Question

Can somebody please help me? IMPORTANT

Answer 1

`\begin{cases} (x-1)^2-(x+2)^2=9y\\\\ (y-3)^2-(y+2)^2=5x \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-1)^2-(x+2)^2=9y\implies (x^2-2x+1)-(x^2+4x+4)=9y \\\\\\ (x^2-2x+1)-x^2-4x-4=9y\implies -6x-3=9y\implies -3(2x+1)=9y \\\\\\ 2x+1=\cfrac{9y}{-3}\implies 2x+1=-3y\implies 2x=-3y-1\implies x=\cfrac{-3y-1}{2} \\\\[-0.35em] ~\dotfill\\\\ (y-3)^2-(y+2)^2=5x\implies (y^2-6y+9)-(y^2+4y+4)=5x \\\\\\ (y^2-6y+9)-y^2-4y-4=5x\implies -10y+5=5x`

`\stackrel{\textit{substituting from above}}{-10y+5=5\left( \cfrac{-3y-1}{2} \right)}\implies -10y+5=\cfrac{-15y-5}{2} \\\\\\ -20y+10=-15y-5\implies 10=5y-5\implies 15=5y \\\\\\ \cfrac{15}{5}=y\implies \boxed{3=y} \\\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{-3y-1}{2}}\implies x=\cfrac{-3(3)-1}{2}\implies \boxed{x=-5}`