How do solve the following linear system?: $y = 8 x - 3, 3 y = 15 - 6 x$ $y = 8 x - 3 , 3y = 15 -6x$ ?

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Answer 1 · 2018-06-10T06:22:33

$x = 0.6, y = 3.8$ $x=0.6,y=3.8$

Multiply the first equation by 3

$3 y = 24 x - 9$ $3y=24x-9$

Subtract the second equation

$0 = 30 x - 18$ $0=30x-18$

Add 18

$30 x = 18$ $30x=18$

Divide by 30

$x = \frac{3}{5}$ $x=3/5$

Substitute this answer into $3 y = 15 - 6 x$ $3y=15-6x$

$3 y = 15 - \frac{18}{5}$ $3y=15-18/5$

$3 y = \frac{57}{5}$ $3y=57/5$

Divide by 3

$y = \frac{19}{5}$ $y=19/5$

How do solve the following linear system?: y=8x−3,3y=15−6x y = 8 x - 3 , 3y = 15 -6x ?