# How do you solve the system 2x+4y=6 and 2x+y= -3 by substitution?

Apr 30, 2018

x=-3, y=3

#### Explanation:

$2 x + 4 y = 6$ eq 1
$2 x + y = - 3$ eq 2
From eq 2, make y the subject:
$y = - 3 + 2 x$
Substitute y from above in eq 1
$2 x + 4 \left(- 3 + 2 x\right) = 6$
$2 x - 12 + 8 x = 6$
$2 x - 8 x = 6 + 12$
$- 6 x = 18$
$x = - \frac{18}{6}$
$x = - 3$
Now, substitute x=-3 in eq 1 or eq 2. Lets use eq 1.
$2 \left(- 3\right) + 4 y = 6$
Now solve for y
$- 6 + 4 y = 6$
$4 y = 6 + 6$
$4 y = 12$
$y = \frac{12}{4}$
$y = 3$
$2 \left(- 3\right) + 3 = - 6 + 3 = - 3$