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AddemF
 2 years ago
Best ResponseYou've already chosen the best response.1I'm not sure what you mean by the "linear combination" method, but here's my best guess. View the x coefficients as a vector (4,3) and the y coefficients as a vector (5,1) and likewise the solutions as a vector (13,4). We want to find some combination of the first two vectors which produces the last. To do that you either have to just think about or turn this into a matrix problem. If you just try thinking about it, you see that you want a negative number in the second coordinate and so you probably want a negative value of x since it has a larger second coordinate. (If this reasoning is opaque I can try to elaborate.) From here we just try stuff. Would x=3 work? Well, in that case you get 3*(4,3)=(12,9) and we want to add some multiple of (5,1) to get (13,4), but to do that you need y=5 which makes (12,9)+(25,5)=(13,4). So these are the values of x and y. However, that method is sloppy and annoyingbetter to do this using a matrix \[\left(\begin{matrix}4 & 5 & 13\\ 3 & 1 & 4\end{matrix}\right) \rightarrow \left(\begin{matrix} 1 & 4 & 17 \\3 & 1 & 4\end{matrix}\right)\rightarrow \left(\begin{matrix}1 & 4 & 17 \\ 0 & 11 & 55 \end{matrix}\right)\rightarrow \left(\begin{matrix}1 & 4 & 17\\ 0 & 1 & 5\end{matrix}\right)\] From this last matrix we can see that x+4y=17 and y=5. Therefore x+4(5)=17 and so x=1720=3. These are the same values we found the other way.

ShikhaDessai
 2 years ago
Best ResponseYou've already chosen the best response.0multiply the 2nd equation by 5, so as to make the numerical value(coefficient) of 'y' same....then subtract equation 2 from one...n we get  11x = 33.....so x = 3....then substitute the value of 'x' in any of the equations to get the value of 'y'.....
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