![\begin{array}{rrrrr} 10x&-&18y&=&2\\ -5x&+&9y&=&-1 \end{array}~\hfill \implies ~\hfill \stackrel{\textit{second equation }\times 2}{ \begin{array}{rrrrr} 10x&-&18y&=&2\\ 2(-5x&+&9y&)=&2(-1) \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{rrrrr} 10x&-&18y&=&2\\ -10x&+&18y&=&-2\\\cline{1-5} 0&+&0&=&0 \end{array}\qquad \impliedby \textit{another way of saying \underline{infinite solutions}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%20-5x%26%2B%269y%26%3D%26-1%20%5Cend%7Barray%7D~%5Chfill%20%5Cimplies%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsecond%20equation%20%7D%5Ctimes%202%7D%7B%20%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%202%28-5x%26%2B%269y%26%29%3D%262%28-1%29%20%5Cend%7Barray%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Brrrrr%7D%2010x%26-%2618y%26%3D%262%5C%5C%20-10x%26%2B%2618y%26%3D%26-2%5C%5C%5Ccline%7B1-5%7D%200%26%2B%260%26%3D%260%20%5Cend%7Barray%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Banother%20way%20of%20saying%20%5Cunderline%7Binfinite%20solutions%7D%7D)
if we were to solve both equations for "y", we'd get

notice, the 1st equation is really the 2nd in disguise, since both lines are just pancaked on top of each other, every point in the lines is a solution or an intersection, and since both go to infinity, well, there you have it.
For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

Where:
m: It is the slope of the line
b: It is the cut-off point with the y axis

We have the following points:

Substituting we have:

Thus, the equation is of the form:

We substitute one of the points and find "b":

Finally, the equation is of the form:

Answer:

Answer: i think its either 3:5 or 1:3 or there just not similar
Step-by-step explanation:
The answer is 8 plz make me brainlist
4 days at the beach.*7 or 5* days of visiting museums.
*If you take away 4 from 14 you get 10 and halve it you get 5 but if you halve 14 you get 7*