Question

1.A,b
A=(b+200)-56
A:2=37+b

2.Sa se determine patru nr. A caror suma este cu 42 mai mare decat al patrulea număr,iar primele 3 nr. Sunt consecutive pare

Answer 1

1)
A=(b+200)-56
A=b+144
A:2=37+b
(b+144)/2=37+b
2(37+b)=b+144
74+2b=b+144
2b-b=144-74
b=70
A=70+200-56
A=214
2)
a+b+c+d=d+42
a+b+c=42
a=b-2=c-4
b=a+2
c=b+2
c=a+2+2
a+a+2+a+4=42
3a+6=42
3a=42-6
3a=36
a=36/3
a=12
b=12+2=14
c=14+2=16
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