Solve:
2(4-4)+3r-4/2+4r-12+13s+5r+5(9-r)-6r+2(-4)-5(2s)+2(9+3r+2-3s = 57
you have two variables, and unmatched parentheses.
You want to solve for r or s?
Oh, I forget to close a bracket. It's (9+3r)+2-3s
still, you want r or s?
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.
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.
Could you explain how you got rid of the other numbers?
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
...