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

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Answer 1 · 2016-11-21T14:53:00

$y = 0$ and $x = 3$

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

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

$2x = 6 - 8y$

$(2x)/2 = (6 - 8y)/2$

$x = 3 - 4y$

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

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

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

$-15 + 40y = -15$

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

$40y = 0$

$(40y)/40 = 0/40$

$y = 0$

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

$x = 3 - 4*0$

$x = 3 - 0$

$x = 3$

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