Week of January 27, 2020

CRM
Solving Inequalities

Sometimes we need to solve Inequalities like these:

Symbol	Words	Example
>	greater than	x + 3 > 2
<	less than	7x < 28
≥	greater than or equal to	5 ≥ x − 1
≤	less than or equal to	2y + 1 ≤ 7

Solving

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

Something like:		x < 5
or:		y ≥ 11

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

Simplify:

x > 10

Solved!

How to Solve

Solving inequalities is very like solving equations ... we do most of the same things ...

... but we must also pay attention to the direction of the inequality.

Direction: Which way the arrow "points"

Some things can change the direction!

< becomes >

> becomes <

≤ becomes ≥

≥ becomes ≤

Safe Things To Do

These things do not affect the direction of the inequality:

• Add (or subtract) a number from both sides
• Multiply (or divide) both sides by a positive number
• Simplify a side

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

3x < 10

But these things do change the direction of the inequality ("<" becomes ">" for example):

Multiply (or divide) both sides by a negative number
Swapping left and right hand sides

Example: 2y+7 < 12

When we swap the left and right hand sides, we must also change the direction of the inequality:

12 > 2y+7

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra), like this:

Solve: x + 3 < 7

If we subtract 3 from both sides, we get:

x + 3 − 3 < 7 − 3    

x < 4

And that is our solution: x < 4
In other words, x can be any value less than 4.

What did we do?

We went from this: To this:				x+3 < 7 x < 4

And that works well for adding and subtracting, because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5
If we subtract 5 from both sides, we get:

12 − 5 < x + 5 − 5    

7 < x

That is a solution!
But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

x > 7

Do you see how the inequality sign still "points at" the smaller value (7) ?
And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number:

Solve: 3y < 15

If we divide both sides by 3 we get:

3y/3 < 15/3

y < 5

And that is our solution: y < 5

Negative Values

When we multiply or divide by a negative number we must reverse the inequality.

Why?

Well, just look at the number line!

For example, from 3 to 7 is an increase,
but from −3 to −7 is a decrease.

−7 < −3	7 > 3

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Solve: −2y < −8

Let us divide both sides by −2 ... and reverse the inequality!

−2y < −8

−2y/−2 > −8/−2

y > 4

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Solve: bx < 3b

It seems easy just to divide both sides by b, which gives us:

x < 3

... but wait ... if b is negative we need to reverse the inequality like this:

x > 3

But we don't know if b is positive or negative, so we can't answer this one!

To help you understand, imagine replacing b with 1 or −1 in the example of bx < 3b:

• if b is 1, then the answer is x < 3
• but if b is −1, then we are solving −x < −3, and the answer is x > 3

The answer could be x < 3 or x > 3 and we can't choose because we don't know b.

So:

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

Solve: x−32 < −5

First, let us clear out the "/2" by multiplying both sides by 2.
Because we are multiplying by a positive number, the inequalities will not change.

x−32 ×2 < −5 ×2  

x−3 < −10

Now add 3 to both sides:

x−3 + 3 < −10 + 3    

x < −7

And that is our solution: x < −7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Solve:

−2 < 6−2x3 < 4

First, let us clear out the "/3" by multiplying each part by 3.
Because we are multiplying by a positive number, the inequalities will not change:

−6 < 6−2x < 12

Now subtract 6 from each part:

−12 < −2x < 6

Now multiply each part by −(1/2).
Because we are multiplying by a negative number, the inequalities change direction.

6 > x >−3

And that is the solution!
But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

−3 < x < 6

Source:mathisfun.com

Algebra 1

Solving Literal Equations

t first glance, these exercises appear to be much worse than our usual solving exercises, but they really aren't that bad. We pretty much do what we've done all along for solving linear equations and other sorts of equation; the only substantial difference is that, due to all the variables, we won't be able to simplify our work as we go along, nor as much as we're used to at the end. Here's how solving literal equations works:

• Solve A = bh for b

This is the formula for the area A of a rectangle with base b and height h. They're asking me to solve this formula for the base b.

