How do you solve the system #x=y-7# and #x+8y=2# using substitution?

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Answer 1 · 2018-03-05T15:50:15

#color(red)(x = -6, y = 1)#

#x = y - 7# Eqn (1)

#x + 8y = 2# Eqn (2)

Substituting value of x in terms of y in Eqn (2),

#(y - 7) + 8y = 2#

#9y - 7 = 2#

#9y = 2 + 7 = 9# or #y = 1#

Substituting value of y in Eqn (1),

#x = 1 - 7 = -6#