2. The students will be able to write lines in slope intercept form and point slope form.

3. The students will be able to solve systems of equations using ordered pairs, graphing and substitution.

4. The students will be able to solve systems of equations by adding, subtracting and multiplying.

Some rules for finding slope:

1. Always find the change in y first and let that be your numerator,

then find the change in x and let that be your denominator.

2. When counting UP, that is a positive number.

When counting DOWN, that is a negative number.

2. (-3, -7) (2,4)

3. (0,2) (3,5)

4. (5,0) (-1, 4)

Now, pick your own sets of 2 points. Use graph paper to plot these points and connect the dots. Compute the slope for each line that you have drawn. Do 5 examples this way using the graph paper given.

Let's review the definition of slope:

Compute the slope of the following lines.

Example #2

We call this "0 slope" We call this "No slope" (Notice, here we cannot divide by 0!)

What can we call lines with a positive slope?

These lines go up or "increase"

What can we call lines with a negative slope?

These lines go down or "decrease"

An easier way to write the slope formula is:

Now, graph the line that passes through the given point and has the given slope.

Explore the following:

1. Plot the following points

A(-10, 0) B(6, 12) C(-4, -5) D(12, 7)

Next, draw the segments AB and CD. What do you notice?

Find the slopes of each line segment.

2. Investigate the following diagram:

What pairs of segments appear to be parallel?

Find the slope of each segment. What is true?

4. Now plot the following points:

A(-1, 3) B(5, -6) C(6, 1) D(-3, -5)

Draw the segments AB and CD. What do you notice about these segments?

Find the slopes of these line segments. What is their relationship to each other?

What do you get when you multiply these slopes together?

5. Investigate the following diagram:

What pairs of lines appear to be perpendicular?

Find the slope of each segment. What is true?

What do you get when you multiply these slopes together?

* If two lines are perpendicular, the have a slope that is the

Use Geometer's Sketchpad to compute the slopes of the following points and find the slope of a line parallel and the slope of a line perpendicular to the given line.

1. A(6, 3) and B(4, 6)

2. C(-1, -5) and D(3, -2)

3. E(0, 5) and F(-1, 6)

4. G(-4, -2) and H(-5, -2)

Now, using Algebra Xpresser to find 3 lines parallel to each other and 2 lines perpendicular through the parallel lines. Give each slope and two points that lie on each line.

The standard form that equations are written in is called slope-intercept form of the line where y=mx + b. In this form,

Find the slope and y-intercept of the following lines.

Now, using Algebra Xpresser, graph the above lines.

Determine whether the 2 lines are parallel, perpendicular or neither.

Now, plug these equations into your TI-82 calculator. Check to see of you were correct about your assumptions of slope.

1. For example, write an equation for a line with slope=4 and y-intercept that is 2.

Given: m=4 and b=2

Therefore, the equation in y=mx + b form is y=4x + 2.

Now, what is the slope and a point on the line were given?

2. Write an equation for a line whose slope is 5 and contains the point (3, -2).

Given: m=5, x=3 and y=-2.

The form for the equation is y=mx + b and we are looking for the

y-intercept.

So, -2 = 5(3) +b

-2 = 15 +b

b= -17

Therefore, the equation for the line is y=5x - 17.

What happens if you are given 2 points?

3. Write the equation for the line that passes through the points

(4, 3) and (-1, -5).

Given two points, we can easily find the slope of this line.

Now, we have slope also x=4 and y=3 (or we could have picked the other point to use.)

Using the information that we now know about writing the equation of lines using the given slope, point, and/or y-intercept, try some on your own.

An ordered pair is a solution of a

State whether the ordered pair is a solution of the system of equations.

1. (2, -3)

2x - y = 10

x + 2y = -5

Solution:

2(2) - (-3) = 10

4 + 3 = 10

7 = 10

2+2(-2) = -5

2 - 4 = -5

-2 = -5.

So, (2, -3) is not a solution for this system of equations.

2. (2, 5)

-4x + y = -3

2x + y = 9

Solution:

-4(2) + 5 = -3

-8 + 5 = -3

-3 = -3

2(2) + 5 = 9

4 + 5 = 9

9 = 9.

