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g180101
4x+5y=13 and 3x+y=-4 using linear combination method
I'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 annoying--better 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=17-20=-3. These are the same values we found the other way.
multiply 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'.....