How do you solve the following system of equations?: $- x - 6 y = 2, 4 x + y = 8$ $-x- 6y = 2 , 4x+y=8$ ?

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Answer 1 · 2017-11-06T22:37:05

$y = - \frac{16}{23}$ $y=-16/23$

$x = \frac{50}{23}$ $x=50/23$

$- x - 6 y = 2$ $-x-6y=2$
$4 x + y = 8$ $4x+y=8$

Make $y$ $y$ the subject in the second equation:

$y = 8 - 4 x$ $y=8-4x$

Then inject that as $y$ $y$ in the first equation:

$- x - 6 (8 - 4 x) = 2$ $-x-6(8-4x)=2$

Expand and solve for $x$ $x$

$- x - 48 + 24 x = 2 \to 23 x = 50$ $-x-48+24x = 2 -> 23x=50$

$x = \frac{50}{23}$ $x=50/23$

Place that $x$ $x$ value into the rearranged equation to find $y$ $y$

$y = 8 - \frac{200}{23}$ $y=8-200/23$

$y = \frac{184}{23} - \frac{200}{23}$ $y=184/23 - 200/23$

$y = - \frac{16}{23}$ $y=-16/23$

How do you solve the following system of equations?: −x−6y=2,4x+y=8-x- 6y = 2 , 4x+y=8?