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

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Answer 1 · 2016-11-21T14:53:00

$y = 0$ $y = 0$ and $x = 3$ $x = 3$

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

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

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

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

$x = 3 - 4 y$ $x = 3 - 4y$

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

$- 5 (3 - 4 y) - 20 y = - 15$ $-5(3 - 4y) - 20y = -15$

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

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

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

$40 y = 0$ $40y = 0$

$\frac{40 y}{40} = \frac{0}{40}$ $(40y)/40 = 0/40$

$y = 0$ $y = 0$

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

$x = 3 - 4 \cdot 0$ $x = 3 - 4*0$

$x = 3 - 0$ $x = 3 - 0$

$x = 3$ $x = 3$

How do you solve the system of Equations 2x+8y=6,−5x−20y=−152x+8y=6, -5x-20y=-15?