Solve:

Question

2(4-4)+3r-4/2+4r-12+13s+5r+5(9-r)-6r+2(-4)-5(2s)+2(9+3r+2-3s = 57

Answer 1

you have two variables, and unmatched parentheses.

You want to solve for r or s?

Answer 2

Oh, I forget to close a bracket. It's (9+3r)+2-3s

Answer 3

still, you want r or s?

Answer 4

I believe it is for both, since it says two of them on my paper. It could be that I just add a number to substitute the variable.

Answer 5

ah. never mind. After you clear away all the noise, you are left with

7r+43 = 57
7r = 14
r = 2

the value of s does not matter; it can be anything.

Answer 6

Could you explain how you got rid of the other numbers?

Answer 7

sure.

Note that 4-4=0, so that term goes away. Also, 4/2 = 2. Then you have

3r- 4/2 +4r-12+13s+5r+5(9-r)-6r+2(-4)-5(2s)+2(9+3r)+2-3s = 57

remove the parentheses:

3r-2+4r-12+13s+5r+45-5r-6r-8-10s+18+6r+2-3s = 57

Now collect all the r and s stuff, and collect all the constants:

3r+4r+5r-5r-6r+6r + 13s-10s-3s + -2-12+45-8+18+2 = 57

All the s's cancel out, and you then have

7r + 43 = 57
...

Answer 8

Thank you very much! :)

Answer 9

sure thing. When you encounter complicated math problems, just take a deep breath and realize that since you only know a few simple rules for working with the stuff, the big mess can always be broken down into simple pieces that you can do easily, using the rules you know.

Answer 10

I'll definitely keep that in mind. Thanks once again.