Question

How do you solve `4(2y-1)=4(y+3)` for y?

Answer 1

Less thought, more mechanical: first remove the parentheses on both sides by distributing the multiplication. Then collect all the terms involving `y` on one side (say the left) and the terms without `y` on the right. Finally, divide both sides by the coefficient of `y`. (the number in front of `y`).

It looks like this:

`4(2y-1)=4(y+3)`

`8y-4=4y+12`

`8y-4y=12+4` (subtract `4y` from both sides and add `4` to both sides)

`4y=16`

`(4y)/4=16/4`

`y=4`

Check your answer.

More thought, looking at the equation:

Instead of distributing the `4`'s on both sides, you could start by dividing both sides by `4` (multiplying by `1/4`. This keeps the numbers smaller. It looks like this:
`4(2y-1)=4(y+3)`

`1/4 [4(2y-1)] = 1/4 [4(y+3])`

#1/4*4 = [1/4 *4] (y+3)

`2y-1=y+3`

`2y-y=3+1` (subtract `y` and add `1` on both sides)

`y=4`

Check your answer.

Answer 2

First multiply `4` to get rid of the brackets:
`8y-4=4y+12`
Now, isolate the `y`s on one side and numbers on the other of the `=` sign (remembering to change sign in crossing the `=` sign);
`8y-4y=+4+12`
Notice the change of sign of `4y` and `-4`;
now we can add and subtract:
`4y=16`
The `4` is multiplying the `x` and can go to the right dividing to get:
`x=16/4=4`

hope it helps!