Problem 23 - Intermediate Algebra

Question

For Problems \(1-28\), (a) graph each system so that approximate real number solutions (if there are any) can be predicted, and (b) solve each system using the substitution method or the elimination-by-addition method. (Objectives 1 and 2) $$ \left(\begin{array}{l} y=-x^{2}-3 \\ y=-2 x^{2}+1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solutions are \((2, -7)\) and \((-2, -7)\).

Step 1: Understanding the Equations

We have two equations, namely \( y = -x^2 - 3 \) and \( y = -2x^2 + 1 \). These equations represent two parabolas that open downward because the coefficients of \(x^2\) are negative.

Step 2: Graph the Equations

Plot both equations on the coordinate plane. For the first equation \( y = -x^2 - 3 \), the vertex is at (0, -3) and it extends downward. For the second equation \( y = -2x^2 + 1 \), the vertex is at (0, 1) and it is narrower compared to the first one, extending downward as well.

Step 3: Identify Intersection Points

Look for the points where the two parabolas intersect. By visually inspecting the graph, estimate the approximate points of intersection. The points where they intersect are the approximate solutions.

Step 4: Using Substitution to Solve

Set the equations equal to each other since they are both equal to \( y \). This gives us \(-x^2 - 3 = -2x^2 + 1\). Solve for the variable \(x\) by first rearranging the equation to \(-x^2 + 2x^2 = 1 + 3\). This simplifies to \(x^2 = 4\).

Step 5: Solve for x-values

Take the square root of both sides to find \(x\), giving us \(x = 2\) or \(x = -2\). So, the solutions for \(x\) are \(x = 2\) and \(x = -2\).

Step 6: Solve for Corresponding y-values

Substitute \(x = 2\) and \(x = -2\) back into one of the original equations, such as \(y = -x^2 - 3\), to find the corresponding \(y\) values. For \(x = 2\), \(y = -(2)^2 - 3 = -4 - 3 = -7\). For \(x = -2\), \(y = -(-2)^2 - 3 = -4 - 3 = -7\).

Step 7: Conclusion - List the Solutions

The full solutions to the system are the points \((2, -7)\) and \((-2, -7)\). These points represent the intersections of the graphs of the two equations.

Key Concepts

Graphing Systems of Equations

Graphing systems of equations is an effective visual method for solving and understanding the relationships between different equations. When graphing, each equation is represented as a line or curve on a coordinate plane.
For instance, consider the system given by \( y = -x^2 - 3 \) and \( y = -2x^2 + 1 \). Both are quadratic equations, which graph as parabolas.

• The graph of \( y = -x^2 - 3 \) is a downward-opening parabola with a vertex at \((0, -3)\).
• The graph of \( y = -2x^2 + 1 \) is also a downward-opening parabola, narrower than the first, with a vertex at \((0, 1)\).

By graphing, you can find the intersection points, which are the solutions to the system. These points are where both equations have the same \( y \)-value for the same \( x \)-value. Identifying these graphically gives you approximate solutions that can be further verified algebraically.

Substitution Method

The substitution method is a strategic way to find exact solutions to systems of equations. You start by solving one of the equations for one variable, then substitute this expression into the other equation.
To solve the system using substitution:

• Since both given equations are equal to \( y \), set \(-x^2 - 3 = -2x^2 + 1\).
• Rearrange this to \(-x^2 + 2x^2 = 1 + 3\), which simplifies to \(x^2 = 4\).

Taking the square root of both sides results in \(x = 2\) or \(x = -2\).
Next, plug these \( x \)-values back into one of the original equations (such as \( y = -x^2 - 3 \)) to find corresponding \( y \)-values, resulting in the solutions \((2, -7)\) and \((-2, -7)\).
This method eliminates one variable, simplifying the problem to an equation with a single variable.

Elimination Method

The elimination method is another powerful algebraic technique used to solve systems of equations. It involves adding or subtracting equations to eliminate a variable. Although it was not used explicitly in this problem, understanding it can expand your problem-solving toolkit.
Here's a brief overview of how you would use elimination:

• Align both equations, such that similar terms are in columns.
• Multiply each equation by constants if necessary, so that adding or subtracting them will cancel out one of the variables.
• Solve the resulting equation for the remaining variable.

Once one variable is eliminated, and you have a single equation in the other variable. Solve for it, and substitute back into either original equation to find the second variable.
This method is particularly useful when the coefficients of one variable are already the same or can easily be made the same.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, generally expressed in the form \( y = ax^2 + bx + c \). The graphs of quadratic functions are parabolas.
Key features include:

• Vertex: The highest or lowest point of the parabola; for \( y = ax^2 + bx + c \), this can be found at \( x = -\frac{b}{2a} \).
• Direction: If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards.
• Axis of Symmetry: A vertical line through the vertex dividing the parabola into two mirror-image halves, at \( x = -\frac{b}{2a} \).

In our given system, both equations are quadratic, thus their intersections help identify the solutions to both equations. Understanding quadratics provides insight into the behavior and solutions of these systems of equations.