Question 23 of 25

Solve the system of equations:

[tex]\[
\begin{array}{l}
y = 2x + 1 \\
y = x^2 + 2x - 8
\end{array}
\][/tex]

A. [tex]\((-3, 5)\)[/tex] and [tex]\((3, 2)\)[/tex]

B. [tex]\((0, 1)\)[/tex] and [tex]\((2, 5)\)[/tex]

C. [tex]\((-4, 0)\)[/tex] and [tex]\((2, 0)\)[/tex]

D. [tex]\((-3, -5)\)[/tex] and [tex]\((3, 7)\)[/tex]



Answer :

To solve the system of equations:

[tex]\[ \begin{cases} y = 2x + 1 \\ y = x^2 + 2x - 8 \end{cases} \][/tex]

we can follow these steps:

1. Set the equations equal to each other:

Since both expressions are equal to [tex]\( y \)[/tex], we can set them equal to each other:

[tex]\[ 2x + 1 = x^2 + 2x - 8 \][/tex]

2. Simplify and solve for [tex]\( x \)[/tex]:

Subtract [tex]\( 2x \)[/tex] and 1 from both sides to set the equation to zero:

[tex]\[ 0 = x^2 - 8 - 1 \][/tex]

Simplify further:

[tex]\[ 0 = x^2 - 9 \][/tex]

3. Factor the quadratic equation:

The equation [tex]\( x^2 - 9 = 0 \)[/tex] can be factored as a difference of squares:

[tex]\[ (x - 3)(x + 3) = 0 \][/tex]

4. Solve for [tex]\( x \)[/tex]:

Set each factor equal to zero:

[tex]\[ x - 3 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]

Solving these gives:

[tex]\[ x = 3 \quad \text{or} \quad x = -3 \][/tex]

5. Substitute [tex]\( x \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]:

We can use the first equation [tex]\( y = 2x + 1 \)[/tex].

For [tex]\( x = 3 \)[/tex]:

[tex]\[ y = 2(3) + 1 = 6 + 1 = 7 \][/tex]

For [tex]\( x = -3 \)[/tex]:

[tex]\[ y = 2(-3) + 1 = -6 + 1 = -5 \][/tex]

6. Write the solutions as ordered pairs:

Thus, the solutions to the system of equations are:

[tex]\[ (3, 7) \quad \text{and} \quad (-3, -5) \][/tex]

Comparing to the given options:

A. [tex]\((-3,5)\)[/tex] and [tex]\((3,2)\)[/tex]

B. [tex]\((0,1)\)[/tex] and [tex]\((2,5)\)[/tex]

C. [tex]\((-4,0)\)[/tex] and [tex]\((2,0)\)[/tex]

D. [tex]\((-3,-5)\)[/tex] and [tex]\((3,7)\)[/tex]

The correct choice is therefore:

D. [tex]\((-3, -5)\)[/tex] and [tex]\((3, 7)\)[/tex]