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use the subtitution method to solve. x^2 + 2y = 9 x - y + 3 = 0 can someone help me?
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from equation (2): y=x+3 -> (3) substitute for y with x+3, you get: \[x^2+2(x+3)=9 \implies x^2+2x-3=0\] factorize it to get: \[(x+3)(x-1)=0 \implies x=-3, x=1\] substitute for each value of x in equation 3\[x=-3 \implies y=-3+3=0\] \[x=1 \implies y=1+3=4\]
therefore the problem has two solutions: \[(x,y)=(-3,0) , (x,y)=(1,4)\]
Ok I get it. My answer was really close to that. Thank you!
you're welcome :).. feel free to fan me ;)
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