How do you solve the following system: $3 x - 2 y = 9, 6 x + y = 2$ $3x - 2y = 9, 6x + y = 2$ ?

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Answer 1 · 2016-04-08T07:42:10

The solution for the system of equations is:
$x = \frac{13}{15}$ $color(green)(x = 13/15$

$y = \frac{- 16}{5}$ $color(green)(y = (-16)/ 5$

$3 x - 2 y = 9$ $3x - 2y = 9$ , multiplying by $2$ $2$
$6 x - 4 y = 18$ $color(green)(6x) - 4y = 18$ .............equation $(1)$ $(1)$

$6 x + y = 2$ $color(green)(6x) + y =2$ .................equation $(2)$ $(2)$

Solving by elimination.

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

$cancelcolor(green)(6x) - 4y = 18$
$-cancelcolor(green)(6x) - y =-2$

$-5y = 16$

$color(green)(y = (-16)/ 5$

Finding $x$ from equation $1$

$3x - 2y = 9$

$3x = 9 + 2y$

$3x = 9 + 2 * (-16/5)$

$3x = 9 -32/5$

$3x = 45/5 -32/5$

$3x = 13/5$

$x = 13/(5 * 3)$

$color(green)(x = 13/15$

How do you solve the following system: 3x - 2y = 9, 6x + y = 23x−2y=9,6x+y=23x - 2y = 9, 6x + y = 2?