what is 3(y-2)=1-3y over 5?

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

Multiply both sides by 5.
15(y - 2) = 1 - 3y

Distribute.
15y - 30 = 1 - 3y

Isolate 'y'.
18y = 31

Divide.
y = 31 / 18

3(y - 2) = 1 - 3y
3*y - 3*2 = 1 - 3y
3y + 3y = 1 + 6
6y = 7
y = 7/6

3(y-2)=1-3y/5
3y-6=1/5-3y/5 [Collect The Like Terms]
3y+3y/5=1/5+6 [Make Same Denominator Then Add]
18y/5=31/5 [Cross Multiplication]
90y=155 [Divide By 5]
18y=31 [Divide By 18]
y=31/18

multiply by 5 to get rid of the fraction and get

15(y-2) = 1-3y

add 3y to gather the 'y's together

18y - 30 = 1

add 30 to isolate y

18y = 31

divide by 18 to solve for y

y = 31/18

done

3y - 6 = 1-3y/5
6y = 6 1/5
y= 1.03 (recurring) OR 1 1/6



3 (y-2) = 1-3y / 5
3y - 6 = 1-3y / 5
6y - 6 =1/5
6y = 6 + 1/5
y = 1.03 (recurring) OR 1 1/6

Or if you meant (3(y-2)=1-3y)/5 It is 31/18

15y-30=1-3y
18y=31
y=31/18

Use distributive property to solve the left side and you get 3y-6

3y-6=1-3y/5

Multiply both sides by 5 to clear out that 5 in the denom.

5(3y-6)=1-3y

15y-30=1-3y

18y=31

y=31/18 or 1 13/18