introduction

IntroductionA system of equations is a set of equations with the same unknowns. It is likely that you have solved a system of equations or a quadratic-linear system in the past. In this lesson, we will learn to solve systems that include at least one polynomial by graphing.

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2.4.1: Solving a Linear-Polynomial System of Equations

Key Concepts• The real solutions of a system of equations can often

be found by graphing. Recall that the graph of an equation is the set of all the solutions of the equation plotted on a coordinate plane.

• The points of intersection of the graphed system of equations are the ordered pairs where graphed functions intersect on a coordinate plane. These are also the solutions to the system. The solution or solutions to a system of equations with f(x) and g(x) occurs where f(x) = g(x) and the difference between these two values is 0. Also recall that the solutions to a system of equations are written using braces: {}.

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Key Concepts, continued• A system may have an infinite number of solutions, a

finite number of solutions, or no real solutions. • A system with a finite number of points of intersection

is called an independent system. • It is possible that a system of equations does not

intersect. The solution to such a system of equations is the empty set, denoted by { }.

• A system of equations could also intersect at every point. This type of system is known as a dependent system.

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Key Concepts, continuedSolutions to Systems of Equations with Polynomials

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• If the equations overlap, they have an infinite number of solutions.

• This is an example of a dependent system.


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• If the equations intersect, they have a finite number of solutions.

• This is an example of an independent system.


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• If the equations do not intersect, they have no solutions.

• The solution to this system is the empty set.

Key Concepts, continued• When graphing a system of equations with

polynomials, it is helpful to recall various forms of equations and their general shapes.

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Key Concepts, continuedGraphs of General Equations

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Type ofequation

General equation

Graph

Linear equation m is the slope and b is the y-intercept

y = mx + b

Key Concepts, continued

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Type ofequation

General equation

Graph

Polynomial equation a1 is a rational number, an ≠ 0, and n is a nonnegative integer and the highest degree of the polynomial

y = anxn + an – 1xn – 1 + + a2x2 + a1x + a0


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Type of equation

General equation

Graph

Quadratic equation x is the variable, a, b, and c are constants, and a ≠ 0

y = ax2 + bx + c

Key Concepts, continued• To graph a system of equations, create a table of

values and plot each coordinate on the same coordinate plane. It may be necessary to find additional points to determine where the point(s) of intersection may occur. Each point of intersection represents a solution to the system.

• Solutions to a system of equations can also be identified in a table by finding points such that the y-value of each function is the same for a particular x-value.

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Key Concepts, continued• Using a graphing calculator to graph both equations is

often helpful. Most calculators have a trace feature to approximate where the equations intersect, or can display a table of values for each equation so you can analyze the values. Some calculators will find intersection points for you.

• To confirm a solution to a system, substitute the coordinates for x and y in both of the original equations. If the intersection point is approximated, the results will be nearly equal.

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Key Concepts, continuedSolving Systems of Equations Algebraically

• When solving a quadratic-linear system, if both functions are written in the form of a function (such as “y =” or “f(x) =”), set the equations equal to each other.

• When you set the equations equal to each other, you are replacing y in each equation with an equivalent expression, thus using the substitution method to find a solution.

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• Recall that you can then solve by factoring the equation

or by using the quadratic formula. The quadratic

formula states that the solutions of a quadratic equation

of the form ax2 + bx + c = 0 are given by

. A quadratic equation in this form

can have no real solutions, one real solution, or two

real solutions. 14


Common Errors/Misconceptions• not identifying all points of intersection

• incorrectly graphing each equation

• incorrectly entering the equations into the calculator

• not understanding that solutions must satisfy every function in a system

• believing that every system has at least one solution

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Guided Practice

Example 2

Use a graph to estimate the real solution(s), if any, to the

system of equations . Verify that any

identified coordinate pairs are solutions.

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Guided Practice: Example 2, continued

1. Create a graph of the two functions, f(x) and g(x).You can use what you have learned about graphing equations to graph each equation on the same coordinate plane, or you can use a graphing calculator.

To graph the system on your graphing calculator, follow the directions appropriate to your calculator model that begin on the next slide.

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Guided Practice: Example 2, continuedOn a TI-83/84:

Step 1: Press the [Y=] button.

Step 2: Type the first function into Y1, using the

[X, T, θ, n] button for the variable x. Press [ENTER].

Step 3: Type the second function into Y2, using the [^] button for powers. Press [ENTER].

Step 4: Press [ZOOM]. Arrow down to 6: ZStandard. Press [ENTER].

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Guided Practice: Example 2, continuedOn a TI-Nspire:

Step 1: Press the [home] key.

Step 2: Arrow over to the graphing icon and press [enter].

Step 3: Type the first function next to f1(x), using the [X]

button for the letter x or the [x2] button for a

square. Press [enter].

