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).

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!!!

Algebra Honors ( Periods 4 & 7)

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!!!

Algebra Honors ( Periods 4 & 7)

Chapter 3 -5 Arithmetic Sequences as Linear Functions

An arithmetic sequence is an ordered list of numbers ( called terms) where there is a common difference ( d) between consecutive terms. the common difference can either be positive ( increasing)  or negative ( decreasing)

Because an arithmetic sequence has a constant difference, it is a linear function. There is a formula using this common difference to find the equation of any arithmetic sequence:

a_n = a₁ + ( n-1)d

Notice that it is saying that any term in the sequence, a_n,  can be found by adding 1 less than the number of terms of the common difference to the first term, a_1.   Why 1 less than the number of the terms you want?

Take the following arithmetic sequence:

…17, 21, 25,29, 33….

If it’s the 2^nd term, there is only 1 common difference of 4 between the 2 terms. If it is the 3^rd term, there would be 2 common differences of 4 between the 3  terms, etc.

n is always positive because it represents the number of terms and that can’t be negative.

Where  a_n  represents any term in the sequence and a₁ represents the first term in a sequence, n represents the number of the terms in a sequence and d represents the common difference between consecutive terms in a sequence.

This is called an EXPLICIT FORMULA You can explicitly find any term number in the sequence as long as you know the 1^st term and the common difference. For example if the the common difference is 4 and the first term is 1 and you are trying to find the 27^th term

a_n = a₁ + ( n-1)d

a₂₇ = 1 + ( 27-1)4

a₂₇ = 1 + ( 26)4

a₂₇ = 1 + ( 104

a₂₇ = 105

You can also find the next term in the sequence if you know the
RECURSIVE FORUMULA

This is a Function Rule that tells you what the relationship between consecutive terms is. For example if the common difference is 4 and your last term is 101 the next term is 105—without knowing any of the other preceding terms. The terms in the sequence are shown as a list with 3 periods (called an ellipsis) at the end showing that it continues infinitely.

For example, the arithmetic sequences for the above example was

1, 5,9, 13, …

Graphing the terms of an arithmetic sequence shows that it is a linear function

a_n = 1 + ( n-1)4

Simplify

a_n = 1 + ( n-1)4

a_n = 1 + 4n -4

a_n = 4n -3

Now just substitute y for a_n   and x for n

y = 4x – 3

Notice that d is now the slope and also notice that the domain of the sequence is the natural numbers (the counting numbers) because you can’t have a negative term number!

Graph 3 points using 1, 2, 3 for the first 3 terms

(1, 1) ( 2,5) (3, 9)

Algebra (Period 1)

Chapter 4-4 Parallel and Perpendicular Lines

Parallel Lines:

2 lines that are parallel to each other have the SAME SLOPE

Example:

y = 2x – 10

y = 2x + 4

Are parallel because they both have a slope of 2

BE CAREFUL:
to be parallel they must have the SAME SLOPE BUT DIFFERENCE y-INTERCEPTS

If they have the same m and the same b then they are COLLINEAR ( or they COINCIDE) they are the same line and NOT parallel

Perpendicular Lines

2 lines that are perpendicular to each other have SLOPES that are:

Opposite signs
and
Reciprocals

y = 2x – 10 is perpendicular to y = (-1/2)x + 4

This is another twist to our Mystery Line Puzzles…

If you know that the mystery line is parallel  or perpendicular to another given line they you know the mystery line’s slope!!

EXAMPLE 1:

 Your mystery line has a point of (2, -5) and is PARALLEL to the line y = 2x + 4

So you know that the mystery line’s slope
because it is the same as the given line  (m = 2)

Substitute the slope and the point given for the mystery line and solve for b OR use Point Slope Form and solve easily!
This is why I like Point Slope: y + 5 = 2(x – 2)

y + 5 = 2x – 4

y = 2x – 9

If I used Slope Intercept form or y = mx + b

-5 = 2(2) + b

-5 = 4 + b

b = -9

Plug everything back in

y = 2x – 9

Same results—but I think that it takes longer (students also tend to FORGET to put everything together)