If they'd asked me to solve 3 = 2b for b, I'd have divided both sides by 2 in order to isolate (that is, in order to get by itself, or solve for) the variable b. I'd end up with the variable b being equal to a fractional number.

In this case, I won't be able to get a simple numerical value for my answer, but I can proceed in the same way, using the same step for the same reason (namely, that it gets b by itself). So, following the same reasoning for solving this literal equation as I would have for the similar one-variable linear equation, I divide through by the "h":

$\small{ A = bh }$

$\small{ \dfrac{A}{h} = \dfrac{bh}{h} }$

$\small{ \bm{\color{purple}{ \dfrac{A}{h} = b }}}$

The only difference between solving the literal equation above and solving the linear equations you first learned about is that I divided through by a variable instead of a number (and then I couldn't simplify, because the fraction was in letters rather than in numbers). Because we can't simplify as we go (nor, probably, at the end), it can be very important not to try to do too much in one's head. Write everything out completely; this will help you end up with the correct answers.

---

• Solve d = rt for r

This equation is the "uniform rate" equation, "(distance) equals (rate) times (time)", that is used in "distance" word problems, and solving this for the specified variable works just like solving the previous equation.

The variable they want has a letter multiplied on it; to isolate the variable, I have to divide off that letter. So I'll solve for the specified variable r by dividing through by the t:

$\small{ d = rt }$

$\small{ \dfrac{d}{t} = \dfrac{rt}{t} }$

$\small{ \bm{\color{purple}{ \dfrac{d}{t} = r }}}$

$Source:$https://www.purplemath.com/modules/solvelit.htm

Week of January 21, 2020

CRM

equation vs. expression

Expression - Phrase, Equation - Sentence       If you've searched the topic "equation vs. expression" on the net & found this page, you will probably read the following four items as mostly the same. Ten is five less than a number. A number is less than five. a number less than five five less than a number       Please now reconsider the above items and determine which are sentences (subject, verb, complete idea) and which are phrases (no subject, no verb). For assistance from an English point of view, see http://www.richmond.edu/~writing/wweb/fragment.html       The mathematical term expression is equivalent to an English phrase. The most common mathematical statements or sentences, are called equations and inequalities. The reader is asked to review each of the links in this paragraph before reading further and to pay particular attention to the mathematical terms relation and relation symbol, the "verbs of mathematical statements," as they relate to each of the other words.   Expression vs. Equation       The algebra student, or algebraically able individual, is expected to know the difference between an expression and a statement because each serves a different purpose and each is handled in a certain way. Equation Expression An equation is a SENTENCE. One solves an equation. An equation HAS a relation symbol. Ex. Ten is five less than a number. 10 = x - 5 A number is less than five. x < 5 An expression is a PHRASE, a sentence fragment. One simplifies an expression. An expression HAS NO relation symbol. Ex. a number less than five x five less than a number x - 5   Equation, Not Expression Computation Errors Expression, Not Equation Computation Errors Source:https://www.google.com/search?q=expressions+or+equations&rlz=1C1GCEA_enUS861US861&oq=expressions+or+equations&aqs=chrome..69i57.7626j0j8&sourceid=chrome&ie=UTF-8

Week of January 13,2020

CRM and Algebra2

Factoring in Algebra

Factors

Numbers have factors:

And expressions (like x²+4x+3) also have factors:

Factoring

Factoring (called "Factorising" in the UK) is the process of finding the factors:

Factoring: Finding what to multiply together to get an expression.

It is like "splitting" an expression into a multiplication of simpler expressions.

