anonymous
  • anonymous
4x+5y=13 and 3x+y=-4 using linear combination method
Linear Algebra
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i need help
anonymous
  • anonymous
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.
anonymous
  • anonymous
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'.....

Looking for something else?

Not the answer you are looking for? Search for more explanations.