Problem 32 - Intermediate Algebra

Question

For each of the following systems, (a) use your graphing calculator to show that there are no real number solutions, and (b) solve the system by the substitution method or the elimination-by-addition method to find the complex solutions. (a) \(\left(\begin{array}{l}y=x^{2}+1 \\ y=-3\end{array}\right)\) (b) \(\left(\begin{array}{l}y=-x^{2}+1 \\ y=3\end{array}\right)\) (c) \(\left(\begin{array}{r}y=x^{2} \\ x-y=4\end{array}\right)\) (d) \(\left(\begin{array}{l}y=x^{2}+1 \\ y=-x^{2}\end{array}\right)\) (e) \(\left(\begin{array}{l}x^{2}+y^{2}=1 \\ x+y=2\end{array}\right)\) (f) \(\left(\begin{array}{l}x^{2}+y^{2}=2 \\ x^{2}-y^{2}=6\end{array}\right)\)

Step-by-Step Solution

Answer

Systems (a), (b), (d), (e), and (f) have no real solutions, only system (c) has real solutions.

Step 1: Solve System (a) Using Substitution

For system (a), we have the equations:\[ y = x^2 + 1 \ y = -3 \] Since both are equal to \( y \), we can set the equations equal to each other: \[ x^2 + 1 = -3 \] Solve for \( x \) by subtracting 1 from both sides to get: \[ x^2 = -4 \] Since the right side is negative, there are no real solutions for \( x \). The solutions in terms of complex numbers are \( x = \pm 2i \).

Step 2: Solve System (b) Using Substitution

For system (b), we have the equations:\[ y = -x^2 + 1 \ y = 3 \] Set them equal to each other: \[ -x^2 + 1 = 3 \] Subtract 1 from both sides: \[ -x^2 = 2 \] This gives \( x^2 = -2 \) when both sides are multiplied by -1. Thus, no real solutions exist. The complex solutions are \( x = \pm i\sqrt{2} \).

Step 3: Solve System (c) Using Elimination

For system (c), we have:\[ y = x^2 \] \[ x-y=4 \] Substitute \( y = x^2 \) into the second equation: \[ x - x^2 = 4 \] Rearrange to form a quadratic equation: \[ x^2 - x - 4 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -4 \):\[ x = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2} \] These are real solutions. Find \( y \) using \( y = x^2 \).

Step 4: Solve System (d) Using Substitution

For system (d), we have:\[ y = x^2 + 1 \] \[ y = -x^2 \] Set them equal: \[ x^2 + 1 = -x^2 \] This simplifies to \( 2x^2 = -1 \), so \( x^2 = -\frac{1}{2} \). No real solutions exist, complex solutions are \( x = \pm i/\sqrt{2} \).

Step 5: Solve System (e) Using Substitution

Given equations: \[ x^2 + y^2 = 1 \] \[ x + y = 2 \] Express \( y \) from the second equation: \( y = 2 - x \).Substitute into the first equation:\[ x^2 + (2 - x)^2 = 1 \]Expand and simplify: \[ x^2 + 4 - 4x + x^2 = 1 \]\[ 2x^2 - 4x + 3 = 0 \]Using the quadratic formula, we find \( x \) values are complex, since the discriminant is negative. Hence, no real solutions.

Step 6: Solve System (f) Using Elimination

Equations are: \[ x^2 + y^2 = 2 \] \[ x^2 - y^2 = 6 \] Add both equations: \[ 2x^2 = 8 \] \( x^2 = 4 \) gives \( x = \pm 2 \).Substitute \( x = 2 \) in any equation, no solutions due to contradiction in \( y \). Hence only complex solutions exist.

Key Concepts

Systems of Equations

In mathematics, a system of equations is a collection of two or more equations with a set of variables. The solution to a system is the set of values that satisfy all equations in the system simultaneously. Systems can be

• Linear: where each equation is a first-degree equation
• Non-linear: includes equations like quadratic, exponential, etc.

For instance, the system \( y = x^2 + 1 \) and \( y = -3 \) consists of two equations, one quadratic and one linear. Solving this system involves finding the values of \( x \) and \( y \) that make both equations true.
In some cases, such as these exercises, systems may not have solutions in the set of real numbers and instead have complex solutions, indicating the intersection of graphs doesn’t occur on the real plane.

Substitution Method

The substitution method is a technique for solving systems of equations by solving one equation for one variable and then substituting this expression into the other equation(s). This method is particularly useful when one of the equations is already solved for one variable. Let's break it down:

• Step 1: Solve one of the equations for one variable, if not already done.
• Step 2: Substitute this expression into the other equation(s), replacing the variable.
• Step 3: Solve the resulting equation for the remaining variable.
• Step 4: Back-substitute to find the other variable.

Using the substitution method, complex solutions may arise when the equations fundamentally don’t intersect in the set of real numbers. For example, solving \( y = x^2 + 1 \) and \( y = -3 \) gives a complex solution for \( x = \pm 2i \).

Elimination Method

The elimination method, also known as the addition method, involves eliminating one of the variables by adding or subtracting equations. This method is particularly effective when equations are in standard form, and coefficients can be easily manipulated. Here's a simple explanation:

• Step 1: Arrange the equations so that one variable aligns.
• Step 2: Add or subtract equations to eliminate one variable.
• Step 3: Solve the resulting equation for the remaining variable.
• Step 4: Use the found value to substitute back and find the other variable.

For example, with the system \( x^2 + y^2 = 2 \) and \( x^2 - y^2 = 6 \), adding them eliminates \( y^2 \), allowing an easy solve for \( x^2 \). Complex solutions often arise when contradictions or negative discriminants occur during solving.

Quadratic Equations

Quadratic equations are polynomial equations of degree two and are pivotal in many systems of equations problems. A standard quadratic equation takes the form \( ax^2 + bx + c = 0 \). Here's how to solve it:

• The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) allows finding both real and complex roots.
• The discriminant \( b^2 - 4ac \) determines the nature of the roots: real if it's positive, or complex if negative.

For example, in solving systems like \( y = x^2 \) and \( x - y = 4 \), converting the second equation into a quadratic form \( x^2 - x - 4 = 0 \) involves using the quadratic formula. Quadratics can lead to complex solutions when intersections occur outside the real number plane.