Problem 85 - College Algebra with Modeling and Visualization

Question

Solve each system. $$ \begin{aligned} &(y-1)^{2}=x+1\\\ &(y+2)^{2}=-x+4 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((0, 0)\) and \((3, -1)\).

Step 1: Simplify the First Equation

The first equation is \((y-1)^{2} = x+1\). We can expand \((y-1)^2\) to get it in standard quadratic form:\[(y-1)^2 = y^2 - 2y + 1\]Substituting this back, we have:\[y^2 - 2y + 1 = x + 1\] Now, simplify the equation to solve for \(x\):\[x = y^2 - 2y + 1 - 1\]Thus, \(x = y^2 - 2y\).

Step 2: Simplify the Second Equation

In the second equation, \((y+2)^{2} = -x+4\), expand \((y+2)^2\):\[(y+2)^2 = y^2 + 4y + 4\]Substitute back into the equation:\[y^2 + 4y + 4 = -x + 4\]Simplify to solve for \(x\):\[x = -y^2 - 4y\]

Step 3: Set Equations for x Equal

We now have two equations for \(x\):1. \(x = y^2 - 2y\)2. \(x = -y^2 - 4y\)Set them equal to solve for \(y\):\[y^2 - 2y = -y^2 - 4y\]

Step 4: Solve for y

Simplify the equation from Step 3:\[y^2 + y^2 - 2y + 4y = 0\]\[2y^2 + 2y = 0\]Factor out a \(2y\):\[2y(y + 1) = 0\]Thus, the solutions for \(y\) are \(y = 0\) and \(y = -1\).

Step 5: Solve for x using y = 0

Plug \(y = 0\) into either equation for \(x\), for instance, \(x = y^2 - 2y\):\[x = 0^2 - 2(0) = 0\]This gives \((x, y) = (0, 0)\).

Step 6: Solve for x using y = -1

Plug \(y = -1\) into \(x = y^2 - 2y\):\[x = (-1)^2 - 2(-1) = 1 + 2 = 3\]This gives \((x, y) = (3, -1)\).

Step 7: Conclusion: List the Solutions

The solutions to the system are the points where the values satisfy both equations. From Steps 5 and 6, the solutions are:\((x, y) = (0, 0)\) and \((x, y) = (3, -1)\).

Key Concepts

Quadratic Equations

A quadratic equation is a polynomial equation of degree two. It generally takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In the context of systems of equations, quadratics frequently appear as one of the equations we need to solve. When dealing with quadratics:

• We often encounter terms such as \(x^2\) or \(y^2\), which signify the quadratic nature.
• Quadratic equations are pivotal in finding intersection points when graphing curves such as parabolas.
• Solving quadratic equations can involve various methods like factoring, completing the square, or using the quadratic formula.

In this exercise, the quadratic comes from expanding the expressions \((y-1)^2\) and \((y+2)^2\) to form two quadratic equations. Understanding these expansions is key to simplifying and solving the system.

Equation Solving

Equation solving involves finding the values of the variables that make the equation true. It requires strategic manipulation to isolate the variable of interest. Here are some crucial points:

• Identify if you have linear or quadratic equations to determine your solving approach.
• For this system, each equation is simplified first, allowing us to isolate \(x\) and create a base for comparison.

Once isolated, set like terms equal to one another. In this exercise, solving for \(y\) took precedence and was done by equating the two expressions for \(x\), eventually leading to a "simpler" single equation. Solutions to equations need to be checked back in the original system to assure they satisfy both conditions.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying expressions. This skill is fundamental in solving equations, simplifying certain types of expressions, and making substitution steps more straightforward.

• Rearranging terms by addition, subtraction, multiplication, or division helps in isolating variables.
• Factoring, which involves rewriting an expression as a product of its factors, can simplify complex expressions.

In this system of equations, manipulation was used to expand square terms and simplify expressions for \(x\). Factoring allowed us to solve for \(y\) by finding the values that satisfy \(2y(y + 1) = 0\). Each step of manipulation should lead closer to isolating the variable or solving the equation completely.