[tex] \left \{ {{x-2y=0} \atop {4x+3y-55=0}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {4x+3y-55=0}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {4*2y+3y-55=0}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {8y+3y-55=0}} \right. [/tex] -> [tex] \left \{ {{x=2y} \atop {11y=55}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {y=55:11}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {y=5} \right. -\ \textgreater \ \left \{ {{x=2*5} \atop {y=5}} \right. -\ \textgreater \ \left \{ {{x=10} \atop {y=5}} \right. [/tex] -> asta e metoda substitutiei. A doua metoda ,metoda reducerii -> [tex] \left \{ {{x-2y=0} \atop {4x+3y=55}} \right. [/tex] inmultim prima ecuatie cu (-4) si rezulta [tex] \left \{ {{-4x+8y=0} \atop {4x+3y=55}} \right. [/tex] 4x si -4x se reduc -> 8y+3y=55 ,11y=55 | impartim la 11 -> y =5 Acum inlocuim in prima ecuatie x-2y=0 ,y=5 -> x-2*5 =0 x-10=0 x= 0 +10 -> x =10 .IN concluzie x=10 si y =5 .
