Answer:
It must be a solution for both to be a solution to this system
Im going to go with B.The intersection point (x,y) of the two lines.
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)
For A(–3, 2), B(0, 7), and C(0, 2), find all locations of a fourth point, D, so that a parallelogram is formed using A, B, C, D
ki77a [65]
The answer will be a (-3,2)
This is due to the coordination grid and draw
To be honest I don’t know