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

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Answer 1 · 2018-07-13T01:23:05

$y = 10$ $y=10$
$x = 6$ $x=6$

$2 x - y = 2$ $2x-y=2$ --- (1)
$- 3 x + y = - 8$ $-3x+y=-8$ --- (2)

From (1),
$2 x - y = 2$ $2x-y=2$
$y = 2 x - 2$ $y=2x-2$ --- (3)

Sub (3) into (2)
$- 3 x + (2 x - 2) = - 8$ $-3x+(2x-2)=-8$
$- 3 x + 2 x - 2 = - 8$ $-3x+2x-2=-8$
$- x - 2 = - 8$ $-x-2=-8$
$x = 6$ $x=6$ --- (4)

Sub (4) into (3)
$y = 2 x - 2$ $y=2x-2$
$y = 2 (6) - 2$ $y=2(6)-2$
$y = 10$ $y=10$

How do you solve the following linear system: 2x−y=2,−3x+y=−82x-y=2, -3x+y=-8 ?