How do you solve #4x + 2y = 12# and #x + 2y = 6#?

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Answer 1 · 2016-05-02T15:55:58

The solution for the system of equations is:
#color(blue)(x =2 #
#color(blue)(y = 2 #

#4x + color(blue)(2y) = 12#........equation #(1)#
#x + color(blue)(2y) = 6#..............equation #(2)#

Solving by elimination:

Subtracting equation #2# from #1# eliminates #color(blue)(2y#

#4x + cancelcolor(blue)(2y) = 12#
#- x - cancelcolor(blue)(2y) = - 6#

#3x = 6#

#x = 6/3#

#color(blue)(x =2 #

Finding #y# from equation #1#:
#4x + 2y = 12#

#2y = 12 - 4x#

#2y = 12 - 4 * color(blue)(2 #

#2y = 12 - 8 #

#2y = 4 #

#y = 4/2 #

#color(blue)(y = 2 #