How do you solve #y = -x+2# and #y= 3x-2#?

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Answer 1 · 2015-08-10T09:41:42

The answer is #x = 1# ; #y = 1#

Since

#{(y = -x + 2), (y = 3x - 2) :}#

We shall have

#-x +2 = 3x - 2#

#-x +2 - 3x + 2 = 0#

#- 4x + 4 = 0#

#- 4x = -4 implies x = (-4)/(-4) = 1#

Substitute #x = 1# in the first equation to get

#y = -1 + 2#

#y = 1#