To solve the system of equations:
[tex]\[
\begin{cases}
y = x^2 + 3x + 3 \\
y = x^2 + x + 2
\end{cases}
\][/tex]
Firstly, we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations.
Step 1: Set the equations equal to each other.
Since both equations equal [tex]\(y\)[/tex], we can set the right-hand sides of the equations equal to each other:
[tex]\[
x^2 + 3x + 3 = x^2 + x + 2
\][/tex]
Step 2: Simplify the resulting equation.
Subtract [tex]\(x^2 + x + 2\)[/tex] from both sides:
[tex]\[
(x^2 + 3x + 3) - (x^2 + x + 2) = 0
\][/tex]
[tex]\[
3x + 3 - x - 2 = 0
\][/tex]
Combine like terms:
[tex]\[
2x + 1 = 0
\][/tex]
Step 3: Solve for [tex]\(x\)[/tex].
Isolate [tex]\(x\)[/tex]:
[tex]\[
2x = -1
\][/tex]
[tex]\[
x = -\frac{1}{2}
\][/tex]
Step 4: Substitute [tex]\(x = -\frac{1}{2}\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex].
We will use the first equation for this:
[tex]\[
y = \left(-\frac{1}{2}\right)^2 + 3\left(-\frac{1}{2}\right) + 3
\][/tex]
[tex]\[
y = \frac{1}{4} - \frac{3}{2} + 3
\][/tex]
[tex]\[
y = \frac{1}{4} - \frac{6}{4} + \frac{12}{4}
\][/tex]
[tex]\[
y = \frac{1 - 6 + 12}{4}
\][/tex]
[tex]\[
y = \frac{7}{4}
\][/tex]
[tex]\[
y = 1.75
\][/tex]
The solution to the system of equations is [tex]\(\left(-\frac{1}{2}, 1.75\right) \)[/tex].
Step 5: Compare the solution to the given options.
The provided options are:
A. [tex]\((-0.5, 0.2)\)[/tex]
B. [tex]\((-0.5, 1.7)\)[/tex]
C. [tex]\((-0.1, 1.2)\)[/tex]
D. [tex]\((0.2, 1.5)\)[/tex]
We can see that the solution [tex]\((-0.5, 1.75)\)[/tex] is closest to the choice B. [tex]\((-0.5, 1.7)\)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{\text{B}}
\][/tex]
