How do you solve $x = y + 2$ $x=y+2$ and $2 x - 3 y = 12$ $2x-3y= 12$ ?

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Answer 1 · 2018-06-05T13:22:41

$x = - 6, y = - 8$ $x = -6, y = -8$

[1] $x = y + 2$ $x = y+2$
[2] $2 x - 3 y = 12$ $2x-3y=12$

Equation [1] gives the expression for $x$ $x$ . This can be substituted into equation [2]:

$2 (y + 2) - 3 y = 12$ $2(y+2) - 3y = 12$

Now we can solve for $y$ $y$ :

$2 y + 4 - 3 y = 12$ $2y + 4 - 3y = 12$

$4 - y = 12$ $4 - y = 12$

$- y = 8$ $-y = 8$

[3] $y = - 8$ $y = -8$

Now we take our result in [3] and substitute back into either [1] or [2]. Let's use [1]:

$x = (- 8) + 2$ $x = (-8) + 2$

$x = - 6$ $x = -6$

Hence, the solution is $x = - 6$ $x = -6$ and $y = - 8$ $y = -8$ .

How do you solve x=y+2x=y+2x=y+2 and 2x-3y= 122x−3y=12 2x-3y= 12?