Math 6A ( Periods 2 & 7)

 The Distributive Property 3.4  

Definition in words: To multiply a sum of difference by a number, multiply each number in the sum of difference by the number outside the parentheses. Then evaluate.

Distributive Property of Multiplication with Respect to Addition:

For any real  numbers a, b, and c,

a( b + c) = ab + ac

and

(b +c)a = ba +bc

Distributive Property of Multiplication with Respect to Subtraction:

For any real  numbers a, b, and c,

a( b - c) = ab - ac

and

(b - c)a = ba – bc

If you are given the problem 23 x 6 normally you would stack them and multiply but what if you consider

6(23) = 6(20+ 3) = 6 (20) + 6(3) 120

How about 8(26) = 8(20 + 6) = 8(20) + 8(6) = 160 + 48 = 208

But you could have changed it to 8 ( 25 + 1) = 8(25)  = 8(1) = 200 + 8 = 208

What should we do with 20(19)? well… 20(15 + 4) or 20(10 + 9) work but are they easy to multiply—especially in your head?

Why not try 20(20-1) = 20(20) – 20(1) = 400 – 20 = 380

20(39) = 20(40 – 1) = 20(40) – 20(1) = 800 – 20 = 780

25(39) has all sorts of possibilities

25(40 -1) = 25(40) – 25(1) = 1000 – 25 = 975

25( 36 + 3) = 25(36) + 25(3) = 900 + 75 = 975

We talked about the “trick” of multiplying any number by 25… if you forgot… come in and ask again. I would love to go over this. Hint: get out quarters and figure out how many dollars you have with all those quarters…

15(47) ?  again lots of possibilities to make the multiplication easier

15(50-3) = 15(50) – 15(3)

or change 15(47) = 47(15) { what property allows you to change the order?}

Then 47(10 + 5) = 47(10) + 47(5)

How do you multiply by 5 quickly?  Multiply by 10 and take half. Why does that work?

Now the book gives the following example:

3(7 + 2) = 3(7) + 3(2) = 21 + 6 = 27

My question, why would we even want to do that? It does not make it easier—in fact it is just too many steps. So you need to start to realize WHEN the DP (Distributive Property) helps you to make the multiplication easier!

8(53) could be an example when it makes sense because 8(50 + 3) = 8(50) + 8(3) and both of those are easy to multiply—even in your head.. and then add them! 424

What about mixed numbers? We were taught that if you had

 

You needed to change the mixed number to an improper fraction

But watch the Distributive Property in action:

We then reviewed some of the problems similar to our Apples & Bananas WS

4(n + 5) = 4n +4(5) = 4n + 20

9( 6 + x + 2) = 9(6) + 9x + 9(2) = 54 + 9x + 18

but then we combined like terms to get 9x + 72

Why not combine like terms inside the HUGS first?

9( 6 + x + 2) = 9(x + 8) = 9x + 72

Combining LIKE TERMS

Like terms are terms that have the same variable(s) raised to the exact same sxponent(s)

Constants are like terms

5x+ 19 + 2x + 2 becomes

7x + 21

Jose is x years old. His brother is 2 years older than Jose. Their aunt, Maria is three times as old as Felipe. Write and simplify an expression that represents Maria’s age

Jose x

Felipe x + 2

Maria 3( x + 2)

3(x + 2) = 3x + 6

Art Museum question:

	Museum	Exhibit
Child (under 5)	Free	Free
Student	$8	$x
Regular	$12	$4
Senior	$10	$3

A class of 30 students visit an art museum and a special exhibit. Use the DP to write and simplify an expression for the cost.

each student would be 8 + x so for 30 students you would have

30(8 + x) = 240 + 30x

We estimate reasonable values for x—and had a lively discussion about why we picked the price for the student’s exhibit fee.

Then we evaluated for the price of $2

30(8 + 2) = 30(10) = 300

and we also substituted in for 240 + 30x  = 240 + 30(2) = 260 + 60 = 300

We found that the values were the same $300

Extension:

Factoring Expressions:

When you factor an expression, you can factor out any common factor!

Factor 20 -12

Find the GCF of 20 and 12  … the book says by listing the factors… we use inverted division or the box method

12 = 2²∙3

20 = 2²∙5

So the GCF(20,12) = 4

Write each term of the expression as the product of the GCF and the remaining factor(s)

20 – 12 = 4(5) – 4(3)

=4(5 -3)

14x -98

What is the GCF?  Look only at the numbers this time

14 = 2∙7

98 = 2∙7∙7

The GCF( 14, 98) = 2∙7 = 14

so 14x -98 = 14 (x) – 14(7)

14(x -7)

Algebra Honors ( Period 4)

Inequalities Involving Absolute Value 5-5

The inequality │x│<  means that the distance between x and 0 is less than 3

Graph:

so x > -3 and x < 3

The set builder notation or solution set is {x │ -3 < x < 3}

When solving absolute value inequalities there are two cases to consider:

Case 1 The expression inside the absolute value symbols is nonnegative

Case 2  The expression inside the absolute value symbol is negative

The solution is the intersection of these two cases.

