Problem 86 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

The solution is \((x, y) = (-4, 1)\).

Step 1: Solve First Equation for x

Let's start with the first equation: \((y+1)^2 = -x\). We'll solve for \(x\) by isolating it. We get:\[-x = (y+1)^2\]Then, multiply both sides by -1 to obtain:\[x = -(y+1)^2\].

Step 2: Solve Second Equation for x

Now, consider the second equation: \(-(y-1)^2 = x + 4\). We also solve for \(x\):Subtract 4 from both sides:\[x = -(y-1)^2 - 4\].

Step 3: Set Equations for x Equal

Since both expressions represent \(x\), these equations can be set equal to each other:\[-(y+1)^2 = -(y-1)^2 - 4\].

Step 4: Simplify and Solve for y

Simplify the equation:First, expand both sides:- Expand \((y+1)^2\): \[(y+1)^2 = y^2 + 2y + 1\]- Expand \((y-1)^2\): \[(y-1)^2 = y^2 - 2y + 1\]Substitute these into the equation:\[-(y^2 + 2y + 1) = -(y^2 - 2y + 1) - 4\]Simplify:\[-y^2 - 2y - 1 = -y^2 + 2y - 1 - 4\]Remove \(-y^2\) from both sides:\[-2y - 1 = 2y - 5\]Add 2y to both sides:\[-1 = 4y - 5\]Add 5 to both sides:\[4 = 4y\]Divide by 4:\[y = 1\].

Step 5: Calculate x using y = 1

Substitute \(y = 1\) back into one of the equations for \(x\). Using \(x = -(y+1)^2\):Substitute \(y = 1\):\[x = -(1+1)^2 = -(2)^2 = -4\].

Step 6: Write Final Solution

We have found that the solution to the system is \((x, y) = (-4, 1)\).

Key Concepts

Solving Equations

A system of equations is a set of two or more equations that you deal with all at once. These equations usually share the same variables, and our goal is to find the values of these variables that satisfy all equations in the system.When solving this system of equations, \( (y+1)^2 = -x \) and \( -(y-1)^2 = x + 4 \), the approach involves expressing one variable in terms of the others. We start with manipulating each equation to isolate one variable, typically "x" or "y".

• In the first equation, we rearranged it to express \( x \) as \( -(y+1)^2 \).
• Similarly, the second equation was rearranged to express \( x \) as \( -(y-1)^2 - 4 \).

This technique allows us to equate the two expressions since both represent the same variable, leading to a new equation only in terms of the other variable, typically "y" in our case. This is a classic step in solving systems of equations, helping to simplify the problem to a single-variable equation that can be solved using standard algebraic methods.

Algebraic Methods

Algebraic methods provide a logical sequence of actions that help to solve an equation or a system of equations. These methods make use of rules and operations such as addition, subtraction, multiplication, division, and factoring, to simplify equations and isolate variables. Let's break the process down further:

• **Expanding Expressions**: We use the distributive property to open up brackets, as seen when turning \( (y+1)^2 \) into \( y^2 + 2y + 1 \) and \( (y-1)^2 \) into \( y^2 - 2y + 1 \).
• **Combining Like Terms**: This involves consolidating terms that share the same variable components, simplifying them into a more manageable form. This step becomes essential when aligning all terms to one side of the equation to simplify the problem.
• **Isolating Variables**: To simplify, we move terms around within the equation, changing their signs accordingly, until one side of the equation stands alone as just the variable, such as rearranging terms to solve for \( y \) in the equation resulted from setting two expressions for \( x \) equal.

Additionally, algebraic methods require attention to balance; whatever operation is performed on one side of an equation must also be done to the other, to maintain equality. Mastery of these moves helps simplify complex algebraic problems.

Quadratic Equations

Quadratic equations are a form of polynomial equations where the highest power of the variable is two, often written in the standard form of \( ax^2 + bx + c = 0 \). These kinds of equations are prevalent in algebra and solving them can involve multiple techniques such as factoring, using the quadratic formula, or completing the square.When dealing with our system of equations, the expressions \( (y+1)^2 \) and \( (y-1)^2 \) involved squares of linear terms, making them quadratic in nature.

• **Square Expansions**: As shown, expanding \( (y+1)^2 \) results in a quadratic expression \( y^2 + 2y + 1 \).
• **Equal Setting**: By equating two such quadratic expressions, and then simplifying them, you essentially solve a simplified quadratic equation to find the value of variables such as \( y \).

Recognizing quadratic equations within a set of algebraic manipulations is crucial, as they dictate the methods we may use to find the correct solutions. Knowing how to handle these forms equips students with the skill needed to tackle a wide range of problems in mathematics.