How do you solve the system of equations#-8x+5y=-8# and #-16x-y=-16# by elimination?

2 Answers
Cem Sentin
Jul 31, 2018

#x=1# and #y=0#

Explanation:

#2*(-8x+5y)-(-16x-y)=2*(-8)-(-16)#

#-16x+10y+16y+y=-16+16#

#11y=0#, so #y=0/11=0#

Hence,

#-8x+5*0=-8#

#-8x=-8#, thus #x=(-8)/(-8)=1#

EZ as pi
Jul 31, 2018

#x=1 and y=0#

Explanation:

The perfect scenario for the elimination method is if one of the variables can be additive inverses.

Multiply the first equation by #-2# to achieve this:

#color(white)(xxxxxxxxxxx)-8x+5y" "=-8" "....A#
#color(white)(xxxxxxxx.xxx)color(blue)(-16x)-y" "=-16" "....B#
#A xx -2:color(white)(xxxxx.x)color(blue)(16x)-10y =+16" "....C#

#B+C: color(white)(xxxxxxxx.x)-11y =0#

#color(white)(xxxxxxxxxxxxxxxxx.x)y =0#

If #y=0# then #-8x =-8#

Which gives #x =1#

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