How do you solve the following system?: $- x - 2 y = - 1, 6 x - y = - 4$ $-x -2y =-1, 6x -y = -4$

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Answer 1 · 2016-01-18T11:40:19

$x = \frac{7}{13}$ $x = 7/13$ and $y = \frac{10}{13}$ $y = 10/13$

By Cramer's rule:

Make a matrix with the coefficient of each columns, in this case

$[(-1,-2),(6,-1)]$

Take the determinant of this

$|(-1,-2),(6,-1)| = 13$

Now replace the first row with the results row

$[(-1,-2),(-4,-1)]$

Now take this determinant

$|(-1,-2),(-4,-1)| = 7$

The ratio of these determinants is the solution for $x$ , so

$x = 7/13$

Replace the $y$ row by the results row

$[(-1,-1),(6,-4)]$

Take this determinant

$|(-1,-1),(6,-4)| = 10$

Like before the ratio is the answer for $y$

$y = 10/13$

How do you solve the following system?: -x -2y =-1, 6x -y = -4−x−2y=−1,6x−y=−4-x -2y =-1, 6x -y = -4