Algebra ( Period 5)

Using Addition With Multiplication to Solve a System 6-4

This is still  second Algebraic method to solve a system but to get the additive inverses of one variable—you will need to multiply ONE or BOTH equations by a factor.

EXAMPLE 1

Multiplying just ONE equation

5x + 6y = -8
2x + 3y = -5

Multiply the bottom by -2 to eliminate y

5x + 6y = -8
-4x -6y = 10

x = 2

EXAMPLE 2

Multiply BOTH Equations

4x + 2y = 8
3x + 3y = 9

To eliminate x, you would need to multiply the top by 3 and the bottom by -4
so you would get 12x and – 12x
OR

Multiply the top by 3 and the bottom by -2 so that you will get 6y and -6y

It is your choice…
 [and you could actual divide the second equation by 3 to start out with
x + y = 3  Making it even easier… but if you did not see that—use the other two…]

I think keeping the numbers as SMALL as possible is usually easier so I will eliminate y

3(4x + 2y) =  3(8)
-2(3x + 3y) = -2(9)

12x + 6y = 24
-6x -6y = -18

6x = 6
x= 1

Algebra ( Period 5)

Using Addition to Solve a System 6-3

The second Algebraic method to solve a system is known as ELIMINATION
You will be eliminating one variable by using the ADDITIVE INVERSE of it in the other equation

4x + 6y = 32
3x -6y = 3

7x + 0 = 35
7x = 35
x = 5

Plug into EITHER equation to find y
4(5) + 6y = 32                   20 +   6y = 32               
6y = 12                      y = 2
So the solution is ( 5, 2)

Sometimes you almost have additive inverses, but you need to multiply ONE EQUATION by -1 ...FIRST

5x + 2y = 6
9x + 2y = 22

Multiply EITHER the Top or the Bottom by -1 (Your choice)

5x + 2y = 6
-9x -2y = -22

-4x + 0 = -16                  

-4x = -16
x = 4

Plug into either ORIGINAL equation

5(4) + 2y = 6   so 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) = 2   

36 – 14 = 22  YES!!!

Algebra (Period 5)

Using Substitution to Solve a System 6-2

There are two Algebraic ways to find the intersection of 2 (or more) linear equations:

·        1,   .Substitution

·       2.     Addition ( also called Elimination)

Substitution
This method works especially well if both equations are solved for the SAME variable (either x OR y)  OR
ONE  equation is solved for a SINGLE variable ( x or y)

You plug one equation into the other—meaning you will substitute it in. If you have ever been on the bench in a game think of how you hope you will be substituted into the game for another play so you can play—or if you are the understudy in a play – or if you can substitute one book for another and get the same number of AR points!

EXAMPLE 1
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 are EQUAL to each other
(Remember: Transitive property of equality)         
 x + 7 = y = 2x + 1
just get rid of the “middle man” and get

x + 7 = 2x + 1              
 solve for x
x= 6

Now plug into whichever ORIGINAL equation seems easier to find the y coordinate
y = x + 7                           
y = (6) + 7 = 13
The intersection is ( 6, 13)
Make sure to give the solution as an ordered pair  (Ordered, Pair)

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                      y = 2(6) = 1
y = 12 + 1                       y = 13

EXAMPLE 2
A system where one equation is solved for one of the variables
y = 2x      AND          5x + 3y =22
2x is the same value as y
Since that is true, anywhere you see “y” you my 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
IT WORKS!!

