Question

daca x plus 2y egal 17 si y - 3z egal 4 , atunci calculați :a) x plus 3y-3z ; b) (3x plus 6y) plus (2y-6z) ;c) 2x plus 5y -3z ;d) (2x plus 4y) ori (x plus 4y -6z)

Answer 1

x+2y=17
y-3z=4
a)x+3y-3z=x+2y+y-3z=17+4=21
b)(3x+6y)+(2y*6z)=3*(x+2y)+2*(y-3z)=3*17+2*4=51+8=59
c)2x+5y-3z=2x+4y+y-3z=2(x+2y)+(y-3z)=2*17+4=34+4=38
d)(2x+4y)*(x+4y-6z)=2*(x+2y)*(x+2y+2y-6z)=2*17-(17+y+y-3z-3z)=34*(17+y-3z+y-3z)=34*(17+4+4)=34*25=850