How do you solve $3 x + y = 3$ $3x+y=3$ and $x + 2 y = - 4$ $x+2y=-4$ ?

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Answer 1 · 2015-08-14T11:32:02

The solutions for the equation are
$x = 2, y = - 3$ $color(blue)(x=2,y=-3$

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

$x + 2 y = - 4$ $x+2y=−4$ , multiplying this equation by $3$ $3$
$3 x + 6 y = - 12$ $color(blue)(3x)+6y=−12$ ....... equation $(2)$ $(2)$

Now we can solve through elimination as term $3 x$ $color(blue)(3x)$ is common to both equations.

Subtracting equation $(2)$ $(2)$ from $(1)$ $(1)$
$cancel(color(blue)(3x))+y=3$
$-cancel(color(blue)(3x))-6y=+12$

$-5y=15$
$color(blue)(y=-3$

Substituting the value of $y$ in equation $1$
$x+2y=−4$
$x+2 * (-3)=−4$
$x-6=−4$
$color(blue)(x=2$

How do you solve 3x+y=33x+y=33x+y=3 and x+2y=-4x+2y=−4x+2y=-4?