Algebra ( Periods 1 & 4)

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

 An overview of all methods

Solving Systems of Equations by Graphing
  we have 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!!!

Algebra Honors ( Period 6)

Using Substitution to Solve a System  6-2
Using Addition to Solve a System  6-3
Using Addition w/ Multiplication to Solve a System 6-4

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! ;) 

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

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

Math 6A ( Period 2 & 5)

Ratio Tables 5-2

Essential Question:  How can you find two ratios that describe the same relationship?

We looked at a recipe or mixture of lemonade and iced tea.

the one from the book called for 1 cup of lemonade for every 3 cups of iced tea

We created a table and thought of various combinations that still kept that same relationship

C. of Lemonade	1	2	3	5	8	10
C. of Iced tea	3	6	9	15	24	30
Total Cups	4	8	12	20	32	40

A mixture contains 13 cups of lemonade, how could we determine how many cups of iced tea would be required? How could we use the given table to find that answer? 
1) we saw that the relationship was 1:3 so we could multiply 3 (cups of iced tea) by 13 to get 39 cups of iced tea.
2) we could use the existing information in the table. We know that 5 + 8 = 13 so if we just add the information for iced tea in those columns ( 15 + 24 =39) we would get 39 cups of iced tea as well.

Two ratios that describe the same relationship are equivalent ratios. You can find equivalent ratios by:

·        adding or subtracting quantities in equivalent ratios

·        multiplying or dividing each quantity in a ratio by the same number.

You can find and organize ratios in a ratio table.

Pens	1	2
Pencils	3		9

using repeated addition:

Pens	1	2	3
Pencils	3	6	9

The equivalent ratios are 1:3; 2:6 and 3:9

Dogs	4		24
Cats	6	12

You can use multiplication to find the missing values

Dogs	4	8	24
Cats	6	12	36

The equivalent ratios are 4:6; 8:12 and 24:36

We discussed how they are all equivalent to 2:3 as well

Using a Ratio Table in a word problem

The nutrition fact labeled on a box of crackers shows that there are 240 milligrams of sodium in every 36 crackers.

You eat 15 crackers. How much sodium do you consume?

The ratio of sodium to crackers is 240 to 36. Create a ratio table to find equivalent ratios with 15 crackers.

Sodium (mg)	240	120	20	100
Crackers	36	18	3	15

The ratio 100 to 15 is equivalent to 240 to 36.

So, you consumed 100 milligrams of sodium.

You eat 21 crackers. How much sodium do you consume?

Notice, you can add the two middle columns in the table above to find the solution to that question.

Since 18 + 3 = 21 120 + 20 = 140   140 milligrams of sodium in in 21 crackers.
You could also use the ratio 20:3 and multiply both by 7 and you will arrive at 140:21 or 140 milligrams of sodium

Math 6A (Period 2 & 5)

Ratios 5.1

A ratio is a comparison of two quantities. Ratios can be part -to -part, part – to -whole, or whole -to-part comparisons

The ratio of a to b can be written in three ways           

 a:b,                 a/b ,           and                           a to b.
Each of these is read “ a to b”

 Examples:

 2 red crayons  to 6 blue crayons
1 red crayon  for every 3 blue crayons
3 blue crayons for each red crayon
3 blue crayons per 1 red crayon
3 blue crayons out of every 4 crayons
2 red crayons out of 8 crayons

A tape diagram is a diagram that looks like a segment of tape. It shows the relationship between two quantities.

The ratio of your monthly allowance to your friend’s monthly allowance is 5:3 The monthly allowance totals $40. How much is each allowance?

To help visualize the problem, express the ration 5:3 using a tape diagram

You 

Your Friend

Because there are 8 parts, you know that one part represents 40÷8 = $5

5 parts represents 5 ∙5 = 25
3 parts represents 3 ∙5 = 15

Garlic bulbs- you separate 42 garlic bulbs into two groups: one for planting and one for cooking. You will plant 3 bulbs for every 4 bulbs that you will use for cooking. Each bulb has about 8 cloves. How many cloves will you plant?  To help visualize, express the ratio 3 for every 4 using tape diagram

Planting

Cooking

The seven parts represent 42 bulbs so each part represents 42÷ 7 = 6 bulbs

There are 3 ∙6 =18 bulbs for planting and 4 ∙6 = 24 bulbs for cooking.

The group of 18 bulbs has 18 ∙ 8 = 144 cloves. So you will plan 144 cloves of garlic.