Step 4: To graph the second function, press [menu] and

arrow down to 3: Graph Type. Arrow right to bring

up the sub-menu, then select 1: Function. Press

[enter].

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(continued)


Step 5: At f2(x), type the second function and press [enter].

Step 6: To change the viewing window, press [menu]. Select 4: Window/Zoom and select A: Zoom – Fit.

The resulting graph is shown on the next slide.

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2. Find the coordinates of any apparent intersections. On your graph, estimate where the two equations intersect. It may be necessary to plot additional points.

To determine the approximate point(s) of intersection on your graphing calculator, use the following directions.

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Step 1: Press [2ND][TRACE] to call up the CALC screen. Arrow down to 5: intersect. Press [ENTER].

Step 2: At the prompt, use the up/down arrow keys to select the Y1 equation, and press [ENTER].

Step 3: At the prompt, use the up/down arrow keys to select the Y2 equation, and press [ENTER].

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(continued)

Guided Practice: Example 2, continuedStep 4: At the prompt, use the arrow keys to move

the cursor close to an apparent intersection, and press [ENTER]. The coordinates of the intersection points are displayed.

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Step 1: Press [menu]. Arrow down to 6: Analyze Graph, then arrow right to 4: Intersection. Press [enter].

Step 2: Use the pointing hand to click on each graphed line. The coordinates of the intersection points are displayed.

Repeat for each intersection as needed.

The points of intersection are (–2, –7), (0, 1), and (2, 9).

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3. Verify that the identified coordinate pairs are solutions to the original system of equations. In order for a coordinate pair to be solution to a system, the coordinates must satisfy both equations of the system.

Substitute the coordinates into each of the equations, then evaluate the equations to see if the coordinates result in a true statement.

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At the point (–2, –7), let x = –2 and f(x) and g(x) = –7.

f(x) = 4x + 1 First equation

(–7) = 4(–2) + 1 Substitute –2 for x and –7 for f(x).

–7 = –7

g(x) = x3 + 1 Second equation

(–7) = (–2)3 + 1 Substitute –2 for x and –7 for g(x).

–7 = –7

The identified point (–2, –7) satisfies both f(x) and g(x).

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At the point (0, 1), let x = 0 and f(x) and g(x) = 1.


(1) = 4(0) + 1 Substitute 0 for x and 1 for f(x).

1 = 1


(2) = (0)3 + 1 Substitute 0 for x and 1 for

g(x).

1 = 1

The identified point (0, 1) satisfies both f(x) and g(x).28



At the point (2, 9), let x = 2 and f(x) and g(x) = 9.


(9) = 4(2) + 1 Substitute 2 for x and 9 for f(x).

9 = 9


(9) = (2)3 + 1 Substitute 2 for x and 9 for

g(x).

9 = 9

The identified point (2, 9) satisfies both f(x) and g(x). 29



The solution set to the system of equations

is {(–2, –7), (0, 1), (2, 9)}, as identified on the graph as

the intersections of the lines f(x) and g(x).

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Guided Practice

Example 3Create a table to approximate the real solution(s), if any,

to the system .

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1. Create a table of values for the two functions f(x) and g(x).

You can create a table of values by hand similarly to how you have created them in the past, or you can use your graphing calculator.

To create a table of values using your graphing calculator, follow the directions appropriate to your calculator model.

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Step 1: Press the [Y=] button. Clear any equations from previous calculations.

Step 2: Type the first function into Y1, using the [X, T, θ, n] button for the variable x. Press [ENTER].

Step 3: Type the second function into Y2. Use the [^] button for powers. After entering a power, press the right arrow button to get out of exponent mode. Press [ENTER].

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(continued)

Guided Practice: Example 3, continuedStep 4: Press [2ND][WINDOW] to call up the

TABLE SETUP screen. Set TblStart to –5 by clearing out any current values and typing [(–)][5][ENTER]. Set ΔTbl to 1, then press [2ND][GRAPH] to call up the TABLE screen. A table of values for both equations will be displayed.

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Step 1: Press the [home] key.

Step 2: Arrow over to the calculator page, the first icon over, and press [enter].

Step 3: Define f(x) by entering “define” and then the function using the buttons on your keypad. Remember to enter a space after “define”. Press [enter].

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(continued)

Guided Practice: Example 3, continuedStep 4: Define g(x) by entering “define” and then

the function using the buttons on your keypad. Remember to press the right arrow key after entering an exponent. Press [enter].

Step 5: Press the [home] key. Arrow over to the spreadsheets page, the fourth icon over, and press [enter].

Step 6: Press [ctrl][T] to switch to a table window. From the list of defined functions that appears, select f(x)f1, and press [enter].

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(continued)

Guided Practice: Example 3, continuedStep 7: Press the right arrow key. From the list of

defined functions that appears, select f(x)g1, and press [enter]. A table of values for both equations will be displayed.

