How do you solve by substitution $x + 3 y = 8$ $x + 3y = 8$ and $2 x - y = 9$ $2x - y = 9$ ?

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Answer 1 · 2018-07-14T20:22:23

$x = 5, y = 1$ $x=5,y=1$

Solving the first equation for $x$ $x$ :
$x = 8 - 3 y$ $x=8-3y$
Substituting this ( for $x$ $x$ )in the second equation

$2 (8 - 3 y) - y = 9$ $2(8-3y)-y=9$

expanding

$16 - 6 y - y = 9$ $16-6y-y=9$

$16 - 7 y = 9$ $16-7y=9$
so

$7 y = 7$ $7y=7$ so $y = 1$ $y=1$

and

$x = 8 - 3$ $x=8-3$

$x = 5$ $x=5$

How do you solve by substitution x+3y=8x + 3y = 8 and 2x−y=92x - y = 9?