Solving Linear-Quadratic Systems

We want to solve the given system of equations using the substitution method. $$\{\begin{array}{ll}y=x^2-6x+9& \text{(I)}\\ y+x=5& \text{(II)}\end{array}$$ The $y\text{-}$variable is isolated in Equation (I). This allows us to substitute its value $x^2-6x+9$ for $y$ in Equation (II). Notice that in Equation (II), we have a quadratic equation in terms of only the $x\text{-}$variable. $$\begin{array}{c}{x}^2-5x+4=0\iff 1x^2+(\text{-}5)x+4=0\end{array}$$ We can substitute $a=1,$ $b=\text{-}5,$ and $c=4$ into the Quadratic Formula. This result tells us that we have two solutions for $x.$ One of them will use the positive sign and the other one will use the negative sign.

$x=\frac{5\pm 3}{2}$
$x_1=\frac{5+3}{2}$	$x_2=\frac{5-3}{2}$
$x_1=\frac{8}{2}$	$x_2=\frac{2}{2}$
$x_1=4$	$x_2=1$

We obtained two values for the $x\text{-}$variable. Now, consider Equation (I). $$\begin{array}{c}y=x^2-6x+9\end{array}$$ We can substitute $x=4$ and $x=1$ into the above equation to find the values for $y.$ Let's start with $x=4.$ We found that $y=1$ when $x=4.$ One solution to the system, which is a point of intersection of the parabola and the line, is $(4,1).$ To find the other solution, we will substitute $1$ for $x$ in Equation (I) again. We found that $y=4$ when $x=1.$ Therefore, our second solution, which is the other point of intersection of the parabola and the line, is $(1,4).$