Question

Solve the system of equations x^2+y^2=16 and y=x^2-4

Answer 1

`\large\boxed{x=0\ and\ y=-4\ or\ x=-\sqrt7\ and\ y=3\ or\ x=\sqrt7\ and\ y=3}`

Explanation:

`\left\{\begin{array}{ccc}x^2+y^2=16\\y=x^2-4&\text{add 4 to both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}x^2+y^2=16&&(1)\\y+4=x^2&\to x^2=y+4&(2)\end{array}\right\\\\\text{subtitute (2) to (1):}\\\\y+4+y^2=16\qquad\text{subtract 16 from both sides}\\\\y^2+y-12=0\\\\y^2+4y-3y-12=0\\\\y(y+4)-3(y+4)=0\\\\(y+4)(y-3)=0\iff y+4=0\ \vee\ y-3=0\\\\y+4=0\qquad\text{subtract 4 from both sides}\\\\\boxed{y_1=-4}\\\\y-3=0\qquad\text{add 3 to both sides}\\\\\boxed{y_2=3}\\\\\text{Put the values of y to (1)}`

`for\ y_1=-4\\\\x^2=-4+4\\\\x^2=0\to\boxed{x=0}\\\\for\ y_2=3\\\\x^2=3+4\\\\x^2=7\to x=\pm\sqrt7`