EXAMPLE 2:

Your mystery line has a point of (2, -5) and is perpendicular to the line y = 2x + 4

So you know that the mystery line’s slope is the opposite sign and the reciprocal of the given line
(since the given line’s slope is 2, the mystery line’s slope must be –½ )

Substitute the slope and the point given on the mystery line and solve for b OR use Point Slope Form and solve easily!  

y + 5 = (-½)(x -2)

y + 5 =  -½x + 1

y =  -½x - 4

If I used Slope Intercept form or y = mx + b

-5 = (-½)(2) + b

-5 = -1 + b

-4 = b

Plug everything back

 y = -½x - 4 

Algebra Honors ( Periods 4 & 7)

Chapter 4-4 Parallel and Perpendicular Lines 

Parallel Lines:

2 lines that are parallel to each other have the SAME SLOPE

Example:

y = 2x – 10

y = 2x + 4

Those two lines are  parallel because they both have a slope of 2

BE CAREFUL:
To be parallel they must have the SAME SLOPE BUT DIFFERENCE y-INTERCEPTS

If they have the same m and the same b then they are COLLINEAR
(or they COINCIDE) they are the same line and NOT parallel

Perpendicular Lines

2 lines that are perpendicular to each other have SLOPES that are:

Opposite signs

and

Reciprocals

y = 2x – 10 is perpendicular to y = (-1/2)x + 4

This is another twist to our Mystery Line Puzzles…

If you know that the mystery line is parallel  or perpendicular to another given line they you know the mystery line’s slope!!

EXAMPLE 1:

Your mystery line has a point of ( 2, -5) and is PARALLEL to the line y = 2x + 4

So you know that the mystery line’s slope
because it is the same as the given line  (m = 2)

Substitute the slope and the point given for the mystery line and solve for b
OR use Point Slope Form and solve easily!
This is why I like Point Slope: y + 5 = 2(x – 2)

y + 5 = 2x – 4

y = 2x – 9

If I used Slope Intercept form or y = mx + b

-5 = 2(2) + b

-5 = 4 + b

b = -9

Plug everything back in

y = 2x – 9

Same results—but I think that it takes longer ( students also tend to FORGET to put everything together)

EXAMPLE 2:

Your mystery line has a point of ( 2, -5) and is perpendicular to the line y = 2x + 4

So you know that the mystery line’s slope is the opposite sign and the reciprocal of the given line
(since the given line’s slope is 2, the mystery line’s slope must be –½ )

Substitute the slope and the point given on the mystery line and solve for b OR use Point Slope Form and solve easily!  

y + 5 = (-½)(x -2)

y + 5 =  -½x + 1

y =  -½x - 4

If I used Slope Intercept form or y = mx + b

-5 = (-½)(2) + b

-5 = -1 + b

-4 = b

Plug everything back

 y = -½x - 4

Algebra (Period 1)

Chapter 4-3 Point Slope Form 

There is another way to solve for a line other than using slope-intercept form  Although most people use the slope- intercept form for all cases, the POINT-SLOPE Form  is actually easier—everything is built into the format!  You do not need to “put it altogether” at the end!

MY FAVORITE!!

It gives you exactly a point and the slope—just by looking at it!

You do not need to do anything BUT graph!!

Point- Slope Form of a line:

y-y₁ = m ( x- x₁)

Example: ( 3, 1) is a point on the line and m = 2

y – 1 = 2( x – 3)

What you have now is point-slope form of the line!

You can graph easily starting at (3, 1) and going up 2 and to the right 1

You can also simplify this and get the slope- intercept form of the line:

y - 1 = 2x – 6

y = 2x – 5

If you are trying to link the Slope-Intercept form to the Point-Slope form of the line:  The point slope version eliminates one step from using the slope intercept form

In the slope intercept form, you plug in the point and the slope and solve for b—
AND THEN rewrite the equation using the intercept that you found.

In point slope form, once you plug in the point and slope, you just simplify and the equation is already done!

