How do you solve the system of equations #-x - 8y = - 16# and #6x + 4y = 8#?

Question

1105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T13:48:10

#(x,y)-=(0,2)#

#-x-8y=-16#

#6x+4y=8#

If
#a_(11)x+a_(12)y=b_(11)#

#a_(21)x+a_(22)y=b_(22)#

Cramer's rule can be applied

#(x,y)=((b_(11)a_(22)-b_(22)a_(12))/(a_(11)a_(22)-a_(21)a_(12)),(a_(11)b_(22)-a_(22)b_(12))/(a_(11)a_(22)-a_(21)a_(12)))#

Thus, comparing

#x=((-16xx4-8xx(-8)))/((-1xx4-6xx-8))= ((-64+64))/((-4+48))=0/44=0#

#y=((-1xx8-6xx(-16)))/((-1xx4-6xx-8))=((-8+96))/((-4+48))=88/44=2#

#(x,y)-=(0,2)#