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8.03 Solving nonlinear systems of equations

Introduction

In Algebra 1, we solved systems of linear equations and linear-quadratic systems graphically and algebraically in lesson  10.06 Linear-quadratic systems  . We will continue to solve systems of nonlinear equations in this lesson.

Solving nonlinear systems of equations

Exploration

A country has a population of 4 million people and grows at a rate of 2 \% each year. The country's food supply can currently feed 6 million people, and the supply increases to meet the hunger needs of 0.10 million people each year. The graph below depicts the relationship between the years and people in the population and people fed in the population.

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\text{Years}
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\text{Population (in millions)}
  1. How could we use the graph of the functions modeled to determine when the country would have a food shortage?

A system of equations is a set of equations that have the same variables. The solution to a system of equations is any ordered pair that makes all of the equations in the system true. For graphs, this will be the point(s) of intersection. Solutions can be found algebraically or graphically.

x
y
The line and parabola have no points of intersection, so the system has no solution.
x
y
The parabolas do not have any points of intersection, so the system has no solution.
x
y
The absolute value graph and curve intersect at one point, so the system has one solution.
x
y
The circle and parabola have one point of intersection, so the system has one solution.
x
y
The line and curve have two points of intersection, so the system has two solutions.
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The absolute value graph and circle intersect at two points, so the system has two solutions.

Systems of equations may be solved by using the substitution method or the elimination method when solving algebraically. However, note that the substitution method is the most reliable method in most cases because it is difficult to eliminate terms when the function types are not the same.

The solution to a system of equations in a given context is viable if the solution makes sense in the context, and is non-viable if it does not make sense.

Examples

Example 1

Consider the equation: \sqrt{x} + 4 = x^{2} + 4x

a

Write the equation as a system of equations.

Worked Solution
Create a strategy

The system of equations is composed of a radical function and a quadratic function.

Apply the idea

\begin{cases} f\left(x\right)= \sqrt{x} + 4 \\ g\left(x\right) = x^{2} + 4x\end{cases}

b

Complete the table of values to determine the solution(s) to the system of equations.

x01234
f\left(x\right)
g\left(x\right)
Worked Solution
Create a strategy

Evaluate f\left(x\right) and g\left(x\right) for the given values of x in the table, then determine where f\left(x\right)=g\left(x\right).

Apply the idea
x01234
f\left(x\right)455.415.736
g\left(x\right)05122132

Using the table of values, the functions are equivalent when x=1, because both functions are 5. The solution to the system of equations is (1,5).

Reflect and check

We can graph the system of equations to see the location of the solution(s).

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y

Example 2

Consider the system of equations: \begin{cases} f\left(x\right)= 3^{x-2}+6 \\ g\left(x\right) = -\dfrac{2}{3}\left|x-4\right| + 15 \end{cases}

a

State the solution(s) to the system of equations.

Worked Solution
Create a strategy

We can use technology to graph the system of equations and determine its solution(s).

Apply the idea
  1. Enter both functions and view the system graphically.

    A screenshot of the GeoGebra graphing calculator showing the graphs of y equals 3 raised to the quantity x minus 2 plus 6 and y equals negative two thirds absolute value of the quantity x minus 4 plus 15. Speak to your teacher for more details.
  2. Choose the Intersect tool.

    A screenshot of the GeoGebra graphing calculator showing the graphs of y equals 3 raised to the quantity x minus 2 plus 6 and y equals negative two thirds absolute value of the quantity x minus 4 plus 15. A set of options under the Basic Tools menu is shown. Speak to your teacher for more details.
  3. Click the curves to label intersection points and read the points of intersection.

    A screenshot of the GeoGebra graphing calculator showing the graphs of y equals 3 raised to the quantity x minus 2 plus 6 and y equals negative two thirds absolute value of the quantity x minus 4 plus 15. Two points of intersection of the two graphs are highlighted. Speak to your teacher for more details.

By viewing a graph of the system of equations, we can see that there are two solutions to the system.

The solutions to the system of equations are approximately (4, 15) and (-9.5, 6).

Reflect and check

By using technology to graph a system of equations, we can use the zoom function to see the number of real solutions to a system.

b

Verify the solution(s) to the system of equations algebraically.

Worked Solution
Create a strategy

Substitute the x-coordinate into both functions and evaluate the functions.

Apply the idea

Given the solutions to the system of equations from the graph, \left(4,15\right), and \left(-9.5,6\right), we can check each solution by evaluating f\left(x\right) and g\left(x\right) at the given x-coordinate to confirm the functions are equivalent at these points.

When x=4:

\displaystyle f\left(x\right)\displaystyle =\displaystyle 3^{x-2}+6Given
\displaystyle f\left(4\right)\displaystyle =\displaystyle 3^{4-2}+6Substitute x=4
\displaystyle =\displaystyle 15Evaluate

And,

\displaystyle g\left(x\right)\displaystyle =\displaystyle -\dfrac{2}{3}\left|x-4\right| + 15Given
\displaystyle g\left(4\right)\displaystyle =\displaystyle -\dfrac{2}{3}\left|4-4\right| + 15Substitute x=4
\displaystyle =\displaystyle 15Evaluate

Since both functions equal 15 when x=4, we can verify that a solution to the system of equations is \left(4, 15\right).

