Question #72d69

1 Answer
Jan 21, 2016

{ (x=2),(y=0):}

Explanation:

Looking at Systems Using Substitution

therefore:

{ (y=x-2),(2x+2y=4):} <=>{ (y=x-2),(cancel(2)x+cancel(2)y=cancel(4)^2):}

{ (y=x-2 " s1"),(x+y=2 " s2") :}

you have already found y in s1 then substitute it in s2

{ (y=x-2),(x+(x-2)=2) :}<=>{ (y=x-2),(2x-2=2) :}

<=>{ (y=x-2" s1"),(x=2" s2") :}

In s2 you found the x value then substitute it in s1

{ (y=2-2),(x=2):}<=>{ (y=0),(x=2):}