Problem 25 - 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) $$ \left(\begin{array}{l} x^{2}+y^{2}=2 \\ x-y=4 \end{array}\right) $$

Step-by-Step Solution

Answer

The system has no real solutions because the quadratic equation has a negative discriminant.

Step 1: Identify the Equations

We are given a system of equations: the first equation is \(x^2 + y^2 = 2\), which represents a circle centered at the origin with a radius of \(\sqrt{2}\). The second equation is \(x - y = 4\), which represents a line.

Step 2: Graph the Equations

To graph these equations, first plot the circle centered at the origin with radius \(\sqrt{2}\). Next, graph the line \(x - y = 4\). This line can be rewritten as \(y = x - 4\) to help with graphing. Observe where the circle and the line intersect to predict the solutions.

Step 3: Substitute y into the First Equation

Since \(x - y = 4\) implies \(y = x - 4\), substitute \(y = x - 4\) into the first equation \(x^2 + y^2 = 2\) to solve for \(x\). The equation becomes \(x^2 + (x-4)^2 = 2\).

Step 4: Expand and Simplify

Expand the equation: \(x^2 + (x-4)^2 = x^2 + (x^2 - 8x + 16) = 2\). Simplify it to get \(2x^2 - 8x + 16 = 2\).

Step 5: Rearrange into Quadratic Form

Rearrange the equation to standard quadratic form: \(2x^2 - 8x + 14 = 0\). Divide the entire equation by 2 to simplify: \(x^2 - 4x + 7 = 0\).

Step 6: Determine Real Solutions

Check the discriminant of the quadratic equation \(x^2 - 4x + 7 = 0\), which is \(b^2 - 4ac\). Here, \(a=1\), \(b=-4\), \(c=7\). So the discriminant is \((-4)^2 - 4(1)(7) = 16 - 28 = -12\).

Step 7: Conclude on Real Solutions

Since the discriminant is negative, the quadratic equation has no real solutions. Thus, the system of equations has no real number solutions.

Key Concepts

Graphing Systems of Equations

When solving systems of equations, graphing can be a helpful way to visualize the solution. In our example, we have two equations: \(x^2 + y^2 = 2\) and \(x - y = 4\).

• The first equation, \(x^2 + y^2 = 2\), represents a circle centered at the origin with a radius of \(\sqrt{2}\).
• The second equation, \(x - y = 4\), can be rewritten as \(y = x - 4\). This is the equation of a line.

To graph these, start by drawing the circle. Its center is at the origin \(0,0\), and the radius extends to all points at a distance of \(\sqrt{2}\) from the origin. This forms a small circle crossing the axes.
Next, graph the line by finding two points that satisfy the equation \(y = x - 4\). For instance, when \(x = 4\), \(y = 0\); and when \(x = 5\), \(y = 1\). Use these points to draw a straight line.
Finally, observe where the circle and the line meet. This point(s) of intersection would be the graphical solution of the system, if any exist.

Substitution Method

The substitution method is a way to solve systems of equations where one equation is rearranged to express one variable in terms of the other. Then, this expression is substituted into the other equation.
For our system, the linear equation is \(x - y = 4\). Rearrange this equation by solving for \(y\) to get \(y = x - 4\).
Now substitute \(y = x - 4\) into the first equation \(x^2 + y^2 = 2\). The equation becomes \(x^2 + (x - 4)^2 = 2\). You've now expressed the system with a single variable, which simplifies solving.
This method is particularly useful when one of the equations is easy to solve for one of the variables, and it reduces the complexity significantly when dealing with linear and nonlinear systems.

Elimination Method

The elimination method, also known as the addition method, involves adding or subtracting equations to eliminate a variable, making the system easier to solve.
In the case of a circle and a line, like our problem \(x^2 + y^2 = 2\) and \(x - y = 4\), traditional elimination may not be straightforward. However, in cases where both equations are linear, you can add or subtract them to remove a variable directly.
For many systems, especially those with nonlinear equations like circles and parabolas, substituting or transforming one part of the system might be more efficient than elimination. In scenarios with simple systems of linear equations, the elimination method is highly effective for finding solutions.

Quadratic Equations

Quadratic equations often appear in systems involving curves, such as circles and parabolas. These equations are typically in the form \(ax^2 + bx + c = 0\).
In our step-by-step solution, after substituting \(y = x - 4\) into the circle's equation, we derived the quadratic equation \(x^2 - 4x + 7 = 0\).
To check for real solutions, compute the discriminant \(b^2 - 4ac\). For this quadratic, where \(a = 1\), \(b = -4\), and \(c = 7\), the discriminant is \(-12\), which is negative.
A negative discriminant indicates that there are no real solutions, meaning the curves do not intersect at real-number points. Thus, it's crucial to apply this check to predict intersection solutions, helping visualize whether real solutions exist.