How do you solve the system #3x + 6y = 30# and #x + 6y = 20#?

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Answer 1 · 2017-04-08T20:49:23

Solve by elimination to find:

#x=5" "# and #" "y=5/2#

Given:

#{ (3x+6y=30), (x+6y=20) :}#

To eliminate the term in #y#, we can subtract the second equation from the first to get:

#2x=10#

and hence:

#x=5#

Substituting this value of #x# in the second equation, we get:

#5+6y=20#

Subtract #5# from both sides to get:

#6y=15#

Divide both sides by #6# to get:

#y = 15/6 = 5/2#