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#