How do you solve the system #4x + y = 6# and #6x + 3y = 6#?

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Answer 1 · 2015-05-15T22:44:32

Try to eliminate #y# by expressing it in terms of #x# in two ways from the two original equations.

With the first equation, subtract #4x# from both sides to get:

#y = 6 - 4x#

With the second equation, first divide through by 3 to get

#2x+y = 2#

Then subtract #2x# from both sides to get

#y = 2 - 2x#

Now we have two expressions both equal to #y#, so we can put them together:

#6 - 4x = y = 2 - 2x#

So #6 - 4x = 2 - 2x#

Add #4x# to both sides to get

#6 = 2 + 2x#

Subtract 2 from both sides to get

#4 = 2x#

Then divide both sides by 2 to get

#2 = x#, that is #x = 2#.

Then looking back at one of our previous equations:

#y = 2 - 2x = 2 - 2*2 = 2-4 = -2#

Answer 2 · 2015-05-15T22:44:48

You first express #y# as function of #x#, then solve for #x#

#4x+y=6->y=6-4x#

Substitute this in the other equation:
#6x+3y=6->6x+3*(6-4x)=6->#
#6x+18-12x=6->-6x=-12->x=2#

Now put this #x# into the first equation:
#y=6-4x=6-4*2=-2#

Answer:
#x=2#
#y=-2#