數學代數負分數問題

1.
4-6(3-y)=-3(-6y+3)
4-6(3)-(-6)(y) = -3(-6y) + (-3)(3)
4-18+6y = 18y + (-9)
-14 +6y = 18y -9
-14 +9 = 18y -6y
-5 = 12y
y = -5/12

(Prove
4-6(3-y)=-3(-6y+3)
4-6(3-(-5/12))=-3(-6(-5/12)+3)
4-18-(-6)(-5/12) = (-3)(5/2) + (-3)(3)
-14 - 2.5 = -7.5 -9
-16.5 = -16.5 (OK))


2.
1/3 - 5y-3/4 = -7y/3
(-12)(1/3) -(-12)(5y) -(-12)(3/4) = (-12)(-7y/3) (<--全部乘-12)
-4 + 60y +9 = 28y
32y = -5
y = -5/32

(Prove
1/3 - 5y-3/4 = -7y/3
1/3 - 5(-5/32)-3/4 = -7(-5/32)/3
1/3 + 25/32 -3/4 = 35/96
32/96 + 75/96 - 72/96 = 35/96
35/96 = 35/96 (OK))

無名氏：唔該哂thx~~~~

定理一：A-(B-C)=A-B+C
定理二：K(A-B)=KA-KB
定理三：-B+A=A-B

Q1) 4-6(3-y)=-3(-6y+3)
根據定理二，6(3-y)=6*3-6*y=18-6y
根據定理三，-6y+3=3-6y
根據定理二，-3(3-6y)=-3*3-(-3)6y=-9+18y=18y-9
Q1)4-(18-6y)=18y-9
根據定理一，4-(18-6y)=4-18+6y=-14+6y
根據定理三，-14+6y=6y-14
Q1)6y-14=18y-9
6y-14+14=18y-9+14
6y=18y-5
0=12y-5
12y-5=0
12y=5
y=5/12

Q2)1/3-5y-3/4=-7y/3
1/3-3/4-5y=-7y/3
(4-9)/12-5y=-7y/3
-5/12-5y=(-7/3)y
-5/12=(-7/3)y+5y
-5/12=(5-7/3)y
-5/12=(8/3)y
(-5/12)/(8/3)=y
y=-5/12/8*3
=-5/32