Question

Să se determine x prin metode mersului înapoi : a). (x+200):2-100=50 ; b). (x-300):3+150=450; c). (x+100)×2+32×15=2800; d). [(x-300)×2-150]+400=1000; e). [(x-160)-260]-360=460; f). (x+362)-162=200; g). (x+381)+42+56=1000?

Answer 1

a)(x+200):2-100=50
(x+200):2=100+50
(x+200):2=150
x+200=150*2
x+200=300
x=300-200
x=100
b)(x-300):3+150=450
(x-300):3=450-150
(x-300):3=300
x-300=300*3
x-300=600
x=600+300
x=900
c)(x+100)*2+32*15=2800
(x+100)*2+480=2800
(x+100)*2=2800-480
(x+100)*2=2320
x+100=2320:2
x+100=1160
x=1160-100
x=1060
d)[(x-300)*2-150]+400=1000
(x-300)*2-150=1000-400
(x-300)*2-150=600
(x-300)*2=600+150
(x-300)*2=750
x-300=750:2
x-300=375
x=300+375
x=675
e)[(x-160)-260]-360=460
(x-160)-260=460+360
(x-160)-260=820
x-160=820+260
x-160=1080
x=1080+160
x=1240
f) (x+362)-162=200
x+362=200+162
x+362=362
x=362-362
x=0
g)(x+381)+42+56=1000
(x+381)+42=1000-56
(x+381)+42=944
x+381=944-42
x+381=902
x=902-381
x=521