Question

exercitiul 5 : gaseste valoare lui a care face adevarata fiecare relatie : a) (a ori 2 -40 ):23= 12      b) [(a-416:4+2)ori 5 +3] ori 2 = 1986 c) 5ori a -2000:50-(1291-1091)=40  

Answer 1

a) (a x2 -40 ):23= 12
(ax2-40)=12x23
ax2-40=276
ax2=276+40
ax2=316
a=316:2
a=158

b) [(a-416:4+2)x 5 +3]x 2 = 1986
(a-104+2)x 5 +3 = 1986:2
(a-106)x 5 +3 = 993
(a-106)x 5= 993-3
(a-106)x 5= 990
a-106= 990:5
a-106= 198
a=198+106
a=304

c) 5x a-2000:50-(1291-1091)=40
5xa-40-200=40
5xa-40=40+200
5xa-40=240
5xa=240+40
5xa=280
a=280:5
a=56