Question

aflati a+2*b+3*c stiind ca a-b=30 și 3b+2c=50

Answer 1

a+2b+3c=?
a-b=30=>b=a-30
3b+2c=50=>2c=50-3b=>c=(50-3b)/2
                                        c=(50-3(a-30))/2=>c=(50-3a+90)/2=>c=(140-3a)/2
Inlocuim
a+2(a-30)+3(140-3a)/2=
=2a/2+4(a-30)/2+3(140-3a)/2=
=[2a+4(a-30)+3(140-3a)]/2=
=(4a+2a+420-9a-120)/2=
=[(420-120)+(2a+4a-9a)]/2=
=[-3a+(-120+420)]/2=
=(300-3a)/2=
=[3(100-a)]/2=
150-3a/2=0
300/2-3a/2=0
(300-3a)/2=0
300-3a=0
3a=300
a=300/3=100

inlocuim si aflam b
b=a-30
b=100-30
b=70
c=(140-3a)/2=(140-300)/2=-160/2=-80
a+2b+3c=?
100+2*70+3*(-80)=100+140-240=240-240=0
a+2b+3c=0