Question

which system of equations can be graphed to find the solution(s) to $x^2 = 2x + 3$?
\\(\

$$\begin{cases}y = x^2 + 2x + 3\\\\y = 2x + 3\\end{cases}$$

\\)
\\(\

$$\begin{cases}y = x^2 - 3\\\\y = 2x + 3\\end{cases}$$

\\)
\\(\

$$\begin{cases}y = x^2 - 2x - 3\\\\y = 2x + 3\\end{cases}$$

\\)
\\(\

$$\begin{cases}y = x^2\\\\y = 2x + 3\\end{cases}$$

\\)

Explanation:

Step1: Understand the Goal

We need to find a system of equations whose intersection points (when graphed) give the solution to \(x^2 = 2x + 3\). The solution to a system of equations \(

$$\begin{cases}y = f(x)\\y = g(x)\end{cases}$$

\) is found where \(f(x)=g(x)\). So we need to split \(x^2 = 2x + 3\) into two functions \(y = f(x)\) and \(y = g(x)\) such that their equality is the original equation.

Step2: Analyze Each Option

• Option 1: \(

$$\begin{cases}y = x^2 + 2x + 3\\y = 2x + 3\end{cases}$$

\)
Setting \(x^2 + 2x + 3=2x + 3\) simplifies to \(x^2=0\), which is not the original equation \(x^2 = 2x + 3\). So this is incorrect.

• Option 2: \(

$$\begin{cases}y = x^2 - 3\\y = 2x + 3\end{cases}$$

\)
Setting \(x^2 - 3=2x + 3\) simplifies to \(x^2-2x - 6=0\), not the original equation \(x^2 = 2x + 3\) (which is \(x^2-2x - 3=0\)). So this is incorrect.

• Option 3: \(

$$\begin{cases}y = x^2 - 2x - 3\\y = 2x + 3\end{cases}$$

\)
Setting \(x^2 - 2x - 3=2x + 3\) simplifies to \(x^2-4x - 6=0\), not the original equation. So this is incorrect.

• Option 4: \(

$$\begin{cases}y = x^2\\y = 2x + 3\end{cases}$$

\)
Setting \(x^2=2x + 3\) (by equating the two \(y\)-expressions) gives the original equation. So the intersection of \(y = x^2\) and \(y = 2x + 3\) will solve \(x^2 = 2x + 3\).

Answer:

\(\boldsymbol{

$$\begin{cases}y = x^2\\y = 2x + 3\end{cases}$$

}\) (the last option, or the option with \(y = x^2\) and \(y = 2x + 3\))