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What is the approximate solution of this system of equations?

[tex]\[
\begin{array}{l}
y = x^2 + 3x + 3 \\
y = x^2 + x + 2
\end{array}
\][/tex]

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]


Sagot :

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]