How do solve the following linear system?: $3 x - 8 y = 6, 2 x - 5 y = 3$ $3x-8y=6, 2x - 5y=3$ ?

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Answer 1 · 2017-06-23T11:35:14

By some arrangement, $x = - 6$ $x=-6$ and $y = - 3$ $y=-3$

Expand the first original equation with $2$ $2$ and the second original equation with $- 3$ $-3$ to get only equation having one unknown:

$6 x - 16 y = 12$ $6x - 16y = 12$
#-6x + 15 y = -9

Add the terms (of the above equations, yielding
$- 16 y + 15 y = 12 - 9$ $-16 y + 15 y = 12-9$

$- y = 3$ $-y=3$

or $y = - 3$ $y=-3$

Now put this value in any equation (for example the first equation):

$3 x - (8 \times - 3) = 6$ $3x - (8times-3) = 6$

$3 x + 24 = 6$ $3x + 24 = 6$

$3 x = 6 - 24$ $3x = 6-24$

$3 x = - 18$ $3x = -18$

$x = - 6$ $x = -6$

This is your answer: $x = - 6$ $x=-6$ and $y = - 3$ $y=-3$

How do solve the following linear system?: 3x−8y=6,2x−5y=3 3x-8y=6, 2x - 5y=3 ?