Question

b este dublul lui y din 1+{2×[3+(4+y):5]-6}×7=29

Answer 1

Sper ca te - am ajutat

Answer 2

 1+{2×[3+(4+y):5]-6}×7=29
 1+{2×[3+(4+y):5]-6}=29/7
 2×[3+(4+y):5]-6=29/7-1 aduc la acelasi numitor
2×[3+(4+y):5]-6=29/7-7/7
2×[3+(4+y):5]-6=22/7
2×[3+(4+y):5]=22/7+6 aduc la acelasi numitor
2×[3+(4+y):5]=22/7+42/7
2×[3+(4+y):5]=64/7
3+(4+y):5=64/7:2
3+(4+y):5=64/7 ori 1/2 pe (64 cu 2 se simplifica)
3+(4+y):5=32/7
(4+y):5=32/7-3 aduc la acelasi numitor
(4+y):5=32/7-21/7
(4+y):5=11/7
4+y=11/7 ori 5
4+y=55/7
y=55/7-4 aduc la acelasi numitor
y=55/7-28/7
y=27/7
b=2y=>b= 2 ori 27/7=>b=54/7