Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Which system of equations can be graphed to find the solution(s) to [tex][tex]$x^2 = 2x + 3$[/tex][/tex]?

A. [tex]\left\{\begin{array}{l}y = x^2 + 2x + 3 \\ y = 2x + 3\end{array}\right.[/tex]
B. [tex]\left\{\begin{array}{l}y = x^2 - 3 \\ y = 2x + 3\end{array}\right.[/tex]
C. [tex]\left\{\begin{array}{l}y = x^2 - 2x - 3 \\ y = 2x + 3\end{array}\right.[/tex]
D. [tex]\left\{\begin{array}{l}y = x^2 \\ y = 2x + 3\end{array}\right.[/tex]

Sagot :

To solve the equation [tex]\(x^2 = 2x + 3\)[/tex] by finding the system of equations that can be graphed, we want to express both sides of the equation as separate functions of [tex]\(x\)[/tex]. Then, we'll see which system of equations involves these functions.

Starting from the given equation:
[tex]\[ x^2 = 2x + 3 \][/tex]

We can rearrange the terms to form:
[tex]\[ x^2 - 2x - 3 = 0 \][/tex]

Now we need to find a system of equations where one equation represents [tex]\(y = x^2\)[/tex] and the other represents [tex]\(y = 2x + 3\)[/tex].

Let's look at the options provided:

1. [tex]\(\left\{\begin{array}{l} y = x^2 + 2x + 3 \\ y = 2x + 3 \end{array}\right.\)[/tex]

2. [tex]\(\left\{\begin{array}{l} y = x^2 - 3 \\ y = 2x + 3 \end{array}\right.\)[/tex]

3. [tex]\(\left\{\begin{array}{l} y = x^2 - 2x - 3 \\ y = 2x + 3 \end{array}\right.\)[/tex]

4. [tex]\(\left\{\begin{array}{l} y = x^2 \\ y = 2x + 3 \end{array}\right.\)[/tex]

We can see that the correct system should include the equation [tex]\(y = x^2\)[/tex] on one side and [tex]\(y = 2x + 3\)[/tex] on the other side.

Upon comparing and analysis of each options:
1. [tex]\(\left\{\begin{array}{l} y = x^2 + 2x + 3 \\ y = 2x + 3 \end{array}\right.\)[/tex] is incorrect because of the left part [tex]\(x^2 + 2x + 3 \neq x^2\)[/tex]
2. [tex]\(\left\{\begin{array}{l} y = x^2 - 3 \\ y = 2x + 3 \end{array}\right.\)[/tex] doesn't match the rearranged [tex]\(x^2 - 2x - 3 \equiv 0\)[/tex]
3. [tex]\(\left\{\begin{array}{l} y = x^2 - 2x - 3 \\ y = 2x + 3 \end{array}\right.\)[/tex] rearranged terms PERFECTLY combines both full equations.
4. [tex]\(\left\{\begin{array}{l} y = x^2 \\ y = 2x + 3 \end{array}\right.\)[/tex] is incomplete, as it misses the component of matching BOTH equations.

Thus, the correct system of equations to graph in order to find the solutions to [tex]\(x^2 = 2x + 3\)[/tex] is:
[tex]\[ \left\{\begin{array}{l} y = x^2 - 2x - 3 \\ y = 2x + 3 \end{array}\right. \][/tex]