IF you don’t have the slope, you will need to compute it with the formula—just like you did with Slope Intercept FORM

The biggest difference is that any point is plugged into this form, while  the Slope- Intercept Form focuses on the ONE specific point on the y axis 

Algebra Honors ( Periods 4 & 7)

Chapter 4-2 Writing Equations in Slope-Intercept Form 

We usually use the slope- intercept form of the line as our  ”template”

We know that y = mx + b so we can substitute that in what we know (what the problem gives us as information) and solve for whatever we are missing

It helps to memorize this little rhyme (Mrs Sobieraj made it up!)

Oh mystery line,
What could you be?
If I could just find you,
 y = mx + b
First I’ll find m,
Then I’ll find b
Then I’ll put it all together
And I will see:
y = mx + b

The rhyme has 3 steps and usually you will have 3 steps or questions to ask yourself:

1) Do I have the slope (m)? If not find it by using the slope formula or counting it if you have the graph—(carefully pick two sets of integer points)

2) Do I have the y- intercept (b) ? If not, find it by plugging in a point and the slope and solving for b or if you have the graph, just read it on the y axis.

3) Remember: Put it all together in ONE equation at the end!

There are FIVE general cases of mystery lines

First Case:
You are given the slope and the y intercept (that is the easiest case)
For example:  you are given m = 3/2 and b = - 7/5

Just plug in to the generic slope intercept equation

y = (3/2)x – 7/5

Second Case:
You have a graph of a line and need to determine the equation
Look at the graph and find 2 easy points to use to find the slope ( make sure they are integers) If the y intercept is not an integer—then follow the FOURTH CASE (below) completely!

Put the information together in y = mx + b form

Third Case:
You are given a point and the slope and need to find the intercept ( b)
Example: ( 3, 1)is a point on the line and m = 2

Plug in the point and the slope and find b

That is, start with y = mx + b
You have a point ( 3, 1) plug it in to that equation:

1 = (2)(3) + b
1 = 6 + b

-5 = b or
b = -5

Now put it altogether with the given slope of m = 2 and the y intercept ( b) which you just found

y = 2x – 5

Fourth Case:
You are given a point and the y intercept and need to find the slope > Let’s use the point ( 3, 1) again but this time you are given b = 2

Again you can use y = mx + b . This time, however you are solving for m ( the slope)
1 = 3m + 2
-1 = 3m

-1/3 = m
m = -1/3

Again, NOW put it all together with the given intercept and the slope you just found
 y = (-1/3)x + 2

Fifth Case:
You are given 2 points on a line and need to find the slope and the y intercept
Example: ( 1, 3) and ( -2, -3) are 2 points on the line

You first need to find the slope using the formula
m = change in y/ change in x

m = (-3 -3)/(-2-1)  or  (3--3)/(1--2)  which really is (3+3)/(1+2) or
6/3 = 2

Now plug the slope in with one ( you get to pick—it will work with either) of the points and find the intercept, b

3 = 2(1) + b

3 = 2+b

b = 1

Finally put it all together

y = 2x +1

Chapter 4-3 Point Slope Form 

There is another way to solve for a line other than using slope-intercept form  Although most people use the slope- intercept form for all cases, the POINT-SLOPE Form  is actually easier—everything is built into the format!  You do not need to “put it altogether” at the end!

MY FAVORITE!!

It gives you exactly a point and the slope—just by looking at it!

You do not need to do anything BUT graph!!

Point- Slope Form of a line:

y-y₁ = m ( x- x₁)

Example: ( 3, 1) is a point on the line and m = 2

y – 1 = 2( x – 3)

What you have now is point-slope form of the line!

You can graph easily starting at (3, 1) and going up 2 and to the right 1

You can also simplify this and get the slope- intercept form of the line:

y - 1 = 2x – 6

y = 2x – 5

If you are trying to link the Slope-Intercept form to the Point-Slope form of the line:  The point slope version eliminates one step from using the slope intercept form

In the slope intercept form, you plug in the point and the slope and solve for b—
AND THEN rewrite the equation using the intercept that you found.

In point slope form, once you plug in the point and slope, you just simplify and the equation is already done!

