How do you solve the following system?: $8 x + y = 3, 4 x + 12 y = - 4$ $8x +y =3 , 4x +12y = -4$

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Answer 1 · 2015-11-11T07:27:16

The solution for the system of equations is:
$x = \frac{10}{23}, y = - \frac{11}{23}$ $color(blue)(x=10/23, y=-11/23$

$8 x + y = 3$ $color(blue)(8x)+y=3$ ....equation $1$ $1$

$4 x + 12 y = - 4$ $4x+12y=−4$ , multiplying this equation by $2$ $2$
$8 x + 24 y = - 8$ $color(blue)(8x)+24y=−8$ ......equation $2$ $2$

Now, we solve by elimination :
Subtracting equation $2$ $2$ from $1$ $1$

$cancelcolor(blue)(8x)+y=3$
$-cancelcolor(blue)(8x)-24y=+8$

$-23y=11$

$y=11/-23$

$y=color(blue)(-11/23$

Now we substitute $y$ in any of the equations to find $x$

$8x+y=3$

$8x -11/23=3$

$8x =3+ 11/23$

$8x =80/23$

$x =cancel80/(23*cancel8)$

$color(blue)(x =10/23$

How do you solve the following system?: 8x +y =3 , 4x +12y = -48x+y=3,4x+12y=−48x +y =3 , 4x +12y = -4