Solving Systems of Equations

GraphingThere are three methods to solving

systems of equations by graphing:

1) Write both equations in slope – intercept form and graph

2) Write both equations in slope-intercept form and graph using the calculator

3) Solve for the x and y intercepts of each equation

Graphing

Solve the following system of equations by graphing:

-6x +3y = -15

-4x +y = -11

Graphing – Method 1

Write both equations in slope – intercept form and graph. To do this, solve each equation for y

-6x +3y = -15

-4x +y = -11


Writing -6x +3y = -15 in slope intercept form:

-6x +3y = -15

+6x +6x

3y = 6x – 15

3 3 3

y = 2x - 5


Writing -4x + y = -11 in slope intercept form:

-4x + y = -11

+4x +4x

y = 4x - 11


Graph both equations using their slope and y –intercept by starting at the y-intercept and using their slope to do rise over run.

Equation 1: y = 2x - 5

y intercept is (0, -5) slope is rise 2, run 1

Equation 2: y = 4x – 11

y intercept is (0, -11) slope is rise 4, run 1


The lines intersect at the point (1,3)

Graphing – Method 2Write both equations in slope-intercept

form and graph using the calculator

-6x +3y = -15-4x +y = -11The equations were already solved for

slope-intercept form in method 1, so:

y = 2x – 5y = 4x – 11

Graphing – Method 2 (TI-84+)

1) Turn the calculator on2) Hit the “Y=” key3) Type in the first

equation next to Y1

4) Use the “X,T,O,n” key to type “X”

5) Hit “Enter”6) Type in the second

equation next to Y2

7) Hit “Enter”


8) Hit the graph button to see the graph


9) If necessary, adjust the graph by changing the zoom

You can zoom in, or out by hitting the zoom button and then selecting option 2 or 3. Once selected, press enter again when you see the graph

Zoom standard goes back to the regular zoom

Graphing – Method 2 (TI-84+)10) When looking at the

graph hit the “CALC” button. Do this by hitting the “2ND” key followed by the “TRACE” key

11) Move down to choice five and select “intersect”

12) Press “Enter” and the calculator will return to the graph.


13) The calculator will prompt you to select the first curve. Use the arrows to put the blinking cursor on one of the lines

14) Hit “Enter”


15) The calculator will prompt you to select the second curve. Use the arrows to put the blinking cursor on the other line (the calculator should have already done this for you)

16) Hit “Enter”


17) The calculator will prompt you to guess the location of the intersection. Use the arrow keys to move the flashing curser close to the intersection

18) Hit “Enter”


19) The calculator will then tell you the intersection. In this case, “X=3, Y=1”

20) Write your answer as an ordered pair (3,1)

Graphing – Method 2 (TI-89)

1) Turn the calculator on

2) Hit the “Y=” key by hitting Diamond + F1

3) Type in the first equation next to Y1

4) Hit “Enter”

5) Type in the second equation next to Y2

6) Hit “Enter”


7) Hit the graph button to see the graph

- Do this by hitting diamond and then F3


8) If necessary, adjust the graph by changing the zoom

You can zoom in, or out by hitting the zoom button (F2) and then selecting option 2 or 3. Once selected, press enter again when you see the graph

Zoom standard (option 6) goes back to the regular zoom

Graphing – Method 2 (TI-89)9) When looking at the

graph select the “Math” menu. Do this by hitting the “F5” key

10) Move down to choice five and select “intersection”

11) Press “Enter” and the calculator will return to the graph.


12) The calculator will prompt you to select the first curve. Use the arrows to put the blinking cursor on one of the lines

13) Hit “Enter”


14) The calculator will prompt you to select the second curve. Use the arrows to put the blinking cursor on the other line (the calculator should have already done this for you)

15) Hit “Enter”


16) The calculator will prompt you to select the lower bound of the intersection. Use the arrow keys to move below or to the left of the intersection

17) Hit “Enter”


18) The calculator will prompt you to select the upper bound of the intersection. Use the arrow keys to move above or to the right of the intersection

19) Hit Enter


20) The calculator will then tell you the intersection. In this case, “X=3, Y=1”

21) Write your answer as an ordered pair (3,1)


