Question

Solve the following system of equations algebraically:
y, equals, x, squared, minus, 2, x, minus, 43
y=
x
2
−2x−43
y, equals, 2, x, plus, 2
y=
2x+2

Answer 1

To solve the system of equations algebraically, we have the following two equations:

1. \( y = x^2 - 2x - 43 \)
2. \( y = 2x + 2 \)

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

\[
x^2 - 2x - 43 = 2x + 2
\]

Next, rearranging the equation gives us:

\[
x^2 - 2x - 43 - 2x - 2 = 0
\]

This simplifies to:

\[
x^2 - 4x - 45 = 0
\]

Now, to solve for \( x \), we can factor the quadratic equation:

\[
(x - 9)(x + 5) = 0
\]

Setting each factor equal to zero gives us the possible values for \( x \):

1. \( x - 9 = 0 \) → \( x = 9 \)
2. \( x + 5 = 0 \) → \( x = -5 \)

Now we can find the corresponding \( y \) values by substituting these \( x \) values back into either equation. We choose the second equation for simplicity:

**For \( x = 9 \)**:

\[
y = 2(9) + 2 = 18 + 2 = 20
\]

**For \( x = -5 \)**:

\[
y = 2(-5) + 2 = -10 + 2 = -8
\]

Thus, the solutions to the system of equations are:

1. \( (9, 20) \)
2. \( (-5, -8) \)

So the final answer is:

\[
(9, 20) \quad \text{and} \quad (-5, -8)
\]