方程式(移項題目)

1) 通分母

[5(2x-1) - 3(3-x)]/15 = 2

2) 15移項變 乘15

[5(2x-1) - 3(3-x)] = 2(15)

3) 展開

10x - 5 - 9 + 3x = 30
13x - 14 = 30

4) -14移項變 + 14

13x = 30 + 14 = 44

5) 13移項變 除13

x = 44/13

2x-1 3-x
------ -- ------- = 2
3 5

2x-1 3-x
------ x 15 -- ------- x 15 = 2 x 15
3 5


5 (2x-1) -- 3(3-x) = 30

10x -- 5 -- 9 + 3x = 30

13x = 30 + 5 + 9

13x = 49

x = 49
----
13

10
x = 3 ----
13

請check 一次題目有冇抄錯

(2x-1)/3 - (3-x)/5=2
[(2x-1)/3 - (3-x)/5]*3*5=2*3*5 (將左邊分數既分母約簡)
(2x-1)/3*3*5-(3-x)/5*3*5=30 (拆括號)
(2x-1)*5-(3-x)*3=30
(10x-5)-(9-3x)-30
10x-5-9+3x=30(拆括號,小心負負得正)
10x+3x-5-9=30(將同項 放左一起)
13x-14=30
13x=44
x=44/13 or 3又5/13

Multiple both sides by 15

then we get

5(2x-1) - 3(3-x) = 30

5x - 5 - 9 + 9x = 30

14x = 44

x = 44/14