Graph by solving for the x and y intercepts of each equation:

-6x +3y = -9

-4x +y = -8

Graphing – Method 3Find the x and y intercepts of the

first equation:-6x +3y = -15

x-intercept, let y=0-6x +3y = -15-6x +3(0) = -15-6x = -15-6 -6x = -15/-6 = 5/2x-int = (5/2,0)

y-intercept, let x=0-6x +3y = -15-6(0) + 3y = -153y = -153 3y = -5y-int = (0,-5)

Graphing – Method 3Find the x and y intercepts of the

second equation:-4x +y = -11

x-intercept, let y=0-4x +y = -11-4x +(0) = -11-4x = -11-4 -4x = -11/-4 = 11/4x-int = (11/4,0)

y-intercept, let x=0-4x +y = -11 -4(0) + y = -11y = -11y-int = (0,-11)


• Graph by plotting the x and y intercepts of each line and connecting them to form the line

• The solution is the intersection: the point (3,1)

Substitution1) Solve one of the equations for a variable2) Substitute the solved equation into the

OTHER equation in place of the variable you solved for

3) Solve the new equation for the remaining variable.

4) Once you find an answer for the first variable, substitute that answer into one of the original two equations and solve for the second variable.

5) Write your answer as an ordered pair.

Substitution

** If both variables cancel out, the lines are either parallel, or they are the same line

- If the lines are parallel then there is no solution, and when the equation is solved, it will result in an answer that is NOT true like 0 = 4

- If the lines are the same, then there is an infinite number of solutions, and when the equation is solved, it will result in an answer that IS true like 0=0 or 5=5.

Substitution – Example 1

Solve the following two equations using substitution:

y = 2x + 1

3x – 2y = -4


1) Solve one of the equations for a variable

- The first equation is already solved for y:

y = 2x + 1

Substitution – Example 12) Substitute the solved equation into the


So, substitute y = 2x + 1 into 3x – 2y = -4

3x – 2(2x + 1) = -4

Substitution – Example 13) Solve the new equation for the

remaining variable.3x – 2(2x + 1) = -43x -2(2x) -2(1) = -43x – 4x – 2 = -4-x – 2 = -4 +2 +2-x = -2-1 -1x = 2

Substitution – Example 14) Once you find an answer for the first

variable, substitute that answer into one of the original two equations and solve for the second variable.

y = 2x + 1 and x = 2, so

y = 2(2) + 1

y = 4 + 1

y = 5



x = 2 and y = 5 so the answer is:

(2,5)

To check your work, substitute your answer into all of the original equations:

Substitute in (2, 5) for x, y

y = 2x + 15 = 2(2) + 15 = 4+15 = 5

Both equations balance, so our answer is a solution


3x – 2y = -4

3(2) – 2(5) = -4

6 – 10 = -4

-4 = -4



3x - 2y = 5

4x + 4y = 20



- Solving the second equation for y is the easiest:

4x + 4y = 20-4x -4x4y = 20 – 4x4 4 4y = 5 - x


2) Substitute the solved equation into the OTHER equation in place of the variable you solved for

So, substitute y = 5 – x into 3x - 2y = 5

3x – 2(5-x) = 5


remaining variable.3x – 2(5-x) = 53x - 2(5) - 2(-x) = 53x - 10 + 2x = 55x – 10 = 5 +10 +105x = 155 5x = 3

Substitution – Example 24) Once you find an answer for the first

variable, substitute that answer into one of the original two equations and solve for the second variable.

- Substitute x = 3 into 3x - 2y = 53(3) – 2y = 59 – 2y = 5-9 -9-2y = -4-2 -2y = 2



x = 3 and y = 2 so the ordered pair is:

(3,2)



-2x + 4y = -12

-x + 2y = 2

Substitution – Example 31) Solve one of the equations for a

variable - Solving the second equation for

x is very easy:-x + 2y = 2 -2y -2y-x = 2 – 2y-1 -1 -1x = -2 + 2y



So, substitute x= -2 + 2y into -2x + 4y = -12

-2(-2+2y) +4y = -12


remaining variable.-2(-2+2y) +4y = -12-2(-2) + -2(2y) + 4y = -124 – 4y + 4y = -124 = -12** Both variables are eliminated, and 4 = -12 is not true. Thus the lines are

parallel and the answer is “no solution”



