To solve the system of equations, we will determine the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations:
[tex]\[
\begin{array}{l}
y = x^2 + x + 5 \\
y = x + 1
\end{array}
\][/tex]
We can set the right-hand sides of the equations equal to each other, as they both equal [tex]\( y \)[/tex]. So, we have:
[tex]\[
x^2 + x + 5 = x + 1
\][/tex]
Next, we will simplify this equation by moving all terms to one side:
[tex]\[
x^2 + x + 5 - x - 1 = 0
\][/tex]
This simplifies to:
[tex]\[
x^2 + 4 = 0
\][/tex]
To solve for [tex]\( x \)[/tex], we subtract 4 from both sides:
[tex]\[
x^2 = -4
\][/tex]
Taking the square root of both sides, we get:
[tex]\[
x = \pm \sqrt{-4}
\][/tex]
The square root of [tex]\(-4\)[/tex] is [tex]\( \pm 2i \)[/tex], where [tex]\( i \)[/tex] is the imaginary unit.
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[
x = \pm 2i
\][/tex]
Now, we find the corresponding [tex]\( y \)[/tex] values using [tex]\( y = x + 1 \)[/tex]:
For [tex]\( x = 2i \)[/tex]:
[tex]\[
y = 2i + 1
\][/tex]
For [tex]\( x = -2i \)[/tex]:
[tex]\[
y = -2i + 1
\][/tex]
The solutions to the system of equations are the pairs [tex]\((2i, 2i + 1)\)[/tex] and [tex]\((-2i, -2i + 1)\)[/tex].
These do not match any of the given options exactly.
Therefore, the correct answer to the question is:
B no solution
