Answer :
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]
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]
