Exercise 35 Page 262

We want to solve the given system of equations using the substitution method. Note that neither of the variables is isolated in either equation, so we need to start by isolating $y$ in Equation (II). The $y\text{-}$variable is now isolated in Equation (II). This allows us to substitute its value, $x-4,$ for $y$ in Equation (I). We can solve Equation (I) by applying the Zero Product Property. This property states that our equation will be satisfied if either of the factors is $0.$ We can see that there are two possible values of $x$ which would make Equation (I) true, $x=0$ and $x=2.$ Now, consider Equation (II). We can substitute $x=0$ and $x=2$ into the above equation to find the corresponding values of $y.$ Let's start with $x=0.$ We have found that $y=\text{-}4$ when $x=0.$ One solution to the system, which is a point of intersection of the given parabola and line, is $(0,\text{-}4).$ To find the other solution, we will substitute $2$ for $x$ in Equation (II). We found that $y=\text{-}2$ when $x=2.$ Therefore our second solution, which is the other point of intersection of the parabola and the line, is $(2,\text{-}2).$