How do solve the following linear system?: $8 x = - 2 y - 10, 3 x + 2 y - 3 = 0$ $8x = -2y - 10, 3x+ 2y - 3 = 0$ ?

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Answer 1 · 2016-03-25T18:02:01

The solution for the system of equations is:
$x = \frac{- 13}{5}$ $color(green)(x = (-13)/5$

$y = \frac{27}{5}$ $color(green)( y = 27/(5)$

$8 x = - 2 y - 10$ $8x= -2y-10$
$8 x + 2 y = - 10$ $8x + color(blue)(2y) = -10$ ..............equation $(1)$ $(1)$

$3 x + 2 y - 3 = 0$ $3x+2y-3=0$
$3 x + 2 y = 3$ $3x+color(blue)(2y) = 3$ .......................equation $(2)$ $(2)$

Solving by elimination.

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

$8x + cancelcolor(blue)(2y) = -10$

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

$5x = -13$

$color(green)(x = (-13)/5$

Finding $y$ from equation $1$

$8x + 2y = -10$

$2y = -10 - 8x$

$2y = -10 - 8 * ( -13/5)$

$2y = -10 + (104/5)$

$2y = -50/5 + (104/5)$

$2y = 54/5$

$y = 54/(5 * 2)$

$color(green)( y = 27/(5)$

How do solve the following linear system?: 8x = -2y - 10, 3x+ 2y - 3 = 0 8x=−2y−10,3x+2y−3=08x = -2y - 10, 3x+ 2y - 3 = 0 ?