How do you solve #x+2y=12# and #x-2y=0#?

1 Answer
Aug 3, 2015

Answer:

The solution for the system of equations is
#color(blue)(x=6,y=3#

Explanation:

#x+color(blue)(2y)=12 # .....equation #(1)#
#x−color(blue)(2y)=0#....equation #(2)#

We can solve the equations through elimination
Adding the two equations results in elimination of #color(blue)(2y#

#x+cancelcolor(blue)(2y)=12 #
#x−cancelcolor(blue)(2y)=0#

#2x=12#
#color(blue)(x=6#

Substituting #x# in equation #1# to find #y#
#x+2y=12 #
#2y=12-6 #
#2y=6 #
#color(blue)(y=3#