How do you solve the following linear system $2 x - 7 y = 4, x + 3 y = 1 ?$ $2x-7y=4, x+3y=1?$ ?

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Answer 1 · 2018-04-04T04:08:42

$y = - \frac{2}{13}$ $y=-2/13$
$x = \frac{19}{13}$ $x=19/13$

$2 x - 7 y = 4$ $2x-7y=4$ --- (1)
$x + 3 y = 1$ $x+3y=1$ --- (2)

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

Sub (3) into (1)

$2 (1 - 3 y) - 7 y = 4$ $2(1-3y)-7y=4$
$2 - 6 y - 7 y = 4$ $2-6y-7y=4$
$- 2 = 13 y$ $-2=13y$
$y = - \frac{2}{13}$ $y=-2/13$ --- (4)

Sub (4) into (3)

$x = 1 - 3 (- \frac{2}{13})$ $x=1-3(-2/13)$
$x = \frac{19}{13}$ $x=19/13$

How do you solve the following linear system 2x−7y=4,x+3y=1?2x-7y=4, x+3y=1? ?