Question

Stiind ca x+3+y+4+z+5=12 determinati 2019+x+y+z

Answer 1

x+3+y+4+z+5 = 12

⇔ x+y+z + 12 = 12

⇔ x+y+z = 0

⇒ 2019+x+y+z = 2019+0 = 2019

Answer 2

Răspuns:

x+3+y+4+z+5=12

x+y+z=12-3-4-5=12-12=0 rezultă

2019+x+y+z=2019