Answer:
C and D
Step-by-step explanation:
I'm guessing the 3 in A and B are exponents, only equations with an highest exponent of 1 is linear
so, what we do is, we multiply the value by a power of 10, that moves the recurring numbers to the left of the dot, in this case we want moved the "15", so we'll need two zeros, so we'll be using 100.
![\bf 0.151515\overline{15}~\hspace{7em}x=0.151515\overline{15} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{array}{|lll|ll} \cline{1-3} &&\\ 100\cdot x&=&15.1515\overline{15}\\ &&15+0.1515\overline{15}\\ &&15+x \\&&\\ \cline{1-3} \end{array}\implies 100x=15+x \\\\\\ 99x=15\implies x=\cfrac{15}{99}\implies \stackrel{simplified}{x=\cfrac{5}{33}}](https://tex.z-dn.net/?f=%5Cbf%200.151515%5Coverline%7B15%7D~%5Chspace%7B7em%7Dx%3D0.151515%5Coverline%7B15%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Clll%7Cll%7D%20%5Ccline%7B1-3%7D%20%26%26%5C%5C%20100%5Ccdot%20x%26%3D%2615.1515%5Coverline%7B15%7D%5C%5C%20%26%2615%2B0.1515%5Coverline%7B15%7D%5C%5C%20%26%2615%2Bx%20%5C%5C%26%26%5C%5C%20%5Ccline%7B1-3%7D%20%5Cend%7Barray%7D%5Cimplies%20100x%3D15%2Bx%20%5C%5C%5C%5C%5C%5C%2099x%3D15%5Cimplies%20x%3D%5Ccfrac%7B15%7D%7B99%7D%5Cimplies%20%5Cstackrel%7Bsimplified%7D%7Bx%3D%5Ccfrac%7B5%7D%7B33%7D%7D)
I think it 5:50 but it could also be 3:50 and 2:50
D. This sign (*) means division