When the absolute value inequality is less than… I think of “less thAND” … it is the intersection of two parts. It is the “YO”  we talked about in class

│m + 2 │ < 11

Rewrite  │m + 2 │ < 11       for both the above cases

m + 2 < 11                   and            -(m+2)  < 11

m  < ; 9                   this one  becomes m + 2 >; -11

m < ; 9     and     m >  -13

or think m + 2 < 11   AND  -11 < m + 2

The solution set is { m │ -13 < m <; 9}

│y -1│ <  -2    WAIT… this can NEVER be TRUE. There is no solution The solution set is the empty set.  {  }  or 

When the absolute value inequality is a greater than… I think of  “greatOR”  it is the union
of two parts..  It is the “DORKY DANCER we talked about in class… going one way and then the other.
│x│ > 3 means that the distance between x and 0 is greater than 3…

Graph:

so x <  -3 OR  x >; 3 The solution set is { x │ x < -3 or x >  3}

We must consider two cases always
Case 1 The expression inside the absolute value symbols is nonnegative
Case 2 The expression inside the absolute value symbols is negative

Solve      │3n + 6│≥ 12

Case 1
3n + 6 is non negative

3n + 6 ≥ 12              n ≥ 2        That was easy

Case 2
3n + 6 is  negative

The book sets it up  -(3n + 6) ≥ 12

That would mean    3n + 6 ≤ -12

n  ≤ -6     so

 n  ≥2   OR   n  ≤ -6

The solutions set is { n │ n  ≥2   OR   n  ≤ -6}

Graph   

Algebra ( Period 5)

Solving Compound  Inequalities 5-4

Inequalities containing AND….  When considered together two inequalities such as

h  ≥  52  AND  h  ≤  72 form a compound inequality.

A compound inequality contains AND is only true if BOTH inequalities are true.

Its graph is where the graphs of the two inequalities overlap. This is called the intersection of the two graphs.

For instance,

x ≥ 3 is graphed

x < 7 is graphed

x ≥ 3  and  x < 7
3 ≤ x < 7

The statement   3 ≤ x < 7 can be read as “x is greater than or equal to 3 and less than 7”  or “x is between 3 and 7 including 3” or “3 is less than or equal to x which is less than 7.”

Solve  -2  ≤ x – 3 < 4 Then graph the solution set.

First express   -2  ≤ x – 3 < 4  using AND

-2 ≤ x -3      AND        x -3 < 4   Write the inequalities separately

Solve both of them!

1 ≤ x   and x < 7

so

the solution set is

{ x│1≤ x < 7}

Now graph the solution set

Inequalities containing OR….  A compound inequality containing or is TRUE if AT LEAST one of the inequalities is TRUE. Its graph is the UNION of the graph of the two inequalities 

The human ear can only detect sounds between the frequencies 20 Hertz and 20,000 Hertz. Write and graph a compound inequalities that describes the frequency of sounds humans cannot hear.

Let f = the frequency

f ≤ 20  OR f ≥ 20,000

Graph:

Notice that the graphs do NOT intersect. Humans cannot hear sounds at a frequency less than 20 Hertz or greater than 20,000 Hertz. The compound inequality is  { f│ f ≤ 20 or   f ≥ 20,000}

Intersections and Unions:

The graphs of compound inequalities containing AND will be an intersection. The graphs of compound inequalities containing OR will be a union.

The Triangle Inequality Theorem states that the sum of the measures of any two sides of  a triangle is greater that the measure of the third side!

Write and solve three inequalities to express the relationships among the measure of the sides of the triangle show here

           

x + 9 > 4      x > -5 (wait a minute—it can’t be negative—or even 0)

x + 4 > 9    x > 5

4 + 9 > x  so   13 > x which means x < 13

Since x > 5 AND x < 13 the solution set includes the whole numbers… 6, 7, 8, 9, 10, 11, 12

The compound inequality is   5< x < 13

Algebra ( Period 5)

Inequalities Involving Absolute Value 5-5

The inequality │x│<  means that the distance between x and 0 is less than 3

Graph:

so x > -3 and x < 3

The set builder notation or solution set is {x │ -3 < x < 3}

When solving absolute value inequalities there are two cases to consider:

Case 1 The expression inside the absolute value symbols is nonnegative

Case 2  The expression inside the absolute value symbol is negative

The solution is the intersection of these two cases.

When the absolute value inequality is less than… I think of “less thAND” … it is the intersection of two parts. It is the “YO”  we talked about in class

│m + 2 │ < 11

Rewrite  │m + 2 │ < 11       for both the above cases

m + 2 < 11                   and            -(m+2)  < 11

m  < ; 9                   this one  becomes m + 2 >; -11

m < ; 9     and     m >  -13

or think m + 2 < 11   AND  -11 < m + 2

The solution set is { m │ -13 < m <; 9}

│y -1│ <  -2    WAIT… this can NEVER be TRUE. There is no solution The solution set is the empty set.  {  }  or

When the absolute value inequality is a greater than… I think of  “greatOR”  it is the union
of two parts..  It is the “DORKY DANCER we talked about in class… going one way and then the other.
│x│ > 3 means that the distance between x and 0 is greater than 3…

