How do you solve $x + 2 y = 12$ $x+2y=12$ and $x - 2 y = 0$ $x-2y=0$ ?

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Answer 1 · 2015-08-03T08:13:30

The solution for the system of equations is
$x = 6, y = 3$ $color(blue)(x=6,y=3$

$x + 2 y = 12$ $x+color(blue)(2y)=12$ .....equation $(1)$ $(1)$
$x - 2 y = 0$ $x−color(blue)(2y)=0$ ....equation $(2)$ $(2)$

We can solve the equations through elimination
Adding the two equations results in elimination of $2 y$ $color(blue)(2y$

$x+cancelcolor(blue)(2y)=12$
$x−cancelcolor(blue)(2y)=0$

$2x=12$
$color(blue)(x=6$

Substituting $x$ in equation $1$ to find $y$
$x+2y=12$
$2y=12-6$
$2y=6$
$color(blue)(y=3$

How do you solve x+2y=12x+2y=12x+2y=12 and x-2y=0x−2y=0x-2y=0?