How do you solve the following system: #4x-y=6, 5x+2y=20 #?

1 Answer
EZ as pi
Jul 15, 2018

#x = 32/13 and y = 50/13#

Explanation:

#" "4xcolor(blue)(-y)=6" ".........A#
#" "5xcolor(blue)(+2y)=20" ".....B#

One of the easiest methods to eliminate one of the variables is to make them into additive inverses. (Their sum is #0#)

Multiplying equation #A# by #2# will achieve this,:

#Axx2:color(white)(xx)8xcolor(blue)(-2y)=12" ".........C#
#color(white)(xxxxxxx)5xcolor(blue)(+2y)=20" ".....B#

#C+B:color(white)(xx)13x" "=32" ".........C#
#C+B:color(white)(x..x)x" "=32/13" "#

If #x=32/13 #, substitute into one of the original equations to find #y#

#5(32/13)+2y=20" ".....B#

#160/13+2y=20#

#" "2y = 20-160/13#

#" "2y = 100/13#

#" "y=50/13#

Check in #A#

#4(32/13) -50/13#
#=78/13#
#=6#

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