When x=-9.5:

\displaystyle f\left(x\right)\displaystyle =\displaystyle 3^{x-2}+6Given
\displaystyle f\left(-9.5\right)\displaystyle =\displaystyle 3^{-9.5-2}+6Substitute x=-9.5
\displaystyle \approx\displaystyle 6Evaluate

And,

\displaystyle g\left(x\right)\displaystyle =\displaystyle -\dfrac{2}{3}\left|x-4\right| + 15Given
\displaystyle g\left(-9.5\right)\displaystyle =\displaystyle -\dfrac{2}{3}\left|-9.5-4\right| + 15Substitute x=-9.5
\displaystyle =\displaystyle 6Evaluate

Since both functions approximately equal 6 when x=-9.5, we can verify that an approximate solution to the system of equations is \left(-9.5, 6\right).

Notice that the substitution demonstrated that \left(4, 15\right) was an exact solution to the equation while \left(-9.5, 6\right) was an approximate solution to the equation.

Example 3

The booster club for the girls' volleyball team at a school is fundraising. The club has a total of \$3500 to purchase hoodies with the school's mascot and colors on them. The club plans to sell the hoodies to students and families.

The number of hoodies that the booster club can buy, represented by the function f\left(x\right)=\dfrac{3500}{x}, depends on the cost per hoodie, x. The number of hoodies that students and families will purchase, represented by the function g\left(x\right)=210-1.75x, depends on the price which the booster club bought the hoodies for.

a

Determine a reasonable constraint on the system of equations.

Worked Solution
Create a strategy

We need to determine the independent and dependent variables in context. From the instructions, we know f\left(x\right) is the number of hoodies purchased and x represents the cost per hoodie. We are also given that g\left(x\right) is the number of hoodies purchased by students and families.

Apply the idea

A reasonable constraint for the system of equations would be that all values of x are greater than zero. Since x represents the cost, we can state that the cost of the hoodies should not be \$0 or less.

Since the number of hoodies being purchased by the booster club and students and their families are the dependent variable in the system, we can also determine that all values of y will be greater than or equal to zero, since we cannot reasonably purchase a negative number of hoodies in context.

b

For what amount per hoodie will the number of hoodies that can be bought by the booster club equal the number of hoodies that will be sold?

Worked Solution
Create a strategy

Solve the system of equations using substitution to determine the amount per hoodie, x, when the number of hoodies bought by the booster club will equal the number of hoodies that will be sold.

Apply the idea

The solution to the system of equations is when f\left(x\right)=g\left(x\right).

We have

\displaystyle \dfrac{3500}{x}\displaystyle =\displaystyle 210-1.75xf\left(x\right)=g\left(x\right)
\displaystyle 3500\displaystyle =\displaystyle 210x-1.75x^{2}Multiply both sides by x
\displaystyle 0\displaystyle =\displaystyle -1.75x^{2}+210x-3500Subtract 3500 from both sides

We can use the quadratic formula to solve the equation:

\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}Quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-210 \pm \sqrt{210^{2}-4(-1.75)(-3500)}}{2(-1.75)}Substitute a, b and c
\displaystyle x\displaystyle =\displaystyle \dfrac{-210 \pm \sqrt{19600}}{-3.5}Simplify

When x=20 and x=100, f\left(x\right)=g\left(x\right). This means that when the cost per hoodie is \$20 or \$100, the number of hoodies bought by the booster club will equal the number of hoodies sold to students and families.

Reflect and check

We can use technology to graph the system of equations and confirm the solutions.

  1. Enter both functions and view the system graphically.

    A screenshot of the GeoGebra graphing calculator showing the graphs of f of x equals 3,500 over x and y equals 210 minus 1.75 x. Speak to your teacher for more details
  2. Choose the Intersect tool.

    A screenshot of the GeoGebra graphing calculator showing the graphs of f of x equals 3,500 over x and y equals 210 minus 1.75 x. A set of options under the Basic Tools menu is shown. Speak to your teacher for more details.
  3. Click the curves to label intersection points and read the points of intersection.

    A screenshot of the GeoGebra graphing calculator showing the graphs of f of x equals 3,500 over x and y equals 210 minus 1.75 x. Two points of intersection of the two graphs are highlighted. Speak to your teacher for more details.

Graphically, we can see that at both prices the booster club would sell out of hoodies. However, at the cheaper price more hoodies are bought and sold, and at the more expensive price fewer hoodies are bought and sold.

Idea summary

Consider graphing systems of equations or solving systems of equations algebraically:

  • Systems with large numbers should be graphed using technology. By graphing a system first, we can verify the number of solutions to the system
  • When solving systems algebraically the easiest approach is using the substitution method in most cases

Outcomes

A.CED.A.3

Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.

A.REI.D.11

Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately.

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