How do you solve by elimination $8x + 3y= -9$ and $-8x + y = 29$ ?

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Answer 1 · 2015-05-13T07:41:15

$8x + 3y= -9$
$-8x + y = 29$
.________

$(cancel(8x)+(-cancel(8x))) +(3y+y) = -9+29$

$4y = 20$

Dividing both sides by 4, we get :

$(4y)/4 = 20/4$

$y = 5$

Substituting $y = 5$ in the second equation, we get:

$-8x + 5 = 29$

Transposing 5 to the other side we get:

$- 8x = 29 - 5$

$-8x = 24$

Dividing both sides by $-8$ will give us:

$(-8x)/-8 = 24/-8$

$x = -3$

The Solution to the above equations is:

$color(green)(x = -3 and y = 5$

Substitute these values of $x and y$ in any of the equations and check if both sides equal each other.

How do you solve by elimination 8x + 3y= -98x + 3y= -9 and -8x + y = 29-8x + y = 29?