So, (2, 5) is a solution for this system of equations.

To solve a system of equations using graphs, graph each equation on the same pair of axes. All points of intersection are

Solve each system of equations by graphing.

3.a. y = 2x - 6

b. y = x - 4

Check your intersection point:

a. -2 = 2(2) - 6

-2 = 4 - 6

-2 = -2

b. -2 = 2 - 4

-2 = -2.

Now we can see that the point (2, -2) is a solution for the system of equations.

4.a. 3x-2y=4

b. 2x+y=5

In order to graph these equations, we need to put them in y=mx+b form.

So,

a. 3x - 2y = 4

-2y = -3x + 4

y = (3/2)x - 2

b. 2x + y = 5

y = -2x + 5.

Now, graph the two lines.

Check your intersection point:

a. 3(2) - 2(1) = 4

6 - 2 = 4

4 = 4

b. 2(2) + 1 = 5

4 + 1 = 5

5 = 5.

So, the point (2, 1) is a solution to the system of equations.

More examples:

Solve the systems of equations by graphing:

1. y = 3x - 1

y = 3x + 2

2. y = -x + 7

y = 2x - 5

3. x + 3y = 6

3x + 2y = 4

4. y = 7 - 3x

6x + 2y = 14

5. y = 4x - 2

x - 4y = 2

One of the methods of solving systems of equations is called

1. Solve one of the equations for one variable in terms of the other.

2. Substitute that expression in the other equation and solve.

3. Substitute that value in one of the original equations and solve.

4. Check in both of the original equations.

Solve by the substitution method:

1. x + y = 6

x - 2y = 3

x + y = 6

y = -x + 6

x - 2(-x+6) = 3

x + 2x - 12 = 3

3x - 12 = 3

3x = 15

x = 5.

y = 1.

a. 5 + 1 = 6

6 = 6

b. 5 - 2(1) = 3

5 - 2 = 3

3 = 3.

So, the point (5, 1) is a solution for this system of equations.

2. 2x - y = 14

x + y = 7

x + y = 7

y = -x + 7.

2x - (-x + 7) = 14

2x + x - 7 = 14

3x - 7 = 14

3x = 21

x = 7.

7 + y = 7

y = 0.

a. 2(7) - 0 = 14

14 = 14

b. 7 + 0 = 7

7 = 7.

So, the point (7, 0) is a solution to this system of equations.

3. x - 2y = -9

-2x + 4y = -13

x - 2y = -9

x = 2y - 9.

-2(2y - 9) + 4y = -13

-4y + 18 +4y = -13

18 = -13 is not true.

So, there is no solution to this system of equations.

4. At the food truck, Joe bought 3 hot dogs and 4 drinks for $10. Cindy paid $5 for 1 hot dog and 3 drinks. Find the cost of each hot dog and each drink.

Let h=hot dog and d=drink

3h + 4d = 10

1h + 3d = 5

1h + 3d = 5

h = -3d + 5.

3(-3d + 5) + 4d = 10

-9d +15 + 4d = 10

-5d + 15 = 10

-5d = -5

d = 1.

h + 3(1) = 5

h + 3 = 5

h = 2.

a. 3(2) + 4(1) = 10

6 + 4 = 10

10 = 10

b. 1(2) + 3(1) = 5

2 + 3 = 5

5 = 5.

So, each hot dog cost $2 and each drink cost $1.

1. -2x + y = 2

4x - 2y = 3

2. 2x + y = 3

-12x - 6y = -18

3. x + y = 5

2x - y = 4

4. -2x + y = -3

3x - 3y = -18

5. Jim bought 4 jars of jelly and 6 jars of peanut butter. Adam bought 3 jars of the same jelly and 5 jars of the same peanut butter. Jim paid $19.32 and Adam paid $15.67. What is the cost of a jar of peanut butter?

6. On a construction site there are 14 cranes and trucks. Three times the number of cranes is 2 more than twice the number of trucks. How many cranes are there on the site?

