Question

Stiind ca x+4*y+2*z=13 si 3*x+2*z=11 determinati x+y+z

Answer 1

$x+4y+2z=13\\3x+2z=11\\(x+4y+2z)+(3x+2z)=13+11\\4x+4y+4z=24\\4*(x+y+z)=24\\x+y+z=\frac{24}{4}\\x+y+z=6\\Sfarsit!$

Answer 2

x+4*y+2*z=13
3*x+2*z=11
x+4*y+2*z+3*x+2*z=13+11
4*x+4*y+4*z=24
4*(x+y+z)=24
 x+y+z=24:4
 x+y+z=6