# Ugh. I'm drawing a blank.

• May 29th 2006, 12:46 AM
Foolish
Ugh. I'm drawing a blank.
Okay. I just need a refresher on this stuff.

6A - 7C = 9

Here's what I'm doing:

6A - 7C + 7C = 9 + 7C
6A = 9 + 7C
A=(9 + 7C)/6

So far so good, right?

Then substituting:

6((9 + 7C)/6) - 7C = 9

Now when I do this I cancel out the "6"'s...maybe that's wrong?
9 + 7C - 7C = 9

7C - 7C = 9 - 9

...okay...I just did some stuff and found that 7C = 9...and that seems to fit... I solved it but...other than just taking a potshot like I did...

How can I go from 7C - 7C = 9 - 9 and get 7C = 9 correctly?

Am I even doing the equation right altogether?

It's been a little while since I've done this math but I wanted to kinda get back up to speed. Thanks in advance for any help!
• May 29th 2006, 02:15 AM
congratulations, all your steps were perfect, however you missed the point.

you have an equation to solve for two variables, however you only have one equation. consider:

\$\displaystyle A + 2B = 0\$

we cannot solve for A and B as exact numbers, for instance whatever A is, we can find a value for B so that the equality is true:

let A = 2
then
2B = -A
2B = -2
B = -1

or if we let A be 7
2B = -7
B = -3.5

so there are infinitely many solutions that satisfy A + 2B = 0
likewise there are infinitely many solutions which satisfy 6A - 7C = 9