How do solve the following linear system?: $3 x - 2 y = - 6, \frac{x}{3} + 3 y = 9$ $3x - 2y = -6 , x/3 + 3y = 9$ ?

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Answer 1 · 2016-07-05T10:42:30

x=0, y=3

First you can multiply all the terms of the second equation by 3 and then find x:

$\frac{x}{3} + 3 y = 9 \leftrightarrow x + 9 y = 27$ $x/3+3y=9 harr x+9y=27$

$x = - 9 y + 27$ $x=-9y+27$

then you can substitute in x the expression $- 9 y + 27$ $-9y+27$ in the first equation:

$3 (- 9 y + 27) - 2 y = - 6$ $3(-9y+27)-2y=-6$

$- 27 y + 81 - 2 y + 6 = 0$ $-27y+81-2y+6=0$

$- 29 y = - 87$ $-29y=-87$

$y = \frac{87}{29} = 3$ $y=87/29=3$

so

$x = - 9 (3) + 27 = 0$ $x=-9(3)+27=0$

How do solve the following linear system?: 3x−2y=−6,x3+3y=9 3x - 2y = -6 , x/3 + 3y = 9 ?