Why does 2(y-2)-(3-y) =1 equal y=8/3?

Question

I don't understand why 2(y-2)-(3-y) =1 equals y=8/3
Please Help!
Thanks,
JJ.

Answer 1

you set up the problem 2(y-2)-(3-y)=1

then you distribute the 2 so you get
2y-4-(3-y)=1

then distribute the - you get
2y-4-3+y=1

move the numbers to the other side of the equal sign
2y+y=1+4+3

combine like terms
3y=8

divide both sides by 3
y=8/3

Answer 2

First factor the 2(y-2)
this is 2y - 2(2) or 2y - 4
Then by order of operations subtract the two binomials
2y - 2 - 3 +y (you get -3+y by factoring the negative sign in front of (3-y)
2y + y -2-3 = 1
3y - 5 = 1
3y = 6
y = 6/3
y = 2
2 is the right answer, 8/3 is wrong. If your teacher told you this answer, they are wrong. Hope this helps

Answer 3

2(y-2)-(3-y) =1 equal y=8/3

2(y-2)-(3-y) =1
First of all, multiply in the 2:
(2y-4)-(3-y) = 1
Expand the brackets:
2y-4 -3 -(-) y =1 --> - and - = +
therefore, 2y-4-3+y=1
Simplfy the equation
3y-7=1
separate y and numbers,
3y = 1+7
3y = 8
y = 8/3

Answer 4

2(y-2) - (3-y) =1
Distributive property 2(y-2) --> 2y-4
2y-4 -(3-y) = 1
Distributive property -(3-y) --> -3 + y
2y - 4 -3 + y =1
Combine like terms
3y -7 =1
Add 7 to both sides
3y = 8
Divide both sides by 3
y = 8/3

Answer 5

2y-4-3+y=1
3y-7=1
3y-7+7=1+7
3y=8
3y/3=8/3
y=8/3

Answer 6

2(y-2)-(3-y)=1
=>2y-4-(3-y)=1
=>2y-4-3+y=1
=>3y-7=1
=>3y=1+7
=>3y=8=>y=8/3