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

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Answer 1 · 2015-06-04T07:50:44

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

Adding equation $(1)$ $(1)$ and $(2)$ $(2)$ we can eliminate $y$ $color(blue)(y$

$3x - cancel(y) = 3$
$2x +cancel(y) = 2$
$5x = 5, color(green) x=1$

Substituting the value of $x$ in equation $(1)$ to find $y$ :
$3x - y = 3$
$3x - 3= y$

$color(green)(y = 3-3.(1) = 0$
the solution for the system of equations is:
$color(green)(x=1 , y=0$

How do you solve the system 3x - y = 33x−y=33x - y = 3 and 2x + y = 22x+y=22x + y = 2?