How do you solve using elimination of #x-15y=3# and #3x+y=21#?

2 Answers
Dec 20, 2015

Answer:

#(x,y)=(159/23, 6/23)#

Explanation:

Given
[1]#color(white)("XXX")x-15y=3#
[2]#color(white)("XXX")3x+y=21#

Eliminate the #x# terms by subtracting #3xx#[1] from [2]

#{: ([2],color(white)("XXX"),,"(",3x,+y,"=",21,")"), (3xx[1],color(white)("XXX"),-,"(",3x,-45y,"=",9,")") :}#

[4]#color(white)("XXX")46y=12#
[5]#color(white)("XXX")y=6/23#

Eliminate the #y# terms by adding #15xx[2]# to [1]

#{: ([1],color(white)("XXX"),,"(",x,-15y,"=",3,")"), (15xx[2],color(white)("XXX"),+,"(",45x,+15y,"=",315,")") :}#

[6]#color(white)("XXX")46x=318#
[7]#color(white)("XXX")x=159/23#

Dec 20, 2015

Answer:

I found:
#x=159/23#
#y=6/23#

Explanation:

We can multiply the first equation by #-3# and add (in column) to the second:
#{(-3xx[x-15y=3]),(3x+y=21):}#
#{(-3x+45y=-9),(3x+y=21):}# add:
#0+46y=12#
So:
#y=12/46=6/23#
use this into the first equation:
#x-15*color(red)(6/23)=3#
#x=90/23+3=159/23#