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

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Answer 1 · 2018-03-11T14:42:40

$x = - \frac{1}{4}$ $x = -1/4$ , $y = \frac{5}{8}$ $y = 5/8$

$- x - 2 y = - 1$ $-x - 2y = -1$ Eqn (1)

$6 x - 4 y = - 4$ $6x - 4y = -4$

$3 x - 2 y = - 2$ $3x - 2y = -2$ Eqn (2) (dividing both sides by 2).

Subtracting Eqn (1) from (2),

$3x cancel(-2y)+ x + cancel(2y) = -2 + 1$

$4x = -1$ or $x = -1/4$

Substituting value of x in Eqn (1),

$1/4 - 2y = -1$

$2y = 5/4$ or $y = 5/8$

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