Problem
ALG-B1-M03-P018 Two Branches After Subtraction
#18
★★★★☆ Level 4 of 5
Solve the system \(x^2+2y=9\), \(y^2+2x=9\).
Subtract the equations and factor the result.
Subtract: \(x^2-y^2+2y-2x=0\), so \((x-y)(x+y-2)=0\).
If \(x=y\), then \(x^2+2x=9\), so \(x=-1\pm\sqrt{10}\). This gives two pairs \((-1+\sqrt{10},-1+\sqrt{10})\), \((-1-\sqrt{10},-1-\sqrt{10})\).
If \(x+y=2\), then \(y=2-x\). Substitution gives \(x^2+4-2x=9\), so \(x^2-2x-5=0\). Hence \(x=1\pm\sqrt6\), \(y=1\mp\sqrt6\).
Both branches must be handled completely.