How do you solve the system of Equations 2x+8y=6, -5x-20y=-15?

Nov 21, 2016

$y = 0$ and $x = 3$

Explanation:

To complete this problem, first solve the first equation for $x$:

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

$2 x = 6 - 8 y$

$\frac{2 x}{2} = \frac{6 - 8 y}{2}$

$x = 3 - 4 y$

Next, substitute $3 - 4 y$ for $x$ in the second equation and solve the $y$:

$- 5 \left(3 - 4 y\right) - 20 y = - 15$

$- 15 + 60 y - 20 y = - 15$

$- 15 + 40 y = - 15$

$- 15 + 15 + 40 y = 15 + 15$

$40 y = 0$

$\frac{40 y}{40} = \frac{0}{40}$

$y = 0$

Finally, we substitute $0$ for $y$ into the solution for the first equation and calculate $x$:

$x = 3 - 4 \cdot 0$

$x = 3 - 0$

$x = 3$