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You have provided a system of two equations in two variables \(x\) and \(y\):

\[
\begin{aligned}
1. & \quad xy - x + y = 1 \\
2. & \quad 2x^2 - xy - y = 0
\end{aligned}
\]

Let’s solve this system step by step.

### Step 1: Rearrange the first equation
From the first equation:

\[
xy - x + y = 1
\]

We can try solving for \(y\) in terms of \(x\). Rearrange the terms:

\[
xy + y = x + 1
\]

Now factor out \(y\) on the left-hand side:

\[
y(x + 1) = x + 1
\]

If \(x + 1 \neq 0\), we can divide both sides by \(x + 1\):

\[
y = 1
\]

So, one possible solution is \(y = 1\). Let's substitute \(y = 1\) into the second equation.

### Step 2: Substitute \(y = 1\) into the second equation
The second equation is:

\[
2x^2 - xy - y = 0
\]

Substitute \(y = 1\):

\[
2x^2 - x(1) - 1 = 0
\]

Simplify:

\[
2x^2 - x - 1 = 0
\]

This is a quadratic equation in \(x\). We can solve this using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(2x^2 - x - 1 = 0\), we have \(a = 2\), \(b = -1\), and \(c = -1\). Substituting these values into the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 8}}{4}
\]
\[
x = \frac{1 \pm \sqrt{9}}{4}
\]
\[
x = \frac{1 \pm 3}{4}
\]

So, we have two solutions for \(x\):

\[
x = \frac{1 + 3}{4} = 1 \quad \text{or} \quad x = \frac{1 - 3}{4} = -\frac{1}{2}
\]

### Step 3: Solutions
- If \(x = 1\), then \(y = 1\).
- If \(x = -\frac{1}{2}\), then \(y = 1\).

Thus, the system of equations has two solutions:

\[
(x, y) = (1, 1) \quad \text{or} \quad (x, y) = \left(-\frac{1}{2}, 1\right)
\]

Would you like further details or explanations on any step?

Here are five related questions to consider:
1. What happens if \(x = -1\) in the first equation?
2. How would the solution change if we modified the constants in the second equation?
3. Can this system be solved using matrices or another algebraic method?
4. How would you interpret the graphical solution of this system?
5. How would you handle the case where \(x + 1 = 0\)?

**Tip:** For quadratic equations, always consider checking if factoring is simpler than using the quadratic formula.

Solve the system of equations: xy - x + y = 1 and 2x^2 - xy - y = 0.

Learn how to solve a system of equations involving xy - x + y = 1 and 2x^2 - xy - y = 0. This problem explores algebraic methods, including the quadratic formula and equation rearrangement. Suitable for students in Grades 9-11.

Math Problem Statement

{ x y − x + y

1 2 x 2 − x y − y

Solution

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