IF you don’t have the slope, you will need to compute it with the formula—just like you did with Slope Intercept FORM

The biggest difference is that any point is plugged into this form, while  the Slope- Intercept Form focuses on the ONE specific point on the y axis 

Algebra ( Period 1)

Chapter 4-2 Writing Equations in Slope-Intercept Form 

We usually use the slope- intercept form of the line as our  ”template”

We know that y = mx + b so we can substitute that in what we know (what the problem gives us as information) and solve for whatever we are missing

It helps to memorize this little rhyme (Mrs Sobieraj made it up!)

Oh mystery line,
What could you be?
If I could just find you,
 y = mx + b
First I’ll find m,
Then I’ll find b
Then I’ll put it all together
And I will see:
y = mx + b

The rhyme has 3 steps and usually you will have 3 steps or questions to ask yourself:

1) Do I have the slope (m)? If not find it by using the slope formula or counting it if you have the graph—(carefully pick two sets of integer points)

2) Do I have the y- intercept (b) ? If not, find it by plugging in a point and the slope and solving for b or if you have the graph, just read it on the y axis.

3) Remember: Put it all together in ONE equation at the end!

There are FIVE general cases of mystery lines

First Case:
You are given the slope and the y intercept (that is the easiest case)
For example:  you are given m = 3/2 and b = - 7/5

Just plug in to the generic slope intercept equation

y = (3/2)x – 7/5

Second Case:
You have a graph of a line and need to determine the equation
Look at the graph and find 2 easy points to use to find the slope ( make sure they are integers) If the y intercept is not an integer—then follow the FOURTH CASE (below) completely!

Put the information together in y = mx + b form

Third Case:
You are given a point and the slope and need to find the intercept ( b)
Example: ( 3, 1)is a point on the line and m = 2

Plug in the point and the slope and find b

That is, start with y = mx + b
You have a point ( 3, 1) plug it in to that equation:

1 = (2)(3) + b
1 = 6 + b

-5 = b or
b = -5

Now put it altogether with the given slope of m = 2 and the y intercept ( b) which you just found

y = 2x – 5

Fourth Case:
You are given a point and the y intercept and need to find the slope > Let’s use the point ( 3, 1) again but this time you are given b = 2

Again you can use y = mx + b . This time, however you are solving for m ( the slope)
1 = 3m + 2
-1 = 3m

-1/3 = m
m = -1/3

Again, NOW put it all together with the given intercept and the slope you just found
 y = (-1/3)x + 2

Fifth Case:
You are given 2 points on a line and need to find the slope and the y intercept
Example: ( 1, 3) and ( -2, -3) are 2 points on the line

You first need to find the slope using the formula
m = change in y/ change in x

m = (-3 -3)/(-2-1)  or  (3--3)/(1--2)  which really is (3+3)/(1+2) or
6/3 = 2

Now plug the slope in with one ( you get to pick—it will work with either) of the points and find the intercept, b

3 = 2(1) + b

3 = 2+b

b = 1

Finally put it all together

y = 2x +1

Math 6 High ( Period 5)

Chapter 4-2, 4-3 and 4-4: Multiply Fractions, Whole Numbers and Mixed Numbers

MULTIPLYING FRACTIONS:

First see if you can SIMPLIFY ... some call it CROSS CANCEL:

Does the numerator of one fraction go into the denominator of the other (or vice versa)?

If so take this OPPORTUNITY to make the math easier and simplify before multiplying.

Next, multiply the numerators and multiply the denominators.

Simplify: Make any improper fractions into mixed numbers.

When it’s a whole number, we usually put the number over 1, but you don’t have to.

What’s the danger of missing a cross cancelling opportunity?

You’ll just have to simplify more at the end!

Multiply Mixed Numbers

Restate mixed numbers into improper fractions

Then multiply the usual way as above

Algebra ( Period 1)

Chapter 4-1 Graphing Equations in Slope-Intercept Form

The most used form of a linear equation: Slope Intercept form

You must restate the equation to get it into the following format:

y = mx + b

where        m= slope     and            b = y-intercept ( where the line crosses the y axis)

The m and b are called the PARAMETERS of the equation.