Graph:

so x <  -3 OR  x >; 3 The solution set is { x │ x < -3 or x >  3}

We must consider two cases always
Case 1 The expression inside the absolute value symbols is nonnegative
Case 2 The expression inside the absolute value symbols is negative

Solve      │3n + 6│≥ 12

Case 1
3n + 6 is non negative

3n + 6 ≥ 12              n ≥ 2        That was easy

Case 2
3n + 6 is  negative

The book sets it up  -(3n + 6) ≥ 12

That would mean    3n + 6 ≤ -12

n  ≤ -6     so

 n  ≥2   OR   n  ≤ -6

The solutions set is { n │ n  ≥2   OR   n  ≤ -6}

Graph   

Algebra Honors (Period 4)

Solving Compound  Inequalities 5-4

Solving  Compound  Inequalities 5-4

Inequalities containing AND….  When considered together two inequalities such as

h  ≥  52  AND  h  ≤  72 form a compound inequality.

A compound inequality contains AND is only true if BOTH inequalities are true.

Its graph is where the graphs of the two inequalities overlap. This is called the intersection of the two graphs.

For instance,

x ≥ 3 is graphed

x < 7 is graphed

x ≥ 3  and  x < 7
3 ≤ x < 7

The statement   3 ≤ x < 7 can be read as “x is greater than or equal to 3 and less than 7”  or “x is between 3 and 7 including 3” or “3 is less than or equal to x which is less than 7.”

Solve  -2  ≤ x – 3 < 4 Then graph the solution set.

First express   -2  ≤ x – 3 < 4  using AND

-2 ≤ x -3      AND        x -3 < 4   Write the inequalities separately

Solve both of them!

1 ≤ x   and x < 7

so

the solution set is

{ x│1≤ x < 7}

Now graph the solution set

Inequalities containing OR….  A compound inequality containing or is TRUE if AT LEAST one of the inequalities is TRUE. Its graph is the UNION of the graph of the two inequalities 

The human ear can only detect sounds between the frequencies 20 Hertz and 20,000 Hertz. Write and graph a compound inequalities that describes the frequency of sounds humans cannot hear.

Let f = the frequency

f ≤ 20  OR f ≥ 20,000

Graph:

Notice that the graphs do NOT intersect. Humans cannot hear sounds at a frequency less than 20 Hertz or greater than 20,000 Hertz. The compound inequality is  { f│ f ≤ 20 or   f ≥ 20,000}

Intersections and Unions:

The graphs of compound inequalities containing AND will be an intersection. The graphs of compound inequalities containing OR will be a union.

The Triangle Inequality Theorem states that the sum of the measures of any two sides of  a triangle is greater that the measure of the third side!

Write and solve three inequalities to express the relationships among the measure of the sides of the triangle show here

           

x + 9 > 4      x > -5 (wait a minute—it can’t be negative—or even 0)

x + 4 > 9    x > 5

4 + 9 > x  so   13 > x which means x < 13

Since x > 5 AND x < 13 the solution set includes the whole numbers… 6, 7, 8, 9, 10, 11, 12

The compound inequality is   5< x < 13

Algebra ( Period 5)

Solving Multi- Step Inequalities 5-3

Again you will use your equation skills

1.      Do Distributive Property first (if necessary) do it carefully

2.      Combine like terms on each side of the “WALL”

3.      “JUMP” the variables to one side of the wall—that is get all the variables on one side of the inequality by using the Additive Inverse Property (add or subtract using the opposite sign of the variable term, constants second by adding or subtracting

4.      Do it just like a 2 step or 1 step equation but remember;  FLIP THE SYMBOL if you multiply or divide by a NEGATIVE to balance

5.      If you end up with the variable on the right side, SWITCH the SIDES and the SYMBOL

REMINDERS:

Set builder notation

Get familiar with the following notation

{x│ x≥ 5} is read “ x SUCH THAT c is greater than or equal to 5”

Checking your solutions is an important set. Many students skip this step! Checking the solutions is especially important with inequalities because the direction of the inequality sign is often changed when writing solutions in set builder notation.

Algebra Honors (Period 4)

Solving Multi- Step Inequalities 5-3

Again you will use your equation skills

1.      Do Distributive Property first (if necessary) do it carefully

2.      Combine like terms on each side of the “WALL”

3.      “JUMP” the variables to one side of the wall—that is get all the variables on one side of the inequality by using the Additive Inverse Property (add or subtract using the opposite sign of the variable term, constants second by adding or subtracting

4.      Do it just like a 2 step or 1 step equation but remember;  FLIP THE SYMBOL if you multiply or divide by a NEGATIVE to balance

5.      If you end up with the variable on the right side, SWITCH the SIDES and the SYMBOL

REMINDERS:

Set builder notation

Get familiar with the following notation

{x│ x≥ 5} is read “ x SUCH THAT c is greater than or equal to 5”

Checking your solutions is an important set. Many students skip this step! Checking the solutions is especially important with inequalities because the direction of the inequality sign is often changed when writing solutions in set builder notation.