mathcounts4ever

Friday, January 29, 2016

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

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

Thursday, January 28, 2016

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!

Tuesday, January 26, 2016

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

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

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)

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)

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).

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).

Friday, January 22, 2016

Algebra ((Period 1)

Chapter 6-1, 6-2, 6-3

An overview of all methods

Solving Systems of Equations by Graphing
Before we stopped and reviewed fraction operations last week ( I didn’t) we had just finished graphing linear functions with two variables in three different forms:

Slope-intercept: Graphing our home base on the y-axis, then counting slope (rise/run) to another point

Standard: Making an x-y table to find the intercepts and graphing those

Point-slope: Graphing the point in the formula (flip the signs!) and then counting slope to another point

Today, we’ll have a SYSTEM of linear functions and will find the solution (coordinate) to where they INTERSECT.

Systems are 2 or more linear functions (lines)

TO SOLVE MEANS TO FIND THE ONE COORDINATE THAT WORKS FOR BOTH FUNCTIONS

THERE ARE 3 POSSIBILITIES FOR THIS:

1. The lines intersect at ONE POINT. We call this a consistent system that is independent.

2. The lines are COLLINEAR, meaning that they intersect at INFINITELY MANY POINTS. We call this consistent and dependent.

3. The lines are PARALLEL, meaning that they NEVER INTERSECT and therefore there is NO POSSIBLE SOLUTIONS. We call these lines inconsistent.

There are 3 ways to find that point:
(Chapter 3-7) 1. Graph both equations: Where the 2 lines intersect is the solution
(Chapter 3-8) 2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation

(NOT GIVEN IN YOUR BOOK) 3. Addition method: Eliminate one of the variables by multiplying the equations by that magical number that will make one of the variables the ADDITIVE INVERSE of the other

I will be doing the following example solved all 3 ways today.

Find the solution to the following system…Find the COORDINATE where the lines intersect:

2x + y = -3 and 2x - y = -5

1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point....You should get (-2, 1)

We’ll also use the GRAPHING CALCULATORS to find this intersection!

What’s the drawback of this method? It takes time…Many times the intersection’s coordinate is not an integer.

Solving systems of linear equations algebraically (without graphing)

There are two methods: SUBSTITUTION (Chapter 3-8) and ADDITION (not in your book)

2. SUBSTITUTION: Isolate whatever variable seems easiest.
I will isolate y in the first equation: 2x + y = -3

y = -2x – 3 (after you subtract 2x from both sides)

Now plug in (-2x – 3) for y in the other equation:

2x - y = -5
2x – (-2x – 3) = -5

2x + 2x + 3 = -5

4x + 3 = -5

4x = -8

x = -2

Now plug in -2 for x in whichever equation looks easier to find y.

I think the first equation looks easier:

2x + y = -3

2(-2) + y = -3

-4 + y = -3

Y = 1

So the coordinate of the intersection is (-2, 1)

This is the same as we found when we graphed.

To really be sure you didn’t make a silly mistake, you should plug in the coordinate in the OTHER equation:

2x - y = -5
2(-2) – (1) = -5????

-4-1=-5 YES!

When would it be best to solve this way?

When one of the equations is already solved for one of the variables, but you can always isolate one of the variables yourself with equation balancing.

A lot of word problems are easier to solve with substitution.

3. ADDITION: Multiply each equation so that one variable will "drop out" (additive inverses….YAY!)
For the problem above, I will eliminate the y because the two y’s are already Additive Inverses, but I could eliminate the x if I wanted to!

This time you “stack” the equations:

2x + y = -3

+ 2x - y = -5
-----------------

4x = -8

x = -2

Plug into whichever equation is easiest to find y as we did in the Substitution method.

When is it best to use this method?

If no variable is already isolated.

I tend to use this method the most ;)

NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!