Solve the equation for “y” means isolate the y on one side of the equal signs

Example:  -3y = -2x  - 6

If you tried graphing randomly, (setting up a small table or t chart) you would problem select the x points of 0, 1, 2.  Only when x = 0 will give you an integer value for y. All the other y values are fractions/ decimals à which makes it difficult to graph accurately!
 

The slope- intercept form provides the explanation for this.

RESTATE -3y = -2x – 6 INTO SLOPE-INTERCEPT FORM:

 Divide BOTH sides by -3:

y = (2/3)(x) + 2

Look at the coefficient for x?
What  x values will give you integer answers for y?

They need to be multiples of 3.

Now look at the graph of  y = (2/3)(x) + 2  
Notice that the +2 at the end is the y-intercept (without doing any work!) YAY!!!

Use the counting method for slope on your graph, you should have counted:
UP  2 and RIGHT 3. The slope therefore is 2/3
Look at the equation—it told you the slope was 2/3 without any work (YAY!!)

Graphing when the line is in Slope- Intercept Form

So if you have the slope- intercept form of the equation it is really easy to graph the line:

1) Graph the intercept on the y axis (That is the positive or negative constant at the end of your equation. Your HOME BASE)

2) Count the next point by using the slope of x coefficient as a fraction (so if you have an integer—place it “over” 1)

For the equation y = (2/3)(x) + 2

1) graph  a point at (0,2)

2) From (0,2) count up 2 and over to the right 3 to find the next coordinate ( 3, 4)

Remember slope is “ y over x” or “Rise over Run”
The numerator is the change in y  and the denominator is the change in x.

If it is positive you are counting up (positive) and to the right (positive) OR
you can count down (negative) and to the left (negative) because when multiplying 2 negatives become positive

If it is negative you are counting down (negative) and to the right (positive) OR
you can count up (positive) and to the left (negative) because then you would have a positive ∙ negative = negative

Example: m = -2/3 and b = -12
The line would be y = (-2/3)x - 12

Restate Standard Form to Slope- Intercept Form

Another example: 3x + 4y = 10 is the STANDARD FORM of a line 
(Notice; x and y are on the same side of the equal sign, x is positive, there are NO fractions, and the constant is alone)
This equation is NOT easy to graph in this form because your y intercept will not be an integer ( In fact it will not be that easy to graph in slope intercept either—but we will get to that later)

Restate into slope intercept

Solve for y
First subtract 3x from both sides:

4y = -3x + 10

Now divide both sides by 4:

y = (-3/4)x + 10/4   or

y = (-3/4)x + 5/2

The slope is the coefficient of the x so

m = -3/4 (so you are sliding down at a little less than a 45 degree angle. Remember a slope of 1 or -1 is  45 degrees)

The y intercept is the constant b = 5/2 ( so the line crosses the y axis at 2½)

Notice that the ‘b’ is a fraction.
When this happens the slope- intercept form MAY NOT be the best form to graph the line. You MUST start at 2½ on the y axis and count down 3 and to the right 4. That is actually HARD to get accurate!

Neither 3 nor 4 is a factor of the constant 10 ( 3x + 4y = 10) so the intercepts will also be fractions.

We need the x term to end up with ½ so that when we add that to the b (which is 5/2) we will get an integer.
So try letting x be 2 because that will cross cancel with the -3/4 slope

y = (-3/4)(2) + 5/2 =  -3/2 + 5/2 = 1
so we just found a coordinate that has just integers ( 2, 1)
Graph that point. Now count the slope from THAT POINT ( instead of the y intercept)

HORIZONTAL LINES
You can think of these lines in Slope- Intercept form as y = mx + b IF your REMEMBER that the slope of a horizontal line = 0 àthe equation is y = (0)x + b
We have already learned that the equation of any horizontal line is y = a constant
Except for the horizontal line y = 0 ( which is the x axis) horizontal lines have no x intercept and therefore no intercepts ( no roots, no solutions, no zeros)

Finding the equation of a line by looking at its graph is easy if you can read the y intercept!

Simply plug in the y intercept as  b and then count the rise over run as the slope!