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How do you solve the system #y-4/5x=1# and #y-x=1# using substitution?

1 Answer

Mar 13, 2018

#x=0#
#y=1#

Explanation:

#y-x=1|+x#
#y=color(red)(1+x)#

#color(red)(y)-4/5x=1#
#color(red)(1+x)-4/5x=1|-1#
#1/5x=0|*5#
#x=color(blue)(0)#

#y-color(blue)(x)=1|+x#
#y-color(blue)(0)=1#
#y=1#

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