Help with algebra Problem.?

Multiply both sides by 14...

2y=7y-35

Then...

-5y=-35

So...

y=7

Edit: You multiply by 14 to get rid of the denominators, using the Least Common Denominator.

y/7 = y/2 - 5/2 now you have to move the y's to one side.

y/7 - y/2 = -5/2 now you have to make the denominators equal so I am going to use 14

2y/14 - 7y/14 = -5/2 2-7 is -5 so now you have

-5y/14 = -5/2 now you multipy both sides by 14 to give you

-5y = -5/2 x 14

-5/2 x 14 is -35

so -5y = -35 divide both sides by -5

y=7

hopefully I did this all right, Its been years since I even looked at problems like this.

edit to your additional comments. Well everyone else sure made it look easier then I did. We all used 14 because it was the LCD least common denominator. You have find the smallest number that both 7 and 2 can evenly go into. For example 2 and 7 both go into 28 but 14 is the smallest number. Like for 4 and 8 you would want to use 16 because both 4 and 8 can me put into 16 evenly

1. Cross Multiply. Multiply the Numerator by the Denominator.

2y= 7(y-5)

2. Distribute the 7.

2y = 7y-35

3. Isolate the Y. Subtract 7y to both sides.

-5y=-35

4. Divide both sides by -5 to get the y alone.

y=7

(y) / 7 = (y-5) / 2

multiply the equation by 14

2y = 7y - 35

- 5y = -35

y = 7

To solve for y:

2y=7y-35

5y=35

y=7

(y) / 7 = (y - 5) / 2

2(y) = 7(y - 5)

5y = 35

y = 7