How do you solve the following system: $x - 2 y = 3, 2 x + 3 y = 4$ $x-2y=3, 2x+3y =4$ ?

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Answer 1 · 2016-02-22T18:50:39

The solution for the system of equations is:

$x = \frac{17}{7}$ $x=17/7$
$y = - \frac{2}{7}$ $y=-2/7$

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

$2 x + 3 y = 4$ $color(blue)(2x) +3y =4$ ........equation $(2)$ $(2)$

Solving by elimination:

subtracting equation $2$ $2$ from $1$ $1$ :

$cancelcolor(blue)(2x)-4y =6$

$-cancelcolor(blue)(2x) -3y =-4$

$-7y = 2$

$y=-2/7$

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

$x=3+2y$

$x=3+2xx(-2/7)$

$x=3- 4/7$

$x=21/7- 4/7$

$x=17/7$

How do you solve the following system: x-2y=3, 2x+3y =4 x−2y=3,2x+3y=4x-2y=3, 2x+3y =4 ?