Example: factor 2y+6

Both 2y and 6 have a common factor of 2:

• 2y is 2 × y
• 6 is 2 × 3

So we can factor the whole expression into:

2y+6 = 2(y+3)

So 2y+6 has been "factored into" 2 and y+3

Factoring is also the opposite of Expanding:

Common Factor

In the previous example we saw that 2y and 6 had a common factor of 2

But to do the job properly we need the highest common factor, including any variables

Example: factor 3y²+12y

Firstly, 3 and 12 have a common factor of 3.
So we could have:

3y²+12y = 3(y²+4y)

But we can do better!
3y² and 12y also share the variable y.
Together that makes 3y:

• 3y² is 3y × y
• 12y is 3y × 4

So we can factor the whole expression into:

3y²+12y = 3y(y+4)

Check: 3y(y+4) = 3y × y + 3y × 4 = 3y²+12y

More Complicated Factoring

Factoring Can Be Hard !

The examples have been simple so far, but factoring can be very tricky.

Because we have to figure what got multiplied to produce the expression we are given!

It is like trying to find which ingredients
went into a cake to make it so delicious.
It can be hard to figure out!

Experience Helps

With more experience factoring becomes easier.

Example: Factor 4x² − 9

Hmmm... there don't seem to be any common factors.
But knowing the Special Binomial Products gives us a clue called the "difference of squares":

Because 4x² is (2x)², and 9 is (3)²,

So we have:

4x² − 9 = (2x)² − (3)²

And that can be produced by the difference of squares formula:

(a+b)(a−b) = a² − b²

Where a is 2x, and b is 3.

So let us try doing that:

(2x+3)(2x−3) = (2x)² − (3)² = 4x² − 9

Yes!

So the factors of 4x² − 9 are (2x+3) and (2x−3):

Answer: 4x² − 9 = (2x+3)(2x−3)

How can you learn to do that? By getting lots of practice, and knowing "Identities"!

Remember these Identities

Here is a list of common "Identities" (including the "difference of squares" used above).

It is worth remembering these, as they can make factoring easier.

a2 − b2	=	(a+b)(a−b)
a2 + 2ab + b2	=	(a+b)(a+b)
a2 − 2ab + b2	=	(a−b)(a−b)
a3 + b3	=	(a+b)(a2−ab+b2)
a3 − b3	=	(a−b)(a2+ab+b2)
a3+3a2b+3ab2+b3	=	(a+b)3
a3−3a2b+3ab2−b3	=	(a−b)3

There are many more like those, but those are the most useful ones.

Advice

The factored form is usually best.

When trying to factor, follow these steps:

"Factor out" any common terms
See if it fits any of the identities, plus any more you may know
Keep going till you can't factor any more

There are also Computer Algebra Systems (called "CAS") such as Axiom, Derive, Macsyma, Maple, Mathematica, MuPAD, Reduce and many more that are good at factoring.

More Examples

Experience does help, so here are more examples to help you on the way:

Example: w⁴ − 16

An exponent of 4? Maybe we could try an exponent of 2:

w⁴ − 16 = (w²)² − 4²

Yes, it is the difference of squares

w⁴ − 16 = (w² + 4)(w² − 4)

And "(w² − 4)" is another difference of squares

w⁴ − 16 = (w² + 4)(w + 2)(w − 2)

That is as far as I can go (unless I use imaginary numbers)

Example: 3u⁴ − 24uv³

Remove common factor "3u":

3u⁴ − 24uv³ = 3u(u³ − 8v³)

Then a difference of cubes:

3u⁴ − 24uv³ = 3u(u³ − (2v)³)

= 3u(u−2v)(u²+2uv+4v²)

That is as far as I can go.

Example: z³ − z² − 9z + 9

Try factoring the first two and second two separately:

z²(z−1) − 9(z−1)

Wow, (z-1) is on both, so let us use that:

(z²−9)(z−1)

And z²−9 is a difference of squares

(z−3)(z+3)(z−1)

That is as far as I can go.

