MTH120 College Algebra Chapter 7.3

 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables

Systems of nonlinear equations and inequalities in two variables involve equations and inequalities where the relationships between the variables are not linear. This means that the equations may include terms with variables raised to powers other than 1 (nonlinear terms). These systems can be more complex to solve than systems of linear equations.

In this section, we'll explore systems of nonlinear equations and inequalities in two variables and how to approach them. Here are some key concepts:

Systems of Nonlinear Equations:

A system of nonlinear equations in two variables typically consists of two or more equations where at least one of the equations contains nonlinear terms. Solving such systems can involve techniques like substitution, elimination, or graphing. In many cases, exact solutions may be challenging to find, and numerical methods or graphing calculators may be used.

Systems of Nonlinear Inequalities:

Systems of nonlinear inequalities involve two or more inequalities with nonlinear expressions. Solving these systems means finding the regions in the coordinate plane that satisfy all the inequalities simultaneously. Graphical methods are often used to visualize and identify these regions.

Real-World Applications:

Nonlinear systems of equations and inequalities are used in various real-world scenarios:

1. Economics: Modeling supply and demand relationships, utility functions, and production costs often involve nonlinear equations and inequalities.

2. Physics: Many physical phenomena involve nonlinear relationships, such as the equations of motion for a pendulum or the behavior of fluids under non-linear conditions.

3. Biology: Population growth, disease modeling, and ecological systems can be described using nonlinear equations and inequalities.

4. Engineering: Design and analysis of non-linear electronic circuits, structural analysis, and fluid dynamics.

Solving systems of nonlinear equations and inequalities may involve analytical methods (finding exact solutions), numerical methods (using algorithms to approximate solutions), or graphical methods (visualizing solutions). The specific approach depends on the complexity of the equations and the problem at hand.

Keep in mind that solving systems of nonlinear equations and inequalities is often more challenging than solving linear systems. There may not be a unique solution, and the solutions may be approximated. Technology, such as graphing calculators or computer software, can be valuable tools in handling these types of problems.

Solving a system of nonlinear equations using substitution involves isolating one variable in terms of the other and then substituting this expression into the other equation(s) to solve for the remaining variable(s). Here's an example:

Example:

Let's solve the following system of nonlinear equations using substitution:

1. ${�}^2+{�}^2=25$
2. $�-2�=3$

Solution:

We'll start by solving equation (2) for $�$:

$�=2�+3$

Now, we can substitute this expression for $�$ into equation (1):

$(2�+3{)}^2+{�}^2=25$

Now, we have a single equation with one variable, $�$, which we can solve for:

$(4{�}^2+12�+9)+{�}^2=25$

Combine like terms:

$5{�}^2+12�+9=25$

Subtract 25 from both sides:

$5{�}^2+12�+9-25=0$

Simplify further:

$5{�}^2+12�-16=0$

Now, you have a quadratic equation in $�$. You can solve it by factoring or using the quadratic formula. In this case, let's use the quadratic formula:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

For this equation, $�=5$, $�=12$, and $�=-16$. Plug these values into the formula:

$�=\frac{-12\pm \sqrt{12^2-4(5)(-16)}}{2(5)}$

Calculate the discriminant (${�}^2-4��$):

$12^2-4(5)(-16)=144+320=464$

Now, continue with the formula:

$�=\frac{-12\pm \sqrt{464}}{10}$

Simplify the square root:

$�=\frac{-12\pm 4\sqrt{29}}{10}$

Divide both the numerator and denominator by 2 to simplify:

$�=\frac{-6\pm 2\sqrt{29}}{5}$

So, we have two possible solutions for $�$:

1. ${�}_1=\frac{-6+2\sqrt{29}}{5}$
2. ${�}_2=\frac{-6-2\sqrt{29}}{5}$

Now that we have the values of $�$, we can find the corresponding values of $�$ using the equation $�=2�+3$:

For ${�}_1$:

${�}_1=2\left(\frac{-6+2\sqrt{29}}{5}\right)+3$

For ${�}_2$:

${�}_2=2\left(\frac{-6-2\sqrt{29}}{5}\right)+3$

These are the solutions to the system of nonlinear equations. You have two pairs of values, $({�}_1,{�}_1)$ and $({�}_2,{�}_2$, that satisfy both equations simultaneously.