Both calculators should yield a table that resembles the following.

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x –5 –4 –3 –2 –1 0 1 2 3

f (x) 1 0.5 0 –0.5 –1 –1.5 –2 –2.5 –3

g(x) 251 65 1 –7 –1 1 5 41 163


2. Estimate the points of intersection.

Compare the y-values for the same x-value in your completed table.

Look for places where the difference between the values for g(x) and f(x) is decreasing. If the difference between g(x) and f(x) is increasing, then any intersection occurs in the other direction. The point of intersection occurs when f(x) is equal to g(x) or the difference is 0.

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Notice that in the table, f(x) equals g(x) when x = –1. Also notice that there appears to be a second intersection point close to the x-values of –3 and –2.

To better approximate the second point of intersection, you must find additional values.

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x –5 –4 –3 –2 –1 0 1 2 3

f(x) 1 0.5 0 –0.5 –1 –1.5 –2 –2.5 –3

g(x) 251 65 1 –7 –1 1 5 41 163

g(x) – f(x) +250 +64.5 +1 –6.5 0 +2.5 +7 +43.5 +166

Guided Practice: Example 3, continuedUse the following directions to determine additional values on your graphing calculator.

On a TI-83/84:

Step 1: Press [2ND][WINDOW] to call up the TABLE SETUP screen. Set the TblStart to –3 and ΔTbl to 0.01, clearing out any previous values. Press [ENTER], then press [2ND][GRAPH] to call up the TABLE screen. A table of values for both equations will be displayed.

Step 2: Scroll through the table of values to determine the approximate solutions.

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Step 1: Press [menu]. Use the arrow keys to select 5: Table, then 5: Edit Table Settings. Set the Table Start to –3 and Table Step to 0.01. Arrow down to OK and press [enter]. You may need to readjust the Table Step setting to get a better approximation of when f(x) = g(x).

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Guided Practice: Example 3, continuedThe following table displays information for values of x close to the point we’re interested in: x = –3.

Since we know that the difference between g(x) and f(x) is smallest for values of x close to –3, review the table near this value. Determine where the values of g(x) and f(x) have the smallest difference.

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x –3.00 –2.99 –2.98 –2.97 –2.96 –2.95 –2.94 –2.93

f(x) 0 –0.005 –0.010 –0.015 –0.020 –0.025 –0.030 –0.035

g(x) 1 0.7327 0.4707 0.2141 –0.0374 –0.2836 –0.5247 –0.7608

g(x) – f(x) +1 +0.7377 +0.4807 +0.2291 –0.0174 –0.2586 –0.4947 –0.7258

Guided Practice: Example 3, continuedAccording to the generated table of additional values, the value of x with the smallest difference between g(x) and f(x) is –2.96. Because the difference in values is so small and because the coordinates are approximate, it is appropriate to use either the f(x) value or the g(x) value when stating the solution.

It can be seen from the tables that –1 and the approximate value of –2.96 are the x-coordinates at which f(x) and g(x) intersect; therefore, the approximate solutions to the system are (–2.96, –0.02) and (–1, –1).

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3. Verify that the identified coordinate pairs are solutions to the original system of equations. In order for a coordinate to be a solution to a system, the coordinate must satisfy both equations of the system. Because our coordinates are estimated, it is quite possible that our results will be nearly equal, but not exactly equal.

Substitute the coordinates into each of the equations, then evaluate the equations to see if the coordinates result in a true statement.

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Guided Practice: Example 3, continuedAt the point (–2.96, –0.02), let x = –2.96 and f(x) and g(x) = –0.02.

First equation

Substitute –2.96 for x and –0.02 for f(x).

–0.02 = –0.02

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Guided Practice: Example 3, continuedg(x) = x4 + 3x3 + 1 Second equation

(–0.02) = (–2.96)4 + 3(–2.96)3 + 1 Substitute –2.96

for x and –0.02

for g(x).

–0.02 ≈ –0.04

The identified point (–2.96, –0.02) nearly satisfies both f(x) and g(x), and can be considered an approximate solution to the system of equations. 47


Guided Practice: Example 3, continuedAt the point (–1, –1), let x = –1 and f(x) and g(x) = –1.

First equation

Substitute –1 for x and –1 for f(x).

–1 = –1

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Guided Practice: Example 3, continuedg(x) = x4 + 3x3 + 1 Second equation

(–1) = (–1)4 + 3(–1)3 + 1 Substitute –1 for x and

–1 for g(x).

–1 = –1

The identified point (–1, –1) satisfies both f(x) and g(x), and is a solution to the system of equations.

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Guided Practice: Example 3, continuedThe approximate solution set to the system

is {(–2.96, –0.02), (–1, –1)}, as

identified in the tables as the intersection of the equations f(x) and g(x).

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Guided Practice: Example 3, continuedGraphing the system shows that our approximated solution set is valid.

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