How do you solve the system $y = 6 x - 11$ $y=6x - 11$ and $- 2 - 3 y = - 7$ $-2 - 3y = -7$ by substitution?

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Answer 1 · 2015-05-25T17:10:42

Given
$[1] y = 6 x - 11$ $"[1] "y=6x-11$
$[2] - 2 - 3 y = - 7$ $"[2] "-2-3y = -7$

from $[1]$ $"[1]"$ we can substitute $6 x - 11$ $6x-11$ for $y$ $y$ in $[2]$ $"[2]"$
$[3] - 2 - 3 (6 x - 11) = - 7$ $"[3] "-2-3(6x-11) = -7$

$[4] - 3 (6 x - 11) = - 5$ $"[4] " -3(6x-11) = -5$

$[5] 6 x - 11 = \frac{5}{3}$ $"[5] " 6x-11 = 5/3$

$[6] 6 x = \frac{38}{3}$ $"[6] " 6x = 38/3$

$[7] x = \frac{19}{9}$ $"[7] " x = 19/9$

Note that $[2]$ $"[2]"$ can be solved directly for $y$ $y$
$[8] - 3 y = - 5$ $"[8] "-3y = -5$

$[9] y = \frac{5}{3}$ $"[9] " y = 5/3$

Query: Was the second equation intended to be
$- 2 x - 3 y = - 7$ $-2x-3y = -7$ ?

How do you solve the system y=6x - 11y=6x−11 y=6x - 11 and -2 - 3y = -7−2−3y=−7-2 - 3y = -7 by substitution?