Question

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{r} y^{2}+y+x^{2}-x-2=0 \\ y+1+\frac{x-2}{y}=0 \end{array}\right. $$

Short answer

(0, -2), (0, 1), (2, 0), (2, -1)

Step-by-step solution

- Simplify the Second Equation

Start by simplifying the second equation: \[ y + 1 + \frac{x-2}{y} = 0 \]Multiply both sides by \( y \) to get rid of the fraction: \[ y^2 + y + (x-2) = 0 \]Simplify further to: \[ y^2 + y + x - 2 = 0 \]

- Compare with the First Equation

Compare the simplified second equation with the first equation: First equation: \[ y^2 + y + x^2 - x - 2 = 0 \]Simplified second equation: \[ y^2 + y + x - 2 = 0 \]Subtract the two equations to eliminate common terms: \[ (y^2 + y + x^2 - x - 2) - (y^2 + y + x - 2) = 0 \]This results in: \[ x^2 - x - x = 0 \]Or: \[ x^2 - 2x = 0 \]

- Solve for x

Factor the simplified equation to solve for \( x \): \[ x^2 - 2x = 0 \]Factor out \( x \): \[ x(x - 2) = 0 \]This results in two solutions for \( x \): \[ x = 0 \] or \[ x = 2 \]

- Substitute x = 0 into the Second Equation

Substitute \( x = 0 \) into the simplified second equation: \[ y^2 + y + 0 - 2 = 0 \]This results in: \[ y^2 + y - 2 = 0 \]Factor the quadratic equation: \[ (y + 2)(y - 1) = 0 \]This results in two solutions for \( y \): \[ y = -2 \] or \[ y = 1 \]

- Substitute x = 2 into the Second Equation

Substitute \( x = 2 \) into the simplified second equation: \[ y^2 + y + 2 - 2 = 0 \]This results in: \[ y^2 + y = 0 \]Factor the quadratic equation: \[ y(y + 1) = 0 \]This results in two solutions for \( y \): \[ y = 0 \]or \[ y = -1 \]

- List All Solution Pairs

Combine the solutions for \( x \) and \( y \) to list all solution pairs: When \( x = 0 \), \( y = -2 \) or \( y = 1 \). When \( x = 2 \), \( y = 0 \) or \( y = -1 \).

Key concepts

Solving Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They can be identified by their highest exponent, which is 2. Quadratic equations often appear in different problems and can be solved using a variety of methods.

• Factoring: This involves expressing the quadratic equation as a product of its factors.
• Quadratic Formula: This uses the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions.
• Completing the Square: This method involves rearranging the equation into a perfect square trinomial.

In our exercise, we used factoring to solve the quadratic equations for \(y\). By recognizing patterns and having practice, you can quickly identify which method to use and solve quadratic equations effectively.

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This helps to reduce the system to a single equation with one variable.
Here’s how to effectively apply the substitution method:

• Solve for one variable: Pick one of the two equations and solve it for one of the variables.
• Substitute into the other equation: Take the expression obtained in the first step and substitute it into the second equation.
• Solve the resulting equation: This will give you the value of one variable.
• Back-substitute: Substitute the value obtained back into one of the original equations to find the value of the other variable.

In our solution, we started by simplifying the second equation and then used substitution to solve for \(x\) and \(y\).

Factoring

Factoring is an essential algebraic method used to solve quadratic equations. In general, factoring a quadratic equation means expressing it as the product of its linear factors. For example, the equation \(x^2 - 2x = 0\) can be factored as \(x(x - 2) = 0\).
Here's a step-by-step way to factor quadratics:

• Look for a common factor: Determine if there is a common factor that can be factored out.
• Factor by grouping: Grouping involves rearranging and grouping terms to factor by grouping.
• Quadratic expressions: Use techniques such as splitting the middle term to factor quadratic expressions.

Using these steps, we factored the equations in our exercise to find the solutions for \(x\) and \(y\). By practicing factoring, you can become adept at solving quadratic equations quickly and efficiently.