Question

How do you solve for `b` in `(b-c+d)/3=a`?

Answer 1

`:. b = 3a + c-d`

Explanation:

`(b-c+d)/3 = a`

Multiply both sides by 3:

`:. 3(b-c+d)/3 = 3a`
We can then cancel the 3
`:. cancel(3)(b-c+d)/(cancel(3)) = 3a`
`:. b-c+d = 3a`

Now add `c-d` to both sides:
`:. b-c+d + c-d= 3a + c-d`

The `c` and `d` will then also cancel
`:. b cancel(-c)+cancel(color(red)d) + cancel(c)-cancel(color(red)d)= 3a + c-d`
`:. b = 3a + c-d`