How do you solve the following linear system: # -x+3y=-9 , 8x-4y=32 #?

2 Answers
Apr 10, 2018

#y=-2#
#x=3#

Explanation:

We have, #3y-x+9=0# and #4(2x-y)=4xx8#
Or, #3(y+3)=x# and #2x-y=8#
Substituting the value of #x# from Equation 1 in Equation 2,we get,
#2(3y+9)-y-8=0#
Or, #6y-y+18-8=0#
Or, #5y=-10#
Or, #y=-2#
Thus, #x# comes out to be #3xx(3-2)# or #3xx1# or #3#

Apr 10, 2018

#x=3#

#y=-2#

Explanation:

Given:

#-x+3y=-9" "......................Equation(1)#
#color(white)("d")8x-4y=32" ".........................Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of "y" - in detail using first principles")#

#color(brown)("Consider "Eqn(1))#

#color(brown)("Multiply both sides by "(-1))#

#+x-3y=+9#

#color(brown)("Add "3y" to both sides")#

#x=9+3y" ".......Equation(1_a)#

#color(brown)("Using "Eqn(1_a)" substitute for "x" in "Eqn(2))#

#color(green)(8color(red)(x)-4y=32 color(white)("dddd")->color(white)("dddd")8(color(red)(9+3y))-4y=32 )#

#color(green)( color(white)("dddddddddddddd.")->color(white)("dddd")72+24y-4y=32 )#

#color(green)( color(white)("dddddddddddddd.")->color(white)("dddd")72+20ycolor(white)("dddd")=32 )#

#color(brown)("Subtract 72 from both sides")#

#color(green)( color(white)("dddddddddddddd.")->color(white)("dddddddd")20ycolor(white)("dddd")=-40 )#

#color(brown)("Divide both sides by 20")#

#color(green)( color(white)("dddddddddddddddd.")->color(white)("dddddddd")ycolor(white)("dddd")=-2 )#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of "x" - not so much detail")#

#color(brown)("Substitute for "y" in "Eqn(1))#

#-x+3y=-9color(white)("dddd")->color(white)("dddd")-x+3(-2)=-9#

#color(white)("ddddddddddddddddddddddddd")->color(white)("dddd") -x=-3#

#x=3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#

#-x+3y color(white)("d")->color(white)("d")-3+(3xx-2) ->9 # as required

#8x-4y color(white)("d")->color(white)("d") 8(3)-4(-2) -> 24+8->32# as req'd

Tony B