Source: https://www.mathsisfun.com/algebra/factoring.html

Week of January 6, 2020

College Readiness Math

WORD PROBLEMS

Examples

Problems

WORD PROBLEMS require practice in translating verbal language into algebraic language. See Lesson 1, Problem 8.  Yet, word problems fall into distinct types. Below are some examples.

Example 1.   ax ± b = c.  All problems like the following lead eventually to an equation in that simple form.

Jane spent $42 for shoes.  This was $14 less than twice what she spent for a blouse.  How much was the blouse?

Solution.   Every word problem has an unknown number. In this problem, it is the price of the blouse.  Always let x represent the unknown number.  That is, let x answer the question.

Let x, then, be how much she spent for the blouse.  The problem states that "This" -- that is, $42 -- was $14 less  than two times x.

Here is the equation:

2x − 14	=	42.
2x	=	42 + 14   (Lesson 9)
	=	56.
x	=	56  2
	=	28.

The blouse cost $28.

Example 2.   There are b boys in the class.  This is three more than four times the number of girls.  How many girls are in the class?

 Solution.   Again, let x represent the unknown number that you are asked to find:  Let x be the number of girls.

(Although b is not known, it is not what you are asked to find.)

The problem states that "This" -- b -- is three more than four times x:

	4x + 3	=	b.
	Therefore,
	4x	=	b − 3
x		=	b − 3    4	.

The solution here is not a number, because it will depend on the value of b.  This is a type of "literal" equation, which is very common in algebra.

Example 3.  The whole is equal to the sum of the parts.

The sum of two numbers is 84, and one of them is 12 more than the other.  What are the two numbers?

 Solution.  In this problem, we are asked to find two numbers. Therefore, we must let x be one of them.  Let x, then, be the first number.

We are told that the other number is 12 more, x + 12.

The problem states that their sum is 84:

  = 84

The line over x + 12 is a grouping symbol called a vinculum.  It saves us writing parentheses.

We have:

2x	=	84 − 12
	=	72.
x	=	72  2
	=	36.

This is the first number.  Therefore the other number is

x + 12 = 36 + 12 = 48.

The sum of 36 + 48 is 84.

Example 4.   The sum of two consecutive numbers is 37.  What are they?

Solution.   Two consecutive numbers are like 8 and 9, or 51 and 52.

Let x, then, be the first number.  Then the number after it is x + 1.

The problem states that their sum is 37:

  = 37

2x	=	37 − 1
	=	36.
x	=	36  2
	=	18.

The two numbers are 18 and 19.

Example 5.  One number is 10 more than another.  The sum of twice the smaller plus three times the larger, is 55.  What are the two numbers?

 Solution.  Let x be the smaller number.

Then the larger number is 10 more:  x + 10.

The problem states:

2x + 3(x + 10)	=	55.
That implies
2x + 3x + 30	=	55.  Lesson 14.
5x	=	55 − 30 = 25.
x	=	5.

That's the smaller number.  The larger number is 10 more:  15.

Example 6.   Divide $80 among three people so that the second will have twice as much as the first, and the third will have $5 less than the second.

Solution.   Again, we are asked to find more than one number.  We must begin by letting x be how much the first person gets.

Then the second gets twice as much, 2x.

And the third gets $5 less than that, 2x − 5.

Their sum is $80:

 

5x	=	80 + 5
x	=	85  5
	=	17.

This is how much the first person gets.  Therefore the second gets

2x	=	34.
And the third gets
2x − 5	=	29.

The sum of 17, 34, and 29 is in fact 80.

Algebra 2

Factoring in Algebra

Factors

Numbers have factors:

And expressions (like x²+4x+3) also have factors:

Factoring

Factoring (called "Factorising" in the UK) is the process of finding the factors:

Factoring: Finding what to multiply together to get an expression.

It is like "splitting" an expression into a multiplication of simpler expressions.