-2x + y = 3

6x + -3y = -9



**Solving the first equation for y is very easy:

-2x + y = 3

+2x +2x

y = 2x + 3



So, substitute y = 2x + 3 into 6x + -3y = -9

6x + -3(2x+3) = -9


remaining variable.6x + -3(2x+3) = -96x + -3(2x) + -3(3) = -96x – 6x – 9 = -9-9 = -90 = 0** Both variables are eliminated, and 0=0

is true. Thus the lines are the same and the answer is “Infinite solutions”

Elimination

The goal of elimination is to add the equations in a system together and have a variable cancel out and be eliminated

Elimination1) Make sure the equations are in the same

form. i.e. both in standard form so that the x’s, y’s, and numbers all line up in columns

2) See if the equations can be added together immediately to eliminate a variable

3) If the equations cannot be added together to eliminate a variable, choose a variable to eliminate and multiply one or more of the equations by an integer in order to have the same coefficient in from of the variable in both equations. (note, one should be positive, one should be negative)

Elimination4) Add the two equations together and

make sure one variable was eliminated (leaving you with one variable)

5) Solve the equation for the variable

6) Substitute back into one of the original equations to solve for the second variable.

7) Write your answer as an ordered pair

Elimination** If both variables are eliminated, the lines

are either parallel, or they are the same line

- If the lines are parallel then there is no solution, and when the equation is solved, it will result in an answer that is NOT true like 0 = 4

- If the lines are the same, then there is an infinite number of solutions, and when the equation is solved, it will result in an answer that IS true like 0=0 or 5=5.

Elimination – Example 1


2x + 5y = -12

-5x - 5y = -3


1) Make sure the equations are in the same form:

2x + 5y = -12 and

-5x - 5y = -3 are both in standard form

** Note how the variables line up!

2) See if the equations can be added together immediately to eliminate a variableYes they can be! (skip step 3)


4) Add the two equations together and make sure one variable was eliminated (leaving you with one variable)

2x + 5y = -12

+ -5x - 5y = -3

-3x + 0 = -15 => -3x = -15



-3x = -15

-3 -3

x = 5

Elimination – Example 16) Substitute back into one of the original

equations to solve for the second variable.

x = 5; 2x + 5y = -12 so,2(5) + 5y = -1210 + 5y = -12-10 -105y = -225 5y = -22/5



x = 5, y = -22/5 so,

(5, -22/5)


Substitute in (5, -22/5) for x, y

2x + 5y = -122(5) + 5(-22/5) = -1210 + -22 + -12-12 = -12



-5x - 5y = -3-5(5) – 5(-22/5) = -3-25 - - 22 = -3-25 + 22 = -3-3 = -3



3x + 5y = 2

2x - 4y = 16


1) Make sure the equations are in the same form:

3x + 5y = 2 and 2x - 4y = 16 are both in standard

form

** Note how the variables line up!


2) See if the equations can be added together immediately to eliminate a variable

No They cannot (multiply equations by a scalar)

3x + 5y = 2

2x - 4y = 16Lcm of 3 and 2 is 6, so I will make 6x in the first

equation by multiplying the whole first equation by 2, and I will make -6 in the second equation by multiplying the whole second equation by -3


(2)3x + (2)5y = 2(2)(-3)2x – (-3)4y = 16(-3)Rewriting the equations:6x + 10y = 4-6x + 12y = -48

* Now the equations can be added to eliminate a variable

4) Add the two equations together and make sure one variable was eliminated (leaving you with one variable)

6x + 10y = 4

+ -6x + 12y = -48

0 + 22y = -44 => 22y = -44




22y = -44

22 22

y = -2

Elimination – Example 26) Substitute back into one of the original

equations to solve for the second variable.

y = -2; 3x + 5y = 2 so,3x + 5(-2) = 23x – 10 = 2 +10 +103x = 123 3x = 4



x = 4, y = -2 so,

(4, -2)


Substitute in (4, -2) for x, y3x + 5y = 23(4) + 5(-2) = 212 + -10 = 22 = 2



2x - 4y = 16

2(4) – 4(-2) = 16

8 + 8 = 16

16 = 16

Applications of Linear Systems

• Write systems of linear equations

• Model real world situations with systems of equations

Steps for Solving Application Problems

1) Read the problem carefully. Highlight given information

2) Determine the unknowns and define variables.

