Question

using elimination or substitution solve this: 4x^2-y^2+12=0 and x+y=3

Answer 1

\[x+y=3 \implies y=3-x\]Substitute that into the first equation to get\[4x^2-(3-x)^2+12=0\]\[4x^2-(x^2-6x+9)+12=0 \implies 3x^2+6x+3=0 \implies x^2+2x+1=0\]\[\implies x=-1\implies y=4 \implies (x,y)=(-1,4)\]