# Math Help - Ugh. I'm drawing a blank.

1. ## 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!

2. 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:

$A + 2B = 0$

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
you had A = (9+7C)/6
let C = 3 then
A = (9+7*3)/6 = 30/6 = 5
so if C = 3, then A = 5
or if C = -9
A = (9 + 7*(-9))/6 = (-54)/6 = -9
so if C = -9, A = -9.
so you cannot solve for one unique solution of A and C, only a relationship between the two, so that whatever let one variable be, you can find a solution for the other one.

3. Aha. I kinda...was hoping for that answer that it couldn't be "solved" without more information. Thank you so much for you help.