How do you solve the following linear system: $3 x - 2 y = 3, - x + y = 1$ $3x-2y=3 , -x+y=1$ ?

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Answer 1 · 2016-01-03T18:34:30

The solution for the system of equations is

$x = 5, y = 6$ $color(blue)(x=5,y=6$

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

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

Solving by elimination.

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

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

$color(blue)(y=6$

Finding $x$ from equation $1$ :
$3x-2y=3$

$3x=3+2y$

$3x=3+2xx6$

$3x=3+12$

$3x=15$

$x=15/3$

$color(blue)(x=5$

How do you solve the following linear system: 3x-2y=3 , -x+y=1 3x−2y=3,−x+y=13x-2y=3 , -x+y=1 ?