The intersection of a parabola and a line refers to the points where a given parabolic curve and a straight line intersect or coincide. Depending on the specific equations for the parabola and the line, there can be zero, one, or two intersection points.

A parabola is defined by a quadratic equation in the form $�=�{�}^2+��+�$, and a line is defined by an equation in the form $�=��+�$, where $�$, $�$, $�$, $�$, and $�$ are constants.

To find the points of intersection between a parabola and a line, you need to set the two equations equal to each other:

$�{�}^2+��+�=��+�$

This equation simplifies to a quadratic equation:

$�{�}^2+(�-�)�+(�-�)=0$

Now, you can use the quadratic formula to solve for $�$, and then find the corresponding $�$ values using the equation of the line.

Let's work through an example:

Example:

Find the points of intersection between the parabola $�={�}^2$ and the line $�=2�-1$.

1. Set the equations equal to each other:

   ${�}^2=2�-1$

2. Rearrange to get a quadratic equation:

   ${�}^2-2�+1=0$

3. Now, apply the quadratic formula to solve for (x:

   $�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

   In this case, $�=1$, $�=-2$, and $�=1$. Plug in these values:

   $�=\frac{-(-2)\pm \sqrt{(-2{)}^2-4(1)(1)}}{2(1)}$

   $�=\frac{2\pm \sqrt{4-4}}{2}$

   $�=\frac{2\pm \sqrt{0}}{2}$

   Since the discriminant (${�}^2-4��$) is zero, there is only one solution for (x:

   $�=\frac{2}{2}=1$

4. Now that you have the $�$-coordinate of the intersection point, you can find the corresponding $�$-coordinate using the equation of the line:

   $�=2�-1$ $�=2(1)-1$ $�=2-1$ $�=1$

So, the point of intersection between the parabola $�={�}^2$ and the line $�=2�-1$ is (1, 1). This is the only intersection point in this case, and it represents the coordinates where the parabola and the line coincide.

The possible types of solutions for the points of intersection between a parabola and a line depend on the specific equations of the parabola and the line. There are three main types of solutions:

1. No Intersection: The parabola and the line do not intersect at any point. This occurs when the parabola and the line are either parallel and do not touch each other or when they are oriented such that they do not intersect within the specified range.

   Example 1: The parabola $�={�}^2$ and the line $�=2�+3$ have no intersection. These two curves are parallel, and their graphs do not overlap in any region.

2. One Intersection: The parabola and the line intersect at a single point. This occurs when the line is tangent to the parabola or when the line intersects the parabola at one point and does not cross it.

   Example 2: The parabola $�={�}^2$ and the line $�=2�-1$ intersect at a single point, which is (1, 1). The line is tangent to the parabola.

3. Two Intersections: The parabola and the line intersect at two distinct points. This occurs when the line intersects the parabola twice within the specified range.

   Example 3: The parabola $�={�}^2$ and the line $�=-�+2$ intersect at two points, which are (-1, 1) and (2, 2). The line crosses the parabola at two distinct points.

The type of solution depends on the relative positions of the parabola and the line in the coordinate plane and their equations. Graphing the equations is a helpful way to visualize the type and location of the intersections.

Remember that the number of solutions can also depend on the range or interval specified for the variables in the equations. In some cases, the parabola and the line may intersect within the given range but not outside it, resulting in fewer or no solutions when considering the entire real number line.

The intersection of a circle and a line refers to the points where the circle and the line in a two-dimensional coordinate plane intersect or coincide. The circle is defined by its center coordinates and radius, while the line is defined by its equation.

Here are the different scenarios for the intersection of a circle and a line:

1. No Intersection: The circle and the line do not intersect at any point. This occurs when the line is located entirely outside the circle or when the line is tangent to the circle.

2. Tangent Intersection: The circle and the line intersect at a single point where the line is tangent to the circle. In this case, the line touches the circle at one point without crossing it.

