Expert: Scott A Wilson - 10/19/2009

Question
Hello Scott! : ) So I'm a little stuck on what to do after a step. I don't expand the brackets, but if you could show me that way too, that would be great.

I'm trying to rearrange it so the subject is on the other side. So it's like, X = ....

[subject = y]
w = a(mx - by)
w/a = mx - by
w/a - mx = by

So... what do I do after that? I have no idea :S
My friend divided something by something then cancelled, but I have no idea what she did.

Here's another one:
[subject = b]
mp - wx = 3(tb + c)
mp - wx/3 = tb +c
mp - wx/3 - c = tb

and yeah ... I get stuck again...
Am I even doing this right? d:

Oh, and one more quick question -
when you're rearranging, do you follow bedmas or bedmas backwards?

Thank you very much ! : )

Answer
This first note is about

[subject = y]
w = a(mx - by)
This is not right: w/a = mx - by;       Right: w/a - mx = by
Subtracting mx from both sides gives w/a - mx = -by,
so by = mx - w/a, so y = (mx-w/a)/b.

This can be checked by putting the solution back in the equation at the top.

When this is done, we get w = a(mx - b((mx-w/a)/b).

Notice that we have a b/b, so cancel these, giving  w = a(mx - (mx-w/a).

Next, carry out the - on the inside parenthesis, giving w = a(mx - mx + w/a).

Next, notice that mx-mx is 0, so w = a(w/a).

Last, notice that a/a is 1, so all that is left is w = w,
which means our solution for y was correct.


Here's another one:
[subject = b]
mp - wx = 3(tb + c)
here: mp - wx/3 = tb +c: should be (mp - wx)/3 = tb + c
here: mp - wx/3 - c = tb; should be (mp - wx)/3 - c = tb

The last thing to do is to divide by b, so we get t = ((mp - wx)/3 - c)/b.

I have never heard of bedmas.