Question

Solve the system of equations:
y =x+2
y=x2+5x+6
O (-2, 0) and (-3, 0)

0 (-2,01

0 (0,2)

O (0, 2) and (2, 4)

Answer 1

To solve the system of equations:

1. \( y = x + 2 \)
2. \( y = x^2 + 5x + 6 \)

We can set the two equations equal to each other since they both equal \( y \):

\[
x + 2 = x^2 + 5x + 6
\]

Now, rearranging the equation:

\[
0 = x^2 + 5x + 6 - x - 2
\]
\[
0 = x^2 + 4x + 4
\]
\[
0 = (x + 2)^2
\]

This means that \( x + 2 = 0 \), which gives:

\[
x = -2
\]

To find the corresponding \( y \) value, we substitute \( x = -2 \) into the first equation:

\[
y = -2 + 2 = 0
\]

So, we have one solution: \( (-2, 0) \).

Now, we'll verify if there are any other solutions by examining the second equation:

Substituting \( x = -2 \) into the second equation:

\[
y = (-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0
\]

Now we check if the parabola \( y = x^2 + 5x + 6 \) intersects the line \( y = x + 2 \) at any other point. The quadratic \( (x + 2)^2 = 0 \) leads us to conclude that the only intersection point occurs at \( x = -2 \).

Thus, the only solution to the system of equations is:

\[
(-2, 0)
\]

From the options given, the correct answer is:

- \( O (-2, 0) \)

The other potential coordinates provided do not satisfy the equations.