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Solve the system of equations.
[tex]\[
\begin{array}{l}
y = x^2 + x + 5 \\
y = x + 1
\end{array}
\][/tex]

A. [tex]\(x = -2, -2\)[/tex]

B. No solution

C. [tex]\((-2, 0)\)[/tex] and [tex]\((2, 0)\)[/tex]

D. [tex]\((-2, -1)\)[/tex] and [tex]\((2, 3)\)[/tex]


Sagot :

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