Example: factor 2y+6

Both 2y and 6 have a common factor of 2:

• 2y is 2 × y
• 6 is 2 × 3

So we can factor the whole expression into:

2y+6 = 2(y+3)

So 2y+6 has been "factored into" 2 and y+3

Factoring is also the opposite of Expanding:

Common Factor

In the previous example we saw that 2y and 6 had a common factor of 2

But to do the job properly we need the highest common factor, including any variables

Example: factor 3y²+12y

Firstly, 3 and 12 have a common factor of 3.
So we could have:

3y²+12y = 3(y²+4y)

But we can do better!
3y² and 12y also share the variable y.
Together that makes 3y:

• 3y² is 3y × y
• 12y is 3y × 4

So we can factor the whole expression into:

3y²+12y = 3y(y+4)

Check: 3y(y+4) = 3y × y + 3y × 4 = 3y²+12y

More Complicated Factoring

Factoring Can Be Hard !

The examples have been simple so far, but factoring can be very tricky.

Because we have to figure what got multiplied to produce the expression we are given!

It is like trying to find which ingredients
went into a cake to make it so delicious.
It can be hard to figure out!

Experience Helps

With more experience factoring becomes easier.

Example: Factor 4x² − 9

Hmmm... there don't seem to be any common factors.
But knowing the Special Binomial Products gives us a clue called the "difference of squares":

Because 4x² is (2x)², and 9 is (3)²,

So we have:

4x² − 9 = (2x)² − (3)²

And that can be produced by the difference of squares formula:

(a+b)(a−b) = a² − b²

Where a is 2x, and b is 3.

So let us try doing that:

(2x+3)(2x−3) = (2x)² − (3)² = 4x² − 9

Yes!

So the factors of 4x² − 9 are (2x+3) and (2x−3):

Answer: 4x² − 9 = (2x+3)(2x−3)

How can you learn to do that? By getting lots of practice, and knowing "Identities"!

Remember these Identities

Here is a list of common "Identities" (including the "difference of squares" used above).

It is worth remembering these, as they can make factoring easier.

a2 − b2	=	(a+b)(a−b)
a2 + 2ab + b2	=	(a+b)(a+b)
a2 − 2ab + b2	=	(a−b)(a−b)
a3 + b3	=	(a+b)(a2−ab+b2)
a3 − b3	=	(a−b)(a2+ab+b2)
a3+3a2b+3ab2+b3	=	(a+b)3
a3−3a2b+3ab2−b3	=	(a−b)3

There are many more like those, but those are the most useful ones.

Advice

The factored form is usually best.

When trying to factor, follow these steps:

"Factor out" any common terms
See if it fits any of the identities, plus any more you may know
Keep going till you can't factor any more

There are also Computer Algebra Systems (called "CAS") such as Axiom, Derive, Macsyma, Maple, Mathematica, MuPAD, Reduce and many more that are good at factoring.

More Examples

Experience does help, so here are more examples to help you on the way:

Example: w⁴ − 16

An exponent of 4? Maybe we could try an exponent of 2:

w⁴ − 16 = (w²)² − 4²

Yes, it is the difference of squares

w⁴ − 16 = (w² + 4)(w² − 4)

And "(w² − 4)" is another difference of squares

w⁴ − 16 = (w² + 4)(w + 2)(w − 2)

That is as far as I can go (unless I use imaginary numbers)

Example: 3u⁴ − 24uv³

Remove common factor "3u":

3u⁴ − 24uv³ = 3u(u³ − 8v³)

Then a difference of cubes:

3u⁴ − 24uv³ = 3u(u³ − (2v)³)

= 3u(u−2v)(u²+2uv+4v²)

That is as far as I can go.

Example: z³ − z² − 9z + 9

Try factoring the first two and second two separately:

z²(z−1) − 9(z−1)

Wow, (z-1) is on both, so let us use that:

(z²−9)(z−1)

And z²−9 is a difference of squares

(z−3)(z+3)(z−1)

That is as far as I can go.