3) Write a system of equations

4) Solve using any method (graphically, substitution, or elimination)

Example 1

A football club publishes a monthly newsletter. The newsletter costs 90 cents for printing and mailing each copy, plus $600 to write. If the newsletter sells for $1.50 per copy, how many copies of the newsletter must the club sell to break even?

Step 1 – Read and Highlight

A football club publishes a monthly newsletter. The newsletter costs 90 cents for printing and mailing each copy, plus $600 to write. If the newsletter sells for $1.50 per copy, how many copies of the newsletter must the club sell to break even?

Step 2 – Define Variables

• Let x = the number of copies printed and sold

• Let y = the amount of money spent or earned

Step 3 – Write a System of Equations

Equation 1: Cost of printing

y = $0.90x + $600

Equation 2: Money Earned

y = $1.50x

Note: y is the same in both equations because we want income to equal expenses

Step 4 - Solve

Method of solving Chosen: Substitution

y = $0.90x + $600

y = $1.50x

$0.90x + $600 = $1.50x

- $0.90x - $0.90x

$600 = $0.60x$0.60 $0.60

1000 = x

Step 4 (Continued)- Solve

Substitute back in to solve for y:y = $1.50x and x = 1000 so,y = $1.50(1000)y = $1500

The club must sell 1000 copies to break even. After 1000 copies they will have spent and earned $1500

Example 2

A store sells two types of ipod touches. The 8gb version sells for $249.99. The 64gb version sells for $329.99. If the store did $8379.70 in sales last year after they sold 30 ipods, how many of each type of ipod did they sell?

Step 1 – Read and Highlight

A store sells two types of ipod touches. The 8gb version sells for $249.99. The 64gb version sells for $329.99. If the store did $8379.70 in sales last year after they sold 30 ipods, how many of each type of ipod did they sell?

Step 2 – Define Variables

• Let x = the number of 8gb ipods sold

• Let y = the number of 64gb ipods sold

Step 3 – Write a System of Equations

Equation 1: Total sales

$8379.70 = $249.99x + $329.99y

Equation 2: Total number of ipods

30 = x + y

Step 4 - Solve

Method of solving Chosen: Substitution

$8379.70 = $249.99x + $329.99y

30 = x + y

Solve the 2nd equation for y

30 = x + y

-x -x

30 – x = y


Substitute the 2nd equation into the first equation:

$8379.70 = $249.99x + $329.99y30 – x = y$8379.70 = $249.99x + $329.99(30-x)$8379.70 = $249.99x + $9899.70 - $329.99x


$8379.70 = $249.99x + $9899.70 - $329.99x$8379.70 = $249.99x - $329.99x + $9899.70$8379.70 = -$80x + $9899.70-$9899.70 - $9899.70-$1520 = -$80x-$80 -$80 19 = x


Substitute back in to solve for y:

30 = x + y and x = 19 so,

30 = 19 + y

-19 -19

11 = y

So the company sold 19 ipods that were 8gb and 11 ipods that were 64gb

Linear Inequalities

A Linear Inequality is used to describe a region of a coordinate plane that is bounded by one or more lines

The solutions of a linear inequality is any point that makes the inequality true.

