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

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Answer 1 · 2015-08-04T05:21:19

The solution for the system of equations is :
$color(blue)(x=0,y=3$

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

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

Solving by elimination, as adding the two equations would result in elimination of $color(blue)(y$

$x+cancelcolor(blue)(y)=3$
$2x−cancelcolor(blue)(y)=−3$

$3x=0$
$color(blue)(x=0$

Substituting $x$ in equation $1$ to find $y$
$x + y=3$

$y=3-x$
$y=3-0$
$color(blue)(y=3$

How do you solve x+y=3x+y=3 and 2x-y= -32x-y= -3?