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

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Answer 1 · 2015-12-29T17:15:30

The solution for the system of equations is:
$x = - \frac{6}{5}$ $color(blue)(x=-6/5$

$y = - \frac{9}{5}$ $color(blue)(y=-9/5$

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

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

Solving by elimination.

Subtracting equation $2$ $2$ from $1$ $1$

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

$- 2 x - 2 y = + 6$ $-color(blue)(2x)-2y=+6$

$- 5 y = 9$ $-5y=9$

$y = - \frac{9}{5}$ $color(blue)(y=-9/5$

Finding $x$ $x$ from equation $2$ $2$ :
$x + y = - 3$ $x+y=-3$

$x = - 3 - y$ $x=-3-y$

$x = - 3 - (- \frac{9}{5})$ $x=-3- (-9/5)$
$x = - 3 + \frac{9}{5}$ $x=-3 +9/5$

$x = - \frac{15}{5} + \frac{9}{5}$ $x=-15/5 +9/5$

$x = - \frac{6}{5}$ $color(blue)(x=-6/5$

How do you solve the following linear system 2x-3y= 3, x+y=-3 2x−3y=3,x+y=−32x-3y= 3, x+y=-3 ?