Graphing Linear Inequalities

1) Write the equation in slope intercept form

2) Replace the inequality symbol with an equal sign

3) Graph the equation use a dashed line for > or < and a solid line for ≤ or ≥

4) Pick test points in each section that the graph bounds (i.e. pick points both above and below the line)

5) Shade the regions in which the test points are true

Example 1

y ≤ -3x + 3

1) Write in slope intercept form

2) Change the inequality to an = sign

3) Graph y = -3x + 3

Example 1y ≤ -3x + 34) Pick test points

above and below the graph

(-2, 1)-2 ≤ -3(1) + 3-2 ≤ 0(2,3)2 ≤ -3(3) +32 ≤ -6

Example 1

y ≤ -3x + 3

5) Shade the region where the test points are true

Graphing Inequalities

1) To shade a graph, first determine whether you need to shade above or below the graph using test points

2) Go to the y= screena) Hit the y= button on the TI-84

b) Hit Diamond + F1 on the TI-89

Graphing Inequalities (TI-84+)3) In the y= screen, type

the equation

4) To change the shading, move your curser to highlight the \ on the far left of the equation. Each time you hit enter will change how the calculator graphs

Graphing Inequalities (TI-84+)

5) Hit enter two times to shade the graph below the line. The \ will have changed to this:


6) Hit enter a third time to shade the graph below the line. The \ will have changed to this:


6) Hit enter a total of 6 times (3 more times) time to graph a dashed line instead of a solid line. The \ will have changed to this:

Graphing Inequalities

1) To shade a graph, first determine whether you need to shade above or below the graph using test points

2) Go to the y= screena) Hit the y= button on the TI-84

b) Hit Diamond + F1 on the TI-89

Graphing Inequalities (TI-89)3) In the y= screen, type

the equation

Graphing Inequalities (TI-89)4) To change the

shading, hit F6 by hitting “2nd” and then “F1”

Graphing Inequalities (TI-89)5) A menu will appear

• Select choice 7 (Above) to shade the region above the line

• Select choice 8 (Below) to shade the region below the line

• Select choice 2 (Dot) to graph a dotted line

Example 2 – Test PointsDetermine whether or not the point

(2, -3) is a solution to a linear inequality y > -3x + 5

Substitute in for y and xy > -3x + 5-3 > -3(2) + 5-3 > -6 +5-3 > -1 => NO, this point is NOT a

solution

Example 3 – Test PointsDetermine whether or not the point

(6, -1) is a solution to a linear inequality y > (2/3)x -2

Substitute in for y and x

y < (2/3)x - 2

-1 < (2/3)(6) - 2

-1 < 4 - 2

-1 < 2 => YES, this point IS a solution

Example 4 – Describing GraphsGiven the graph shown to

the right, which equation describes the graph?

a) y = -2x + 1

b) y < -2x + 1

c) y > -2x + 1

d) y ≥ -2x + 1

e) y ≤ -2x + 1

Example 4 – Describing Graphsa) y = -2x + 1 => Wrong because a

region of the graph is shaded. The equation is just the equation of a line, not a region of a plane.

Example 4 – Describing Graphsb) y < -2x + 1 => Wrong because the

equation is < which means the line should be dashed, not solid

Example 4 – Describing Graphsc) y > -2x + 1 => Wrong because the

equation is > which means the line should be dashed, not solid

Example 4 – Describing Graphsd) y ≥ -2x + 1 => Correct. The line is shaded

and the test point (2,3) means that 3 ≥ -2(2) + 1 so 3 ≥ -4 +1 thus, 3 ≥ -3, which is true so the shaded region is above the graph

Example 4 – Describing Graphse) y ≤ -2x + 1 => Wrong because the

test point (2,3) means that 3 ≤ -2(2) + 1 so 3 ≤ -4 +1 thus, 3 ≤ -3, which is false.

Example 5 –GraphingGraph 4x - 2y ≥ 8

1) Write in Slope-Intercept form

4x - 2y ≥ 8

-4x -4x

-2y ≥ -4x + 8

-2 -2

y ≤ 2x – 4

** Note: The sign changes because we divided by a negative**

Example 5 –Graphingy ≤ 2x – 42) Change the

inequality to an = sign

y = 2x – 43) Graph – the

line is solid because it is ≤

Example 5 –Graphing

Above: (0,0)y ≤ 2x – 40 ≤ 2(0) – 40 ≤ -4 =>False

Below: (0, -5)y ≤ 2x – 4-5 ≤ 2(0) -4-5 ≤ -4 => True

4) Pick test points above and below the graph.

Thus, we shade the region below the line

Example 5 –Graphing

Solving Systems of Equations

Documents

Transcript of Solving Systems of Equations