Question

Rozwiązać równania :
2(2#-3)+5[2*(9#-4)-(#+1)]= 24
#- niewiadoma
2#-(3/4# + 3)+#-1/4= 1i1/4
3/4 i 1/4 to są ułamki zwykłe

Answer 1

2(2#-3)+5[2*(9#-4)-(#+1)]= 24
4# - 9 + 5(18#-8-#-1) = 24
4# - 9 + 90# - 40 -5# - 5 = 24
89# = 24+54
89# = 78
# = 78/89

2#-(3/4# + 3)+#-1/4= 1i1/4
2# - 3/4# + 3 + # - 1/4 = 5/4
3# - 3/4# = 5/4 + 1/4 + 3
12/4# - 3/4# = 6/4 + 12/4
9/4# = 18/4
# = 18/9
# = 2

Answer 2

2(2#-3)+5[2*(9#-4)-(#+1)]= 24
4x-6+5(18x-8)-x-1=24
4x-6+75x-40-x-1=24
4x+75x-x=24+47
78x=71/78
x=71/78


2#-(3/4# + 3)+#-1/4= 1i1/4
2x-3/4x+3+x-1/4=1¼
-2,25x=-1,5/-2,25
x≈0,66

Answer 3

2(2x-3)+5[2×(9x-4)-(x+1)]= 24
4x - 6 + 5[18x - 8 - x - 1] = 24
4x - 6 + 90x - 40 - 5x -5 = 24
4x + 90x - 5x = 24 + 5 + 40 + 6
89x = 75
x = ⁷⁵/₈₉

2x-(¾x + 3)+ x - ¼= 1¼
2x - ¾x - 3 + x - ¼= 1¼
2x - ¾x + x = 1¼ +3 +¼
2¼x = 4½
x = 2