How do you solve the following system?: # 1/6x-3/5y=1 , 4/5x+2/3y = 2 #

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Answer 1 · 2018-07-10T19:23:39

#x=60/19 , y= -15/19#

Multiply the first equation by 10 #=> 10/6x-6y=10#

Multiply the second equation by 9 #=> 36/5x+6y=18#

Add these two equations together to give #133/15x=28#

Divide by #133/15 => x=60/19#

Substitute this into the first equation

#=> 1/6xx60/19-3/5y=1#

#=> 10/19-3/5y=1#

#=>-3/5y=9/19#

#=> y= -15/19#

Answer 2 · 2018-07-10T19:24:29

#x=60/19,y=-15/19#

Solving the first equation for #x#:

#1/6*x=1+3/5*y#
so
#x=6+18/5y#
substituting this in the second equation:

#4/5*(6+18/5y)+2/3y=2#

expanding

#24/5+72/25y+2/3y=2#
adding #-24/5#

#72/25y+2/3y=(10-24)/5#

note that #72/25+2/3=266/75#

so we get

#266/75y=-14/5#

multiplying by #75/266#

we get #y=-15/19#

so #x=6+18/5*(-15/19)#

#x=60/19#