Problem 28 - 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) $$ \left(\begin{array}{l} y=x^{2}-1 \\ x-y=3 \end{array}\right) $$

Step-by-Step Solution

Answer

The system has no real number solutions because the quadratic equation yields complex roots.

Step 1: Understand the System of Equations

The system given involves a quadratic equation and a linear equation. The first equation is a quadratic function: \(y = x^2 - 1\) which is a parabola opening upwards. The second equation is linear: \(x - y = 3\) or rewritten as \(y = x - 3\). Our goal is to find the intersection points of these two curves.

Step 2: Graph the Equations

Graph \(y = x^2 - 1\) and \(y = x - 3\) on the same coordinate plane. The parabolic graph of \(y = x^2 - 1\) denotes a U-shaped curve going upward from its vertex at (0, -1). The linear graph \(y = x - 3\) is a straight line with a slope of 1, crossing the y-axis at -3. Where these two graphs intersect provides the solution to the system.

Step 3: Use Substitution Method to Solve Algebraically

To find the exact points of intersection, substitute \(y = x - 3\) into the first equation \(y = x^2 - 1\). We get:1. \(x - 3 = x^2 - 1\)Solving this equation will provide the x-values of intersection.

Step 4: Simplify and Solve Equation

Rearrange the equation from the previous step: \(x - 3 = x^2 - 1\) becomes \(x^2 - x + 2 = 0\).This is a quadratic equation in standard form. Attempt to solve it by factoring, using the quadratic formula, or completing the square.

Step 5: Apply Quadratic Formula

Since \(x^2 - x + 2 = 0\) does not factor easily, apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -1\), and \(c = 2\). Calculate:\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times 2}}{2 \times 1} \]\[ = \frac{1 \pm \sqrt{1 - 8}}{2} \]\[ = \frac{1 \pm \sqrt{-7}}{2} \]The discriminant \(1-8 = -7\) is negative, indicating no real solutions.

Step 6: Interpret the Result

Since the quadratic equation yields complex solutions, the parabola and the line do not intersect in the real number plane. Therefore, there are no real number solutions to this system.

Key Concepts

Graphing Quadratic Functions

Graphing quadratic functions involves plotting a parabolic curve on the coordinate plane. A parabola is defined by the equation \(y = ax^2 + bx + c\). This curve exhibits some distinctive features:

• Vertex: The turning point of the parabola. For the function \(y = x^2 - 1\), the vertex is at \((0, -1)\), because it is derived by setting \(x = 0\) in the equation.
• Direction: The parabola opens upwards if the coefficient of \(x^2\) (in this case, 1) is positive, and downward if it's negative.
• Symmetry: Parabolas are symmetric with respect to the vertical line through their vertex. This means if you fold the graph along this line, the two halves will match perfectly.

Graphing helps in visualizing solutions where two functions may intersect. In the given exercise, the parabola \(y = x^2 - 1\) represents one of the equations in the system we are trying to solve. To predict solutions graphically, sketch the curve accurately and note points where it might intersect with other lines or curves. Such intersection points are the possible solutions of the system where both equations hold true simultaneously.

Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of equations. It involves expressing one variable in terms of the other using one of the given equations, and then substituting this expression into the other equation. This helps in simplifying the system to a single equation with one unknown, making it easier to solve.Here's a step-by-step guide to applying the substitution method:

• First, solve one of the equations for either \(x\) or \(y\). In the example system, the linear equation \( x - y = 3\) can be rearranged as \(y = x - 3\).
• Substitute this expression \(y = x - 3\) into the quadratic equation \(y = x^2 - 1\). This gives \(x - 3 = x^2 - 1\).
• Simplify and solve the resulting equation (now only in terms of one variable). Here it becomes a quadratic equation \(x^2 - x + 2 = 0\).

The substitution method is particularly effective when one of the equations is easy to solve for one of the variables. While it may not always yield graphically intuitive solutions due to complex numbers or no real solutions, it offers a clear algebraic path to potential solutions.

Elimination by Addition Method

The elimination by addition method, often simply called the elimination method, is another algebraic approach to solving systems of equations. This technique involves adding or subtracting equations to eliminate one of the variables. Once again, this simplifies the system to a single equation in one variable, making it straightforward to solve. Here's how to use the elimination method effectively:

• Align equations so corresponding terms are stacked vertically. Adjust the equations as necessary by multiplying if needed, so that one of the variables has equal coefficients.
• Add or subtract the equations to eliminate one of the variables. By eliminating a variable, you reduce the system to one equation with a single unknown.
• Solve the resulting equation for the remaining variable.
• Substitute the value obtained back into one of the original equations to find the other variable.

Although in this specific exercise the substitution method and graphing are primary, understanding elimination provides versatility in approaching different systems. It is particularly powerful when the coefficients of one variable are easy to eliminate without excessive manipulation. This clear and systematic approach can simplify complex problems and is a fundamental skill in algebra.