Remember you can do anything to an equation as long as you do it to both sides
x+9=18-2x
add 2x to both sides
2x+x+9=18+2x-2x
3x+9=18+0
3x+9=18
minus 9 both sides
3x+9-9=18-9
3x+0=9
3x=9
divide both sides by 3
(3x)/3=9/3
(3/3)x=3
1x=3
x=3
Answer:
1 rubber cube dude give since we'll
let's firstly, convert the mixed fraction to improper fraction, and then subtract.
![\bf \stackrel{mixed}{2\frac{3}{4}}\implies \cfrac{2\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{11}{4}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{11}{4}-\cfrac{2}{3}\implies \stackrel{\textit{our LCD will be 12}}{\cfrac{(3)11-(4)2}{12}}\implies \cfrac{33-8}{12}\implies \cfrac{25}{12}\implies 2\frac{1}{12}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%204%2B3%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B11%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B11%7D%7B4%7D-%5Ccfrac%7B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bour%20LCD%20will%20be%2012%7D%7D%7B%5Ccfrac%7B%283%2911-%284%292%7D%7B12%7D%7D%5Cimplies%20%5Ccfrac%7B33-8%7D%7B12%7D%5Cimplies%20%5Ccfrac%7B25%7D%7B12%7D%5Cimplies%202%5Cfrac%7B1%7D%7B12%7D)
1 and 8 should be the answer I believe.