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

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

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Answer 1 · 2018-04-30T06:58:46

x=-3, y=3

Explanation:

$2 x + 4 y = 6$ $2x+4y=6$ eq 1
$2 x + y = - 3$ $2x+y=-3$ eq 2
From eq 2, make y the subject:
$y = - 3 + 2 x$ $y=-3+2x$
Substitute y from above in eq 1
$2 x + 4 (- 3 + 2 x) = 6$ $2x+4(-3+2x)=6$
$2 x - 12 + 8 x = 6$ $2x-12+8x=6$
$2 x - 8 x = 6 + 12$ $2x-8x=6+12$
$- 6 x = 18$ $-6x=18$
$x = - \frac{18}{6}$ $x=-18/6$
$x = - 3$ $x=-3$
Now, substitute x=-3 in eq 1 or eq 2. Lets use eq 1.
$2 (- 3) + 4 y = 6$ $2(-3)+4y=6$
Now solve for y
$- 6 + 4 y = 6$ $-6+4y=6$
$4 y = 6 + 6$ $4y=6+6$
$4 y = 12$ $4y=12$
$y = \frac{12}{4}$ $y=12/4$
$y = 3$ $y=3$
Check for answer:
use eq 1 or eq 2
$2 (- 3) + 3 = - 6 + 3 = - 3$ $2(-3) + 3 = -6+3=-3$
So the answers x=-3 and y=3 are correct

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

1 Answer

Explanation:

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