How do you solve the following system?: # 1/9x-3/13y=1 , 3/2x - 8y = 2 #

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Answer 1 · 2017-06-08T12:34:32

#color(blue)(y=-0.25,x=0#

#:.1/9x-3/13y=1#----(1)

#:.3/2x-8y=2#----(2)

multiply # (2) xx 2#

#:.3x-16y=4#

#:.-16y=-3x+4#

#:.16y=3x-4#

#:.color(blue)(y=(3x-4)/16#

substitute #color(blue)(y=(3x-4)/16#--- in # (2)#

#:.3/2x-8((3x-4)/16)=2#

#:.3/2x-(24x+32)/16=2#

multiply both sides by 16

#:.24x-24x+32=32#

#:.24x-24x=32-32#

#:.0=0#

#:.color(blue)(x=0#

substitute #color(blue)(x=0# in #(2)#

#:.3/2(color(blue)0)-8y=2#

#:.-8y=2#

multiply both sides by# -1#

#:.8y=-2#

#:.y=-2/8#

#:.color(blue)(y=-0.25# or #color(blue)( -1/4#

check:

substitute #color(blue)(y=-0.25# in #(2)#

#:.3/2x-8(color(blue)(-0.25))=2#

#:.3/2x+2=2#

#:.3/2x=2-2#

#:.3/2x=0#

#:.color(blue)(x=0#