Question

Solve the system y = -x + 7 and y = 0.5(x - 3)2

Answer 1

$\left\{\begin{array}{ccc}y=-x+7\\y=0.5(x-3)^2\end{array}\right\\\Downarrow\\0.5(x-3)^2=-x+7\ \ \ \ |\cdot2\ \ \ /\text{use:}\ (a-b)^2=a^2-2ab+b^2/\\\\x^2-2\cdot x\cdot3+3^2=-2x+14\\\\x^2-6x+9=-2x+14\ \ \ \ |+2x\\\\x^2-4x+9=14\ \ \ \ |-14\\\\x^2-4x-5=0\\\\x^2+x-5x-5=0\\\\x(x+1)-5(x+1)=0\\\\(x+1)(x-5)=0\iff x+1=0\ \vee\ x-5=0\\\\x=-1\ \vee\ x=5$

$\text{substitute the value of x to the first equation}\\\\\text{for}\ x=-1\to y=-(-1)+7=1+7=8\\\\\text{for}\ x=5\to y=-5+7=2\\\\\text{Answer:}\\x=-1,\ y=8\to(-1,\ 8)\\or\\x=5,\ y=2\to(5,\ 2)$