How do you solve the system using the elimination method for x – 3y = 0 and 3y – 6 = 2x?

1 Answer
Jul 3, 2015

Answer:

#{(x=-6),(y=-2):}#

Explanation:

To solve by elimination, let say

#"Equation 1"# is #" "x-3y=0#
and
#"Equation 2"# is #" "3y-6=2x#

Now , to eliminate #y# you would wanna add Equation 1and Equation 2.

To do that you have to add the Left hand Side(#"LHS"#)of each equation.

Then you equate that to the sum of the Right Hand Sides(#"RHS"#) of the two equations.

If you do that correctly then,

#"LHS"=x-3y+3y-6=x-6#

Now, that's how you eliminated #y#

#"RHS"=0+2x=2x#

Now, do #"LHS"="RHS"#

#=>x-6=2x#

#=>-2x+x-6=2x-2x#

#=>-x-6=0#

#=>-x-6+6=6#

#=>-x=6#

#-1xx-x=-1xx6#

#=>color(blue)(x=-6)#

Now, to obtain #y# we want to eliminate #x#

#"Equation 1"# is #" "x-3y=0#

#"Equation 2"# is #" "3y-6=2x#

Multiply both side of #"Equation 1"# by #2# then add the resulting equation with #"Equation 2"#

#"Equation 1"# becomes #2x-6y=0#
Then with #"Equation 2"#

#=>"LHS"=2x-6y+3y-6=2x-3y-6#

#=>"RHS"=0+2x=2x#

Now , #"RHS"="LHS"#

#=>2x-3y-6=2x#

#=>-2x+2x-3y-6=2x-2x#

#=>-3y-6=0#

#=>-3y-6+6=0+6#

#=>(-3y)/(-3)=6/-3#

#=>color(blue)(y=-2)#