3. Two Intersection Points: The circle and the line intersect at two distinct points. These points are the two solutions to the system of equations formed by the equation of the circle and the equation of the line.

Let's look at some examples:

Example 1: No Intersection

Consider the circle with center (0, 0) and radius 2, given by the equation ${�}^2+{�}^2=4$, and the line $�=4�+3$.

In this case, the circle and the line do not intersect. The line is located outside the circle, and their graphs do not overlap.

Example 2: Tangent Intersection

Consider the circle with center (1, 1) and radius 2, given by the equation $(�-1{)}^2+(�-1{)}^2=4$, and the line $�=�$.

In this case, the circle and the line intersect at a single point, which is the point of tangency at (2, 2). The line touches the circle at this point without crossing it.

Example 3: Two Intersection Points

Consider the circle with center (0, 0) and radius 3, given by the equation ${�}^2+{�}^2=9$, and the line $�=2�-1$.

In this case, the circle and the line intersect at two distinct points, which are (-2, -3) and (2, 3). The line crosses the circle at these two points.

The type of intersection depends on the relative positions of the circle and the line and their equations. When solving for the intersection points, you'll need to find the solutions to the system of equations formed by the equation of the circle and the equation of the line.

Solving a system of nonlinear equations using elimination can be a bit more complex than solving linear systems, as the elimination method usually involves adding or subtracting equations to eliminate one variable at a time. Here's an example of how to solve a system of nonlinear equations using elimination:

Example:

Let's solve the following system of nonlinear equations using elimination:

1. ${�}^2+{�}^2=25$
2. $�+�=7$

Solution:

First, let's solve equation (2) for $�$:

$�=7-�$

Now, we can substitute this expression for $�$ into equation (1):

$(7-�{)}^2+{�}^2=25$

Now, we have a single equation with one variable, $�$, which we can solve for:

$49-14�+{�}^2+{�}^2=25$

Combine like terms:

$2{�}^2-14�+24=0$

Now, we have a quadratic equation in $�$. You can solve it by factoring or using the quadratic formula. In this case, let's use the quadratic formula:

$�=\frac{-�\pm \sqrt{{�}^2-4��}}{2�}$

For this equation, $�=2$, $�=-14$, and $�=24$. Plug in these values:

$�=\frac{-(-14)\pm \sqrt{(-14{)}^2-4(2)(24)}}{2(2)}$

Calculate the discriminant (${�}^2-4��$):

$(-14{)}^2-4(2)(24)=196-192=4$

Now, continue with the formula:

$�=\frac{14\pm \sqrt{4}}{4}$

Simplify the square root:

$�=\frac{14\pm 2}{4}$

Now, you have two possible solutions for $�$:

1. ${�}_1=\frac{14+2}{4}=4$
2. ${�}_2=\frac{14-2}{4}=3$

Now that we have the values of $�$, we can find the corresponding values of $�$ using the equation $�=7-�$:

For ${�}_1$:

${�}_1=7-4=3$

For ${�}_2$:

${�}_2=7-3=4$

So, the points of intersection between the two nonlinear equations are (3, 4) and (4, 3). These are the solutions to the system of nonlinear equations.

The possible types of solutions for the points of intersection between a circle and an ellipse can vary based on the specific equations for the circle and the ellipse, as well as their positions in the coordinate plane. There are four main scenarios:

1. No Intersection: The circle and the ellipse do not intersect at any point. This occurs when the two curves are either located entirely outside of each other or when they are oriented such that they do not intersect within the specified range.

2. Tangent Intersection: The circle and the ellipse intersect at a single point where they are tangent to each other. In this case, one curve touches the other at one point without crossing it.

3. Two Intersection Points: The circle and the ellipse intersect at two distinct points. These points are the solutions to the system of equations formed by the equation of the circle and the equation of the ellipse.

4. Overlapping Intersection: The circle and the ellipse intersect at multiple points, forming a set of overlapping or coinciding curves. This occurs when the equations of the circle and the ellipse produce multiple intersection points within a given range.

The type of solution depends on the relative positions of the circle and the ellipse and their equations. The actual solutions can be found by solving the system of equations formed by the equation of the circle and the equation of the ellipse.

