Question

Daca x+2y=17 si y-32=4, atunci calculati:
a. x+3y-3z;
b. 2x+5y-3z;
c. (3x+6y) + (2y-6z);
d. (2x+4y) • (x+4y-6z).

Va rog ajutati-ma!! Multumesc!!

Answer 1

din y-32=4  ⇒  y=36   iar din  x+2*y=17 ⇒ x=17-2*36=17-72= -55 adica: x= -55 ;
a)... x+3y-3z = -55 + 3*(-55) - 3z = -220-3z ;
b)... 2*(-55)+5*36-3z = -110+165-3z=55-3z ;
c)... [3*(-55)+4*36]+(2*36-6z)=-165+144+72-3z=51-3z ;
d)... [2*(-55)+4*36]*(-55+4*36-6z)=(34)*(89-6z)=3026-204z ;