Question

Se dau următoarele relații: x+y+z=112; x+y=98; y=3×z. Aflati x,y,z

Answer 1

98+z=112
z=112-98
z=14
y=3z=3x14
y=42
x+42+14=112
x+56=112
x=112-56
x=56

Answer 2

$x+y+z=112 <=> (x+y)+z=112<=>98+z=112=> \\ =>z=112-98=14. \\ y=3*z<=>y=3*14=42. \\ x+y=98<=>x+42=98=>x=56. \\ \\ Solutie:~x=56,~y=42~si~z=14.$