Question

find the ordered pair solutions for the system of equations.
$\begin{cases}y = x^{2}-2x - 3\\y = x + 1end{cases}$
$(?, )$ and $(, )$
enter the smallest x first.

Explanation:

Step1: Set the equations equal

Since both expressions equal \(y\), we set \(x^{2}-2x - 3=x + 1\).
\[x^{2}-2x-3=x + 1\]

Step2: Rearrange to standard - form

Move all terms to one side to get a quadratic equation: \(x^{2}-2x - x-3 - 1=0\), which simplifies to \(x^{2}-3x - 4 = 0\).
\[x^{2}-3x - 4=0\]

Step3: Factor the quadratic equation

Factor \(x^{2}-3x - 4\) as \((x - 4)(x+1)=0\).
\[(x - 4)(x + 1)=0\]

Step4: Solve for \(x\)

Using the zero - product property, if \((x - 4)(x + 1)=0\), then \(x-4 = 0\) or \(x + 1=0\). So \(x=4\) or \(x=-1\).

Step5: Find the corresponding \(y\) values

When \(x=-1\), substitute into \(y=x + 1\), then \(y=-1 + 1=0\).
When \(x = 4\), substitute into \(y=x + 1\), then \(y=4 + 1=5\).

Answer:

\((-1,0)\) and \((4,5)\)