EXAMPLE 3

What if you have 2 equations and neither one is solved for a single variable?
You can just solved for one of the variables in whichever equation is easier

x – 3y = 15  and 4x – 2y = 20

You would need to pick which variable ( x or y) would be easier to solve in one of the equations.
Generally look for a variable with no coefficient ( or really it has a coefficient of 1)
So for the above system, pick the 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 = -40                                         
y = -4

Substitute is often used to solve word problems.
The perimeter of a rectangle is 40 in The length is 10 less than twice its width. Find the dimension f the rectangle.

let l = the length of the rectangle           Let w = the width of the rectangle

 2l + 2w = 40              l = 2w -10

substitute ( 2w– 10) for l           2( 2w– 10) + 2w = 40

4w -20 + 2w = 40                      6w = 60
w = 10                                       10 inches
Since the width is 10inches substitute that in

l = 2(10) – 10 = 10
The length is also 10 inches.
It is a SQUARE

Algebra ( Period 5)

Graphing Systems of Linear Equations 6-1

If 2 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 are parallel or have the same slope (m) but different y intercepts (b)—we say they are INCONSISTENT

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

We graphed lines and found the intersection point graphing by hand! (… and tried it on the graphing calculators)

The intersection coordinate is called the ‘SOLUTION OF THE SYSTEM’

You can check your coordinates by plugging them into both equations and make sure they work—or using the TABLE function on the calculator and see that for the same x, they both have the same y value

Algebra Period 5)

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 (Period 4)

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 5)

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 (Period 4)

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 5)

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 = 2

Finally put it all together

y = 2x +1

Algebra Honors (Period 4)

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 = 2

Finally put it all together

y = 2x +1

Algebra ( Period 5)

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!

Algebra Honors ( Period 4)

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!

Algebra ( Period 5)

Chapter 3-6 Proportional and Nonproportional Relationships

This is just real world review of concepts we’ve already covered… comparing and contrasting the two types of linear relationships

SAME: both are linear—meaning they graph as lines
both are diagonal
both have a constant rate of change or slope that can be found by finding the rise/run or the difference of the y’s over the difference of the x’s

DIFFERENT:
proportional relationships go through the origin (0,0) and nonproportional do NOT

Nonproportional have a y-intercept other than 0

Proportional relationships: YOU can take any point and divide the y/x and it will equal the same value as diving any other y/x. This value  is the slope—which is the constant rate of change  VS Nonproportional relationships when you divide the y/x of a point it will NOT equal the y/x of another point. This value is NOT the slope and is NOT the CONSTANT RATE OF CHANGE

TO find the equation for anonproportional relationship
This isn’t as easy as f(x) = kx because it does not go through (0,0)

You will need to find the y intercept (0,y)

Say you find the rate of change or slope is 3 for the following 2 points

(2, 12) and (4, 18)

You can graph these two points and count the slope down to the y intercept

You can find a missing number that will make the equation work

y = 3x + ?  will make ( 2, 12) work in the equation

Plug in( 2, 12) to find b

12 = 3(2) + b

12 = 6 +b

 b = 6

You can try it with other points as well…

 It still works b = 6

So the non-proportional equation is y = 3x + 6

6 is the y intercept on the graph

Math 8 ( Period 1)

Chapter 3-5 Graphing Using Intercepts

Another way to graph a line is to find the two intercepts where the line crosses the x-axis and the y-axis and connect them.

If you have an equation, you can find both by setting the opposite variable to 0

To find the x intercept set y = 0 and you will find a coordinate in the form of ( #, 0)

This will be on the x axis because y is 0

To find the y intercept, Set x = 0 and you will find a coordinate in the form of ( 0, #)

This will be on the y axis because x is 0.

You know that Slope- Intecept isolates the y on one side of the equation

Another form of a linear equation is Standard Form

Ax + By= C

A, B< and C are integers—meaning there are NO DECIMALS or FRACTIONS in the equation.

The x and y terms are on the same side and the other side is a CONSTANT

The x term is always positive so A ≥ 0

You can restate Slope Intercept to Standard Form by using equation balancing.

Restate y = 2/3(x) – 10 to Standard Form

Clear the fractions by multiplying each term by 3

3y = 2x – 30

move 2x to the other side—so we need to subtract 2x from both sides

3y – 2x = -10 
or
-2x +3y = -10

Multiply each term by -1 so that the x term is POSITIVE

2x – 3y = 10