{-15\sqrt{5}}{5}\implies~\boxed{\boxed{x=-3\sqrt{5}}}\\\\\\\boxed{\sqrt{2}y=b}~~~\text{si}~~~\boxed{b=6}\\\\\implies~\sqrt{2}y=6\implies~y=\frac{6}{\sqrt{2}}=\frac{6\sqrt{2}}{2}\implies~\boxed{\boxed{y=3\sqrt{2}}}[/tex]
