Answer :
To determine which system of equations can be graphed to find the solutions to the equation [tex]\( x^2 = 2x + 3 \)[/tex], you need to transform the given equation and determine which system reflects that transformation.
1. Start with the given equation:
[tex]\[ x^2 = 2x + 3 \][/tex]
2. Rearrange this equation to set it to zero:
[tex]\[ x^2 - 2x - 3 = 0 \][/tex]
3. The systems of equations given are:
- [tex]\(\left\{\begin{array}{l}y = x^2 + 2x + 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
- [tex]\(\left\{\begin{array}{l}y = x^2 - 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
- [tex]\(\left\{\begin{array}{l}y = x^2 - 2x - 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
- [tex]\(\left\{\begin{array}{l}y = x^2 \\ y = 2x + 3\end{array}\right.\)[/tex]
4. To solve the original equation by finding the intersection points, we need to match the transformed equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex] with the correct system. Notice that if we add [tex]\( y \)[/tex] to both sides of the equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex], it becomes:
[tex]\[ y = x^2 - 2x - 3 \][/tex]
5. Now we compare this result with the systems given:
- In the first system [tex]\(\{y = x^2 + 2x + 3, y = 2x + 3\}\)[/tex], the first equation doesn't match with [tex]\( y = x^2 - 2x - 3 \)[/tex].
- In the second system [tex]\(\{y = x^2 - 3, y = 2x + 3\}\)[/tex], the first equation doesn't match with [tex]\( y = x^2 - 2x - 3 \)[/tex].
- In the third system [tex]\(\{y = x^2 - 2x - 3, y = 2x + 3\}\)[/tex], the first equation [tex]\( y = x^2 - 2x - 3 \)[/tex] matches perfectly.
- In the fourth system [tex]\(\{y = x^2, y = 2x + 3\}\)[/tex], the first equation doesn't match with [tex]\( y = x^2 - 2x - 3 \)[/tex].
So, the correct system of equations that can be graphed 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]}.
Thus, the answer is:
[tex]\[ \boxed{3} \][/tex]
1. Start with the given equation:
[tex]\[ x^2 = 2x + 3 \][/tex]
2. Rearrange this equation to set it to zero:
[tex]\[ x^2 - 2x - 3 = 0 \][/tex]
3. The systems of equations given are:
- [tex]\(\left\{\begin{array}{l}y = x^2 + 2x + 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
- [tex]\(\left\{\begin{array}{l}y = x^2 - 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
- [tex]\(\left\{\begin{array}{l}y = x^2 - 2x - 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
- [tex]\(\left\{\begin{array}{l}y = x^2 \\ y = 2x + 3\end{array}\right.\)[/tex]
4. To solve the original equation by finding the intersection points, we need to match the transformed equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex] with the correct system. Notice that if we add [tex]\( y \)[/tex] to both sides of the equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex], it becomes:
[tex]\[ y = x^2 - 2x - 3 \][/tex]
5. Now we compare this result with the systems given:
- In the first system [tex]\(\{y = x^2 + 2x + 3, y = 2x + 3\}\)[/tex], the first equation doesn't match with [tex]\( y = x^2 - 2x - 3 \)[/tex].
- In the second system [tex]\(\{y = x^2 - 3, y = 2x + 3\}\)[/tex], the first equation doesn't match with [tex]\( y = x^2 - 2x - 3 \)[/tex].
- In the third system [tex]\(\{y = x^2 - 2x - 3, y = 2x + 3\}\)[/tex], the first equation [tex]\( y = x^2 - 2x - 3 \)[/tex] matches perfectly.
- In the fourth system [tex]\(\{y = x^2, y = 2x + 3\}\)[/tex], the first equation doesn't match with [tex]\( y = x^2 - 2x - 3 \)[/tex].
So, the correct system of equations that can be graphed 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]}.
Thus, the answer is:
[tex]\[ \boxed{3} \][/tex]
