Question

Ma ajuta cineva la exercitiul asta : 1,5x+ 0,(3)y= 1,8 ; -3x+0,(6)y=1,6

Answer 1

Răspuns:

$\left \{ {{1,5x+0,(3)y=1,8} \atop {-3x+0,(6)y=1,6}} \right.$

$\left \{ {{\frac{15x}{10}+\frac{3y}{9}=\frac{18}{10}} \atop {-3x+\frac{6y}{9}=\frac{16}{10}} } \right.$

$\left \{ {{\frac{3x}{2}+\frac{y}{3}=\frac{9}{5}} /*30 \atop {-3x+\frac{2y}{3}=\frac{8}{5}} /*15 } \right.$

$\left \{ {{45x+10y=54} \atop {-45x+10y=24}} \right.$ |+

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   /      +20y=74  /:2   10y=37 ⇒ y=37/10= 3 7/10

45x+10·37/10=54     45x+37=54    45x=17   x=17/45

Explicație pas cu pas: