Algebra Honors ( Periods 4 & 7)

Applying Systems of Linear Equations 6-5

METHOD	BEST TIME to USE
Graphing	estimate solutions
Substitution	If one of the variables has a coefficient of +1 or -1
Elimination Add/Sub	If one of the variables has opposite or same coefficient in both equations
Elimination Multiplication	If none of the coefficients are +1 or -1 & neither of the variables can be eliminated by addition or subtraction

What would the best method? Now find the solution..:

2x + 3y = -11

-8x -5y = 9

Answer: Elim (x)   (2, -5)

3x + 4y = 11

2x + y = -1

Answer: Substitution  ( -3, 5)

3x -4y = -5

-3x + 2y = 3

Answer” Elim (+)  ( -1/3, 1)

We are working with word problems. Turn to your textbook Page 367

#5

4 T- shirts  3 pairs of jeans $181

1 T-shirt  2 pairs of jeans $94

Let statements help keep you organized so

Let T = the cost of each of the T-shirts

Let J = the cost of each of the pairs of jeans

4T  + 3J = 181

1T  + 2J = 94

using substitution you find T = 94-2J

substitute in

4(94 – 2J ) + 3J = 181

J = $39.00

T = $16

#14

The Caverns of Sonora have two different tours: The Crystal Palace tour and the Horseshoe Lake Tour. The total length of both tours is 3.25miles. The Crystal Palace Tour is one-half mile less than twice the distance of the Horseshoe Lake Tour.  Determine the length of each tour.

Let C = the length in miles of the Crystal Palace tour

Let H= the length in miles of the Horseshoe Lake tour

 C + H = 3.25

C = 2H -0.5

substitution

2H-0.5 + H = 3.25

3H = 3.75

H = 1.25

so C = 2

Horseshoe Lake Tour is 1.25miles and Crystal Palace Tour is 2 miles

# 15

The breakeven point is the point at which income = expenses

Ridgemont High school is paying $13,200 for the writing and research of their yearbook plus a printing charge of $ 25 per book.  If they sell the books for $ 40 each, how many will they have to sell to break-even?

I would probably just use one equation but since we are in systems, I will set this up using two variables.

Let b = the number of books

Let p = the breakeven point

$13,200 + 25b = p

40b = p

$13,200 + 25b= 40b

880 books

Math 6H ( Period 5)

Integers & Absolute Value 3-1

Chapter 3-1: Integers and Absolute Value

Integers are the whole numbers and their opposites, which are negative.

Whole numbers are 0, 1, 2, 3…

Notice there are NO FRACTIONS OR DECIMALS in the Integer number system

There are negative fractions and decimals but they are not integers.

We use a number line to understand negative numbers.

Negative numbers are to the LEFT of zero.

Lots of real world situations must be expressed as negatives:

Owing money

Temperatures below zero

Losing yardage in a football game

Diving below the surface level of the water

Absolute value is the distance between a number and zero.

Distance is always positive so absolute value is always positive EXCEPT ZERO WHICH IS NEUTRAL.

The absolute value of 3 is 3 because it’s 3 from zero.

The absolute value of -3 is also zero because it’s 3 steps from zero.The absolute value signs are 2 vertical bars surrounding the number:

I 3 I = I -3 I = 3

Opposites are the same distance from zero but one is positive and one is negative.

3 and -3 are opposites.

Compare and Order Integers

First, look at the signs of the numbers: Positive is always > negative

Next, if a number is to the right of another number on the number line, it is >

Therefore, -3 > -10 or -10 < -3

You can write this either way.

0 is > negative numbers

Using the number line really helps!

Algebra Honors ( Periods 4 & 7)

Using Addition w/ Multiplication to Solve a System 6-4

This is the same method as Chapter 6-3, but to get ADDITIVE INVERSES of one variable you’ll need to multiply one or both equations by a factor.

MULTIPLYING JUST ONE EQUATION:

5x + 6y = -8

2x + 3y = -5

--------------------

Multiply either the bottom by -2 to eliminate y:

 5x + 6y = -8

-4x - 6y = 10

--------------------

 x +  0  =  2

x = 2

MULTIPLYING BOTH EQUATIONS:

4x + 2y = 8

3x + 3y = 9

--------------------

To eliminate x, you’d need to multiply the top by 3 and the bottom by -4 so that you’d get 12x and -12x

OR

Multiply the top by 3 and the bottom by -2 so that you’d get 6y and -6y.

It’s your choice!

I think keeping the numbers as small as possible is usually easier, so I’ll choose eliminating y.

3(4x + 2y) =  3(8)

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

--------------------

12x + 6y = 24

-6x - 6y = -18

--------------------

 6x +  0  =  6

x = 1

Algebra (Period 1)

Using Addition w/ Multiplication to Solve a System 6-4

This is the same method as Chapter 6-3, but to get ADDITIVE INVERSES of one variable you’ll need to multiply one or both equations by a factor.

MULTIPLYING JUST ONE EQUATION:

5x + 6y = -8

2x + 3y = -5

--------------------

Multiply either the bottom by -2 to eliminate y:

 5x + 6y = -8

-4x - 6y = 10

--------------------

 x +  0  =  2

x = 2

MULTIPLYING BOTH EQUATIONS:

4x + 2y = 8

3x + 3y = 9

--------------------

To eliminate x, you’d need to multiply the top by 3 and the bottom by -4 so that you’d get 12x and -12x

OR

Multiply the top by 3 and the bottom by -2 so that you’d get 6y and -6y.

It’s your choice!

I think keeping the numbers as small as possible is usually easier, so I’ll choose eliminating y.

3(4x + 2y) =  3(8)

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

--------------------

12x + 6y = 24

-6x - 6y = -18

--------------------

 6x +  0  =  6

x = 1

Algebra Honors ( Periods 4 & 7)

Using Addition to Solve a System  6-3

The second Algebraic method to solve a system is known as ELIMINATION.

You’ll be eliminating one variable by using the ADDITIVE INVERSE of it in the other equation.

Example where you add the two equations together:

4x + 6y = 32

3x – 6y =  3

--------------------

7x + 0 = 35

x = 5

Plug into either equation to find y:

4(5) + 6y = 32

20 + 6y = 32

6y = 12

y = 2

The solution is (5, 2)

Sometimes you’ll ALMOST have additive inverses, but you’ll need to multiply one equation by -1 first:

5x + 2y =  6

9x + 2y = 22

--------------------

Multiply either the top or bottom by -1 (your choice):

5x + 2y =    6

-9x - 2y = -22

--------------------

-4x + 0 = -16

x = 4

Plug into either ORIGINAL equation:

5(4) + 2y = 6

20 + 2y = 6

2y = -14

y = -7

The solution is (4, -7)

Now you can plug this point into the other equation to check that you haven’t made a mistake:

9x + 2y = 22

9(4) + 2(-7) = 22

36 – 14 = 22

22 = 22

Algebra ( Period 1)

Using Addition to Solve a System  6-3

The second Algebraic method to solve a system is known as ELIMINATION.

You’ll be eliminating one variable by using the ADDITIVE INVERSE of it in the other equation.

Example where you add the two equations together:

4x + 6y = 32

3x – 6y =  3

--------------------

7x + 0 = 35

x = 5

Plug into either equation to find y:

4(5) + 6y = 32

20 + 6y = 32

6y = 12

y = 2

The solution is (5, 2)

Sometimes you’ll ALMOST have additive inverses, but you’ll need to multiply one equation by -1 first:

5x + 2y =  6

9x + 2y = 22

--------------------

Multiply either the top or bottom by -1 (your choice):

5x + 2y =    6

-9x - 2y = -22

--------------------

-4x + 0 = -16

x = 4

Plug into either ORIGINAL equation:

5(4) + 2y = 6

20 + 2y = 6

2y = -14

y = -7

The solution is (4, -7)

Now you can plug this point into the other equation to check that you haven’t made a mistake:

9x + 2y = 22

9(4) + 2(-7) = 22

36 – 14 = 22

22 = 22

Algebra Honors ( Periods 4 & 7)

Using Substitution to Solve a System  6-2

There are ALGEBRAIC ways (not graphing…solving equations) to find the intersection of 2 (or more) linear equations.

The two Algebraic ways:

1. Substitution

2. Addition or Elimination

Today we’ll look at substitution.

This method works especially well if both equations are solved for the SAME variable (x OR y)

OR

One equation is solved for a SINGLE variable (x or y)

You’ll plug one equation into the other…meaning you’ll substitute it in.

If you’ve ever been on the bench in a game, think of how you hope you’ll be substituted into the game for another player so you can play.

(or if you’re the understudy in a play or if you can substitute one book for another and get the same number of AR points)

Let’s look at some examples and you’ll see how it works.

A system where both equations are already solved for one variable:

y = x + 7   and   y = 2x + 1

Since both equations are equal to y, they’re equal to each other!
 (transitive property of equality)

x + 7 = y = 2x + 1

So just get rid of the “middle man” y and get:

x + 7 = 2x + 1

Solve for x:

x = 6

Now plug into whichever original equation seems easier to you to find the y coordinate:

y = x + 7

y = 6 + 7 = 13

The intersection is (6, 13)

What if we plug in this point to the other equation? It should work because both equations have (6, 13) as a solution.

y = 2x + 1

13 = 2(6) + 1

13 = 13

A system where one equation is solved for one variable:

y = 2x

5x + 3y = 22

2x is the same value as y.

Since that is true, anywhere you see y, you may use 2x instead.

SUBSTITUTING INTO THE GAME FOR Y IS 2X:

5x + 3(2x) = 22

5x + 6x = 22

11x = 22

x = 2

Now plug into the other equation to find y:

y = 2x = 2(2) = 4

The solution (intersection) is (2, 4)

CHECK:

Plug in (2, 4) into the other equation:

5(2) + 3(4) = 10 + 12 = 22

WHAT IF YOU HAVE 2 EQUATIONS AND NEITHER ONE IS SOLVE FOR A SINGLE VARIABLE?

You can just solve for one variable in whichever equation is easier.

Example:

x – 3y = 15  and 4x -2y = 20

You would need to pick which variable (x or y) would be easier to solve for in which equation.

Generally, look for a variable with no coefficient (really a coefficient of 1).

So for the above system, I’d pick to solve for x in the first equation:

x = 3y + 15

So wherever you see “x” in the other equation, substitute in (3y + 15)

4(3y + 15) – 2y = 20

12y + 60 -2y = 20

10y + 60 = 20

10y = -40

y = -4

Substitution is often used to solve WORD PROBLEMS.

Example:

The perimeter of a rectangle is 40 in.

The length is 10 less than twice its width.

Find the dimensions of the rectangle.

2l + 2w = 40

l = 2w – 10

Substitute (2w – 10) for l:

2(2w – 10) + 2w = 40

4w – 20 + 2w = 40

6w – 20 = 40

6w = 60

w = 10 in.

l = 2w – 10 = 2(10) – 10 = 20 – 10 = 10 in.

IT’S A SQUARE! ;)

Algebra ( Period 1)

Using Substitution to Solve a System  6-2

There are ALGEBRAIC ways (not graphing…solving equations) to find the intersection of 2 (or more) linear equations.

The two Algebraic ways:

1. Substitution

2. Addition or Elimination

Today we’ll look at substitution.

This method works especially well if both equations are solved for the SAME variable (x OR y)

OR

One equation is solved for a SINGLE variable (x or y)

You’ll plug one equation into the other…meaning you’ll substitute it in.

If you’ve ever been on the bench in a game, think of how you hope you’ll be substituted into the game for another player so you can play.

(or if you’re the understudy in a play or if you can substitute one book for another and get the same number of AR points)

Let’s look at some examples and you’ll see how it works.

A system where both equations are already solved for one variable:

y = x + 7   and   y = 2x + 1

Since both equations are equal to y, they’re equal to each other!
 (transitive property of equality)

x + 7 = y = 2x + 1

So just get rid of the “middle man” y and get:

x + 7 = 2x + 1

Solve for x:

x = 6

Now plug into whichever original equation seems easier to you to find the y coordinate:

y = x + 7

y = 6 + 7 = 13

The intersection is (6, 13)

What if we plug in this point to the other equation? It should work because both equations have (6, 13) as a solution.

y = 2x + 1

13 = 2(6) + 1

13 = 13

A system where one equation is solved for one variable:

y = 2x

5x + 3y = 22

2x is the same value as y.

Since that is true, anywhere you see y, you may use 2x instead.

SUBSTITUTING INTO THE GAME FOR Y IS 2X:

5x + 3(2x) = 22

5x + 6x = 22

11x = 22

x = 2

Now plug into the other equation to find y:

y = 2x = 2(2) = 4

The solution (intersection) is (2, 4)

CHECK:

Plug in (2, 4) into the other equation:

5(2) + 3(4) = 10 + 12 = 22

WHAT IF YOU HAVE 2 EQUATIONS AND NEITHER ONE IS SOLVE FOR A SINGLE VARIABLE?

You can just solve for one variable in whichever equation is easier.

Example:

x – 3y = 15  and 4x -2y = 20

You would need to pick which variable (x or y) would be easier to solve for in which equation.

Generally, look for a variable with no coefficient (really a coefficient of 1).

So for the above system, I’d pick to solve for x in the first equation:

x = 3y + 15

So wherever you see “x” in the other equation, substitute in (3y + 15)

4(3y + 15) – 2y = 20

12y + 60 -2y = 20

10y + 60 = 20

10y = -40

y = -4

Substitution is often used to solve WORD PROBLEMS.

Example:

The perimeter of a rectangle is 40 in.

The length is 10 less than twice its width.

Find the dimensions of the rectangle.

2l + 2w = 40

l = 2w – 10

Substitute (2w – 10) for l:

2(2w – 10) + 2w = 40

4w – 20 + 2w = 40

6w – 20 = 40

6w = 60

w = 10 in.

l = 2w – 10 = 2(10) – 10 = 20 – 10 = 10 in.

IT’S A SQUARE! ;)

Algebra Honors ( Periods 4 & 7)

Graphing Systems of Linear Equations 6-1

We just learned how to graph a single linear equation (a line).

We graphed in standard form using intercepts, in slope-intercept form by graphing the y-intercept and then counting the slope to another point, and in point-slope form by graphing a random point and then counting the slope to another point.

If two or more lines are graphed on the same coordinate plane, one of three things will happen:

1. They will intersect exactly once…we say they are CONSISTENT and INDEPENDENT

2. They will never intersect; they’re parallel or have the same slope (m) but different y intercepts (b)…we say they are INCONSISTENT

3. They will intersect infinitely, everywhere; they’re collinear or the same line. They have the same slope AND the same y intercept…we say they are CONSISTENT and DEPENDENT

We’ll graph two lines today and find the intersection point graphing by hand and on during this week on the graphing calculator.

The intersection coordinate is called the “SOLUTION OF THE SYSTEM.”

that is, x marks the spot ;)

You can check your coordinates by plugging them into both equations and make sure they work.

(or you can check on the graphing calculator by using the intersection app or the TABLE function to see that for the same x, they both have the same y value).

Algebra ( Period 1)

Graphing Systems of Linear Equations 6-1

We just learned how to graph a single linear equation (a line).

We graphed in standard form using intercepts, in slope-intercept form by graphing the y-intercept and then counting the slope to another point, and in point-slope form by graphing a random point and then counting the slope to another point.

If two or more lines are graphed on the same coordinate plane, one of three things will happen:

1. They will intersect exactly once…we say they are CONSISTENT and INDEPENDENT

2. They will never intersect; they’re parallel or have the same slope (m) but different y intercepts (b)…we say they are INCONSISTENT

3. They will intersect infinitely, everywhere; they’re collinear or the same line. They have the same slope AND the same y intercept…we say they are CONSISTENT and DEPENDENT

We’ll graph two lines today and find the intersection point graphing by hand and on during this week on the graphing calculator.

The intersection coordinate is called the “SOLUTION OF THE SYSTEM.”

that is, x marks the spot ;)

You can check your coordinates by plugging them into both equations and make sure they work.

(or you can check on the graphing calculator by using the intersection app or the TABLE function to see that for the same x, they both have the same y value).