一條F.1 maths 題,急~~

Very Easy ga!
只有1個terms,所以可以約簡.
↓
5x[2/5(2y+3)]=(y+1)x5 ←有2個terms,所以唔可以就禁乘,要分開!

2(2y+3) =5y+5

(Break 咗括號)
↓
4y+6 =5y+5
4y+6-5 =5y+5-5
4y+1 =5y
4y+1-4y =5y-4y
1 =y
y =1
The answer is: 1
Hope it can help you!
if you have any mats problem,u can send e-mail to me or add my msn:[email protected]

2/5(2y+3)=y+1
(*5) 2(2y+3)=5y+5
4y+6=5y+5
1=y
Y=1

2/5(2y+3)=y+1
5x2/5(2y+3)=(y+1)x5
2(2y+3)=5y+5
4y+6=5y+5
4y-5y=5-6
-y=-1
y=1

2/5x2y+2/5x3=y+1
4/5y+6/5=y+1
6/5-1=y-4/5y
0.2=0.2y
0.2/0.2=y
y=1

2/5(2y+3)=y+1
4/5y+6/5=y+1
4/5y-y=1-6/5
4/5y-5/5y=5/5-6/5
-y/5=-1/5
-y=-1/5*5
-y=-1
y=1

2/5(2y+3)=y+1
2(2y+3) =(y+1)(5)
2(2y+3) =5y+5
4y+6 =5y+5
4y-5y =5-6
-y =-1
y =1

2/5(2y+3)=y+1
10(2y+3)=y+1
20y+30{←10(2y+3),10x2=20+3x10=30}=y+1
20y+30=y+1
y-20y=1-30
-19y=-29
19y=29←2個negetive=positive
19y=29
------ ------ ← faction
19 19
y=29
----- ← faction
19

2/5(2y+3)=y+1
2y+3=5/2(2y+3)
2y+3=5/2y+5/2
2y-5/2y=5/2-3
-1/2y=-1/2
y=1