How do you solve the following system: $x - 2 y = 3, - 6 x + y = - 8$ $x-2y=3, -6x + y = -8$ ?

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Answer 1 · 2016-02-13T17:45:15

The solution for the system of equations is:

$x = \frac{13}{11}$ $x=13/11$
$y = - \frac{10}{11}$ $y=-10/11$

$x - 2 y = 3$ $x-color(blue)(2y)=3$ ....................equation $(1)$ $(1)$

$- 6 x + y = - 8$ $-6x+y=-8$ , multiplying by $2$ $2$
$- 12 x + 2 y = - 16$ $-12x+color(blue)(2y)=-16$ .......equation $(2)$ $(2)$

Solving by elimination

Adding equations $1$ $1$ and $2$ $2$

$x-cancelcolor(blue)(2y)=3$

$-12x+cancelcolor(blue)(2y)=-16$

$-11x =-13$

$x =(-13)/(-11)$

$x =13/11$

Finding $y$ from equation $1$
$x - 2y=3$

$2y=x-3$

$2y=13/11-3$

$2y=13/11-33/11$

$2y=-20/11$

$y=-cancel20/(11xxcancel2)$

$y=-10/11$

How do you solve the following system: x-2y=3, -6x + y = -8 x−2y=3,−6x+y=−8x-2y=3, -6x + y = -8 ?