For a visual representation and to determine the specific type of solution, it is often helpful to graph both the circle and the ellipse in the coordinate plane. This allows you to see how the curves relate to each other and how many intersection points exist.

Keep in mind that the number of solutions can also depend on the range or interval specified for the variables in the equations. In some cases, the circle and the ellipse may intersect within the given range but not outside it, resulting in fewer or no solutions when considering the entire real number line.

Graphing a nonlinear inequality involves representing the solution set of the inequality on a coordinate plane. The solution set often forms a region in the plane, and you need to shade or highlight the area that satisfies the inequality. Here are the general steps for graphing a nonlinear inequality:

Step 1: Understand the Inequality

Make sure you understand the inequality and its components. Identify the variables and determine the type of curve or region the inequality represents. For example, is it a quadratic equation, a circle, an ellipse, or some other curve?

Step 2: Identify Key Points

Identify key points that can help you sketch the curve or boundary of the inequality. These points may include the intercepts, critical points, or points that satisfy the equality of the inequality.

Step 3: Graph the Boundary

Graph the equation that represents the boundary of the inequality. This is typically the equation obtained by replacing the inequality symbol with an equal sign. The boundary separates the solution region from the non-solution region.

Step 4: Determine the Shaded Region

Identify which side of the boundary curve represents the valid solutions. If the inequality is of the form "<," "<=," ">", or ">=" and includes the boundary, you'll shade the region on that side of the curve. If it's of the form ">", "<," and does not include the boundary, you'll shade the opposite side.

Step 5: Label the Regions (Optional)

You can label or shade different regions if the inequality consists of multiple parts. This is common in systems of inequalities.

Step 6: Check Test Points (Optional)

If you're unsure about the shading, you can pick a point within each shaded region and test it in the original inequality. If the point satisfies the inequality, you've correctly shaded the region.

Step 7: Include Labels and Axes

Make sure your graph includes labeled axes, so it's clear which part of the coordinate plane you're working with.

Here's an example:

Example:

Let's graph the inequality $�>{�}^2$ on the coordinate plane.

1. Understand the Inequality: This is a quadratic inequality where $�$ is greater than the square of $�$. We expect the valid solutions to be above the curve of $�={�}^2$.

2. Identify Key Points: Consider the intercepts (0, 0) and (0, 1), and a point (1, 2) that satisfies the equality.

3. Graph the Boundary: Graph the equation $�={�}^2$, which is a parabola opening upwards.

4. Determine the Shaded Region: Since the inequality is $�>{�}^2$, we shade the region above the parabola.

5. Label the Regions: In this case, there's only one shaded region.

6. Check Test Points (Optional): You can pick a point within the shaded region, like (0, 1), and test it in the original inequality to confirm that it satisfies $�>{�}^2$.

7. Include Labels and Axes: Make sure your graph has labeled axes for clarity.

This is a general outline for graphing nonlinear inequalities. The specific shape and orientation of the shaded region depend on the form of the inequality and the corresponding equality.

Graphing a system of nonlinear inequalities involves representing the solution set of multiple inequalities on a coordinate plane. The solution set is the region where all inequalities overlap or intersect. Let's work through an example:

Example:

Graph the system of inequalities:

1. $�\le {�}^2-4$
2. $�\ge -�$

Solution:

To graph this system, you will graph each inequality separately and then identify the region where the solutions of both inequalities overlap.

Graphing the First Inequality:

1. Start by graphing the boundary of the first inequality ($�={�}^2-4$). This is a downward-opening parabola with its vertex at (0, -4).

2. Since the inequality includes "less than or equal to," you should shade the region below the parabola.

Graphing the Second Inequality:

1. Graph the boundary of the second inequality ($�=-�$). This is a straight line with a slope of -1 and a y-intercept of (0, 0).

2. Since the inequality includes "greater than or equal to," you should shade the region above the line.

Combining the Shaded Regions:

Now, look at the shaded regions from both inequalities. The solution to the system of inequalities is the overlapping region of the two shaded areas. In this case, it's the area under the parabola and above the line.

The graph of the solution set for the system of inequalities is the region between the parabola and the line.