Problem
ALG-B1-M03-P008 Subtracting Equations
#8
★★☆☆☆ Level 2 of 5
Solve the system \(x^2+y=12\), \(y^2+x=12\).
Subtract one equation from the other.
Subtracting gives \(x^2-y^2+y-x=0\), so \((x-y)(x+y-1)=0\).
If \(x=y\), then \(x^2+x=12\), so \(x=3\) or \(x=-4\). This gives \((3,3)\), \((-4,-4)\).
If \(x+y=1\), then \(y=1-x\), and \(x^2+1-x=12\). Thus \(x^2-x-11=0\), so \(x=\frac{1\pm3\sqrt5}{2}\), \(y=\frac{1\mp3\sqrt5}{2}\).
Stronger than direct substitution: factorisation is created first.