Question

Rezolvati prin metoda substitutiei si reducerii sistemul de ecuatii:

x-2y=0
4x+3y-55=0

Dau coroana si puncte va rog mult

Answer 1

$\left \{ {{x-2y=0} \atop {4x+3y-55=0}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {4x+3y-55=0}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {4*2y+3y-55=0}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {8y+3y-55=0}} \right.$  -> $\left \{ {{x=2y} \atop {11y=55}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {y=55:11}} \right. -\ \textgreater \ \left \{ {{x=2y} \atop {y=5} \right. -\ \textgreater \ \left \{ {{x=2*5} \atop {y=5}} \right. -\ \textgreater \ \left \{ {{x=10} \atop {y=5}} \right.$  -> asta e metoda substitutiei.    A doua metoda ,metoda reducerii -> $\left \{ {{x-2y=0} \atop {4x+3y=55}} \right.$ inmultim prima ecuatie cu (-4) si rezulta $\left \{ {{-4x+8y=0} \atop {4x+3y=55}} \right.$  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 .