Question

Solve the following system of equations algebraically:
y, equals, x, squared, minus, x, minus, 3
y=
x
2
−x−3
y, equals, 3, x, minus, 3
y=
3x−3

Answer 1

To solve the system of equations algebraically, we have:

1. \( y = x^2 - x - 3 \) (Equation 1)
2. \( y = 3x - 3 \) (Equation 2)

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

\[
x^2 - x - 3 = 3x - 3
\]

Now, we will rearrange this equation to one side:

\[
x^2 - x - 3 - 3x + 3 = 0
\]

Combining like terms:

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

Now, we can factor the equation:

\[
x(x - 4) = 0
\]

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

1. \( x = 0 \)
2. \( x - 4 = 0 \) → \( x = 4 \)

Now, we will substitute these values of \( x \) back into either of the original equations to find the corresponding \( y \) values.

**For \( x = 0 \):**

Using Equation 2:

\[
y = 3(0) - 3 = -3
\]

So one solution is \( (0, -3) \).

**For \( x = 4 \):**

Using Equation 2:

\[
y = 3(4) - 3 = 12 - 3 = 9
\]

So the second solution is \( (4, 9) \).

In conclusion, the solutions to the system of equations are:

\[
(0, -3) \quad \text{and} \quad (4, 9).
\]