中一一元一次方程(急)(要列式)(4題)(15點)

您好，我是 lop，高興能解答您的問題。

(1)

5z + 7 = 3z - 11
5z - 3z = -11 - 7
2z = -18
z = -18 ÷ 2
z = -9

(2)

(4q+2) ÷ 2 = q + 4
4q + 2 = 2(q+4)
4q + 2 = 2q + 8
4q - 2q = 8 - 2
2q = 6
q = 6 ÷ 2
q = +3

(3)

Case 1 :

(10-a) ÷ 3 = (a+10) ÷ 2
6(10-a) ÷ 3 = 6(a+10) ÷ 2
2(10-a) = 3(a+10)
20 - 2a = 3a + 30
20 - 30 = 3a + 2a
-10 = 5a
a = -10 ÷ 5
a = -2

Case 2 :

10 - a÷3 = (a+10) ÷ 2
6(10-a÷3) = 6(a+10) ÷ 2
60 - 2a = 3(a+10)
60 - 2a = 3a + 30
60 - 30 = 3a + 2a
30 = 5a
a = 30 ÷ 5
a = +6

(4)

(1-a) ÷ 8 = (a-12) ÷ 3
24(1-a) ÷ 8 = 24(a-12) ÷ 3
3(1-a) = 8(a-12)
3 - 3a = 8a - 96
3 + 96 = 8a + 3a
99 = 11a
a = 99 ÷ 11
a = +9

1. 5z-3z=-11-7
2z=-18
z=-9


2. 4q+1(因為2/2=1)=q+4
4q-q=4-1
3q=3
q=1

3.10-(a/3)=a+5
10-a/3=a+5
10-5=a+a/3
5=4a/3
5x3=4a
15=4a
a=3.75


4. 1-(a/8)=a-4
1x8-(a/8)x8=ax8-4x8
8-a=8a-32
40=9a
a=4.444444




第4題不太確定.如有大大計到整數,請通知小弟

1. 5z+7 = 3z-11
5z-3z=-11-7
2z = -18
z=-9
2. 4q+2÷2 = q+4
4q+1= q+4
4q-q=4-1
3q=3
q=1