Warm-Up Activity:

Simplify these expressions:

(3x - y -6)- (3x - 2y - 6) = y

(4x + y + 4) + (2x - y - 10) = 6x - 6

1. If the coefficients of one of the variables are opposites,

2. Solve the resulting equation for the remaining variable.

3. Substitute the value for the variable in one of the original equations and solve for the unknown variable.

4. Check the solution in both of the original equations.

Solve the systems of equations by adding or subtracting:

1. 2x + 7y = -5

-5x + 7y = -12

subtract the two equations and solve for resulting variable

2(1) + 7y = -5

2 + 7y = -5

7y = =7

y = -1.

a. 2(1) + 7(-1) = -5

2 - 7 = -5

-5 = -5

b. -5(1) + 7(-1) = -12

-5 - 7 = -12

-12 = -12.

So, the point (1, -1) is the solution to the system of equations.

2. 3x + y = -7

5x - y = -9

solve using addition and solve for resulting variable

5(-2) - y = -9

- 10 - y = -9

-y = 1

y = -1

a. 3(-2) + (-1) = -7

-6 - 1 = -7

-7 = -7

b. 5(-2) - (-1) = -9

-10 + 1 = -9

-9 = -9.

So, the point (-2, -1) is a solution for the system of equations.

More Examples:

Solve by adding or subtracting:

1. x - y = -5

x + y = 15

2. 6x - 3y = 6

8y = -6x - 16

3. x - y = -1

x + y = 11

4. 5x + 4y = 12

4y = 3x - 4

5. x - y = -1

x + y = 9

Solve the systems of equations by multiplying:

1. 3x - 4y = 10

3y = 2x - 7

2(3x -4y = 10) = 6x -8y = 20

3( -2x +3y = -7) = -6x + 9y = -21

3x - 4(-1) = 10

3x + 4 = 10

3x = 6

x = 2.

a. 3(2) - 4(-1) = 10

6 + 4 = 10

10 = 10

b. -2(2) +3(-1) = -7

-4 - 3 = -7

-7 = -7.

So, the point (2, -1) is a solution for this system of equations.

2. 4x = 2y - 12

3y = x + 13

4x - 2y = -12

4(-x + 3y = 13) = -4x + 12y = 52

3(4) = x + 13

12 = x + 13

x = -1

a. 4(-1) = 2(4) -12

-4 = 8 -12

-4 = -4

b. 3(4) = -1 + 13

12 = -1 + 13

12 = 12.

So, the point (-1, 4) is a solution to this system of equations.

3. WORD PROBLEM:

Tim sold 25 movie tickets for a total of $132. If each adult ticket sold for $6 and each children's ticket sold for $4, how many of each kind did he sell?

Let A=Adult tickets and C=Child tickets

So, since Tim sold 25 Adult and Child tickets, we can set up this equation:

A + C = 25

Since, Tim sold Adult tickets for $6 and Child tickets for $4, we can set up this equation:

6A + 4C = $132

Our system of equations is:

A + C = 25

6A + 4C = 132

Now, solve this system by multiplying the first equation by 4:

4(A + C = 25) = 4A + 4C = 100

By subtracting the two equations we get:

16 + C = 25

C = 9.

a. 16 + 9 = 25

25 = 25.

b. 6(16) + 4(9) = 132

96 + 36 = 132

132 = 132.

So, Adult tickets cost $16 and Child tickets cost $9.

More Examples:

Solve the systems of equations by multiplying:

1. x - 5y = 0

3y = 2x - 7

2. 5x + 3y = 12

4x - 5y = 17

3. y = 3x + 8

4x - 2y = -14

4. 7x + 4y = 4

2x + 3y = -10

5. Adult tickets for a benefit breakfast cost $2.50. Children's tickets cost $1.50. If 56 tickets were sold for total sales of $97, how many of each kind were sold?

6. A class of 31 students was divided into seven groups consisting of 4 or 5 students each. How many of each size group were there?

7. Jason spent $130 on six items of clothing consisting of jeans and shirts. If jeans cost $27 each and shirts cost $19 each, how many of each did he buy?

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