How do you solve using elimination of $3 x - 2 y = - 1$ $3x-2y=-1$ and $3 x - 4 y = 9$ $3x-4y=9$ ?

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Answer 1 · 2018-06-08T05:36:27

Multiply the second equation by $- 1$ $-1$ , and add the two equations together to get:
$x = - \frac{11}{3}$ $x=-11/3$ and $y = - 5$ $y=-5$

Multiply the second equation by $- 1$ $-1$ and add the two equations together:

$+ 3 x - 2 y = - 1$ $+3x-2y=-1$
$- 3 x + 4 y = - 9$ $-3x+4y=-9$

$0 + 2 y = - 10$ $0+2y=-10$

This eliminates the $x$ $x$ variable, so we can find $y$ $y$ :
$y = - 5$ $y=-5$

Now substitute $y$ $y$ in either equation to obtain $x$ $x$ :
$3 x - 2 \cdot (- 5) = - 1$ $3x-2*(-5)=-1$
$3 x = - 1 - 10 = - 11$ $3x=-1-10 = -11$

$x = - \frac{11}{3}$ $x=-11/3$

How do you solve using elimination of 3x−2y=−13x-2y=-1 and 3x−4y=93x-4y=9?