]> Chapter 01 Notes

1: Tools of Algebra

1-1: Properties of Real Numbers

Number Sets

Let's begin by looking at the hierarchy of number sets, and their symbols…

Our text refers to the numbers 1, 2, 3, … as the natural numbers, and 0, 1, 2, 3, … as whole numbers. Unfortunately, there is no agreement of whether the natural numbers include zero (some include it; others don't). However, there is a way to clear it up…

The numbers 1, 2, 3, … along with the negatives of those numbers, and zero, make the integers. The standard symbol for this set is $ℤ$ (from the word zählen: to count). Once you have the integers, then the numbers 1, 2, 3, … can be called the positive integers, and the numbers 0, 1, 2, 3, … are the non-negative integers. Relax; there won't be a quiz on this. The important thing is to know what kinds of numbers the integers are, and what the symbol $ℤ$ stands for.

One step up from the integers are fractions made from integers—these are called the rationals, and use the symbol $ℚ$ (for quotient). One way to define the rationals is $ℚ \equiv {\frac{a}{b} | ({a, b} \subset ℤ) \land (b \neq 0)}$ . That reads as "the rational numbers are defined to be all elements a over b, where a and b are integers and b is not equal to zero."

Again—relax; there won't be a quiz on this! The sooner you see these symbols, the easier it will be later. For those who must know, here's that reading again, this time with the associated symbols: "the rational numbers ( $ℚ$ ) are defined to be ( $\equiv$ ) all elements a over b ( $\frac{a}{b}$ ), where a and b are integers (the set containing a and b is a subset of the integers: ${a, b} \subset ℤ$ ) and ( $\land$ ) b is not equal to zero ( $b \neq 0$ )."

One step up from that is the real numbers— $ℝ$ . The real numbers include (almost) any number you can think of, and they will be the primary set of interest in our journey through algebra.

Another way to think of the real numbers is the union of the rationals with the irrationals. An irrational number is a real number that is not rational—it cannot be expressed as a ratio (fraction, quotient) of integers. Also, its decimal representation will either terminate or repeat.

There is one more set above the reals—more on that later.

Properties of Real Numbers

Operations

First of all, there are only two ways to combine (operate on) real numbers—you can add them, and you can multiply them.

"What about subtraction?" you ask…well, there isn't really any such thing. Subtraction is just a short way of writing addition with an opposite: $a - b = a + (- b)$ . The opposite (or negative; or negation; or additive inverse) of a number is some other number that, when added, results in zero. Using symbols: if a and b are opposites, then $a + b = 0$ . The symbol for the opposite of a is -a. Note that a and -a are the same distance from zero on a number line, but on opposite sides—thus, they have the same absolute value (distance from zero): $| a | = | - a |$ .

"What about division?" you ask…again, there is no such thing. Division is just a short way of writing multiplication with a reciprocal. The reciprocal of a number is some other number that, when multiplied, results in 1. Using symbols: if a and b are reciprocals, then $a \cdot b = 1$ . The symbol for the opposite of a is $\frac{1}{a}$ . There is only one real number that does not have a reciprocal!

Properties

What we've seen so far (though I didn't mention it at the time) is that the real numbers have additive and multiplicative identities: zero and one, respectively. Also, we've seen that (almost) every real number has an additive and multiplicative inverse: the opposite and reciprocal, respectively. Furthermore, the reals are closed under both of these operations: add two reals and you get another real; multiply two reals and the result is real.

Both operations are commutative: the order of the operands does not affect the result! $a + b = b + a$ and $a \cdot b = b \cdot a$ .

Both operations are associative: when the same operation is repeated, the operands can be grouped at will without affecting the results. $(a + b) + c = a + (b + c)$ and $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ .

Finally, there is the distributive property: $a \cdot (b + c) = (a \cdot b) + (a \cdot c)$ .

When each of these properties is demonstrated, you should be able to identify it…we'll see some examples of this in class.

1-2: Algebraic Expressions

Definitions

A variable is a symbol (typically a lower case roman letter from the latter end of the alphabet) that represents one or more unknown values. For example: x, y, t. Note that when typed, variables are typically in italics.

A coefficient is (for now) a number multiplied against something—typically, a power of a variable. For example: $3 x$ (the coefficient is 3), $2 y^{5}$ , $t$ (the coefficient is 1), or $5 (x + 2 y)$ (there are two coefficients—5 and 2).

A term is the product of a coefficient and one or more variables to various powers. For example: $3 x^{2}$ , $2 x^{3} y^{4}$ , $x y z^{2}$ . Note that a plain number can be a term, also.

Like terms are terms that are identical with respect to the variables (differing only in the coefficients). For example: $7 x$ and $-5 x$ , or $2 x y^{3}$ and $3 x y^{3}$ .

An expression is a sum of terms (or just one term). For example: $3 x^{2} + 5 x - 7$ , $x^{2} + 4 y^{2}$ , or $5 t$ .

An equation is two expressions connected by equality. For example: $x^{2} + 2 x - 1 = 5 x + 2$ or $4 x^{2} + 3 y^{3} = z$ .

Evaluating and Simplifying

To evaluate an expression, the value(s) of the variable(s) must be given. Plug in these values, and use the order of operations rules to reduce the expression to a number.

To simplify an expression, eliminate any parentheses, then combine like terms.

Examples

[1.] Evaluate $3 x^{2} + 2 x y - y^{3}$ if $x = -2$ and $y = 1$ .

$3 x^{2} + 2 x y - y^{3} = 3 {(-2)}^{2} + 2 (-2) (1) - {(1)}^{3} = 12 - 4 - 1 = 7$

[2.] The volume of a cone is $\frac{1}{3} π r^{2} h$ . Find the volume if $r = 4 cm$ and $h = 8 cm$ .

$\frac{1}{3} π r^{2} h = \frac{1}{3} π {(4 cm)}^{2} (8 cm) = \frac{1}{3} π (16 {cm}^{2}) (8 cm) = \frac{1}{3} π (128 {cm}^{3}) = \frac{128 π}{3} {cm}^{3} \approx 134.041 {cm}^{3}$ . The cone has a volume of approximately 134.041 cm³.

[3.] Simplify: $2 x^{3} + 4 x - 5 + 5 x^{2} + 7 x - 10$ .

Like terms have been given similar marks: $2 x^{3} + \overset{—}{4 x} \underset{══}{- 5} + 5 x^{2} + \overset{—}{7 x} \underset{══}{- 10} = 2 x^{3} + 5 x^{2} + 11 x - 15$ .

[4.] Simplify: ${(x + 2 y)}^{2} - x y$ .

Note that you will have to FOIL out the parentheses first!

${(x + 2 y)}^{2} - x y = (x^{2} + 4 x y + 4 y^{2}) - x y = x^{2} + 3 x y + 4 y^{2}$ .

1-3: Solving Equations

Solving Equations

A value of a variable that, when plugged in, makes an equation into a true statement, is called a solution of the equation.

For simple equations, solving is exactly the opposite of evaluating. For more complicated equations, you may need to do some simplifying at some point in order to solve.

If the equation has more than one variable, you will only solve for one at a time—while doing so, treat the other variables like constants (numbers).

Not every equation has a solution. Some equations have multiple (or even infinite) solutions—more on these cases later.

Writing Equations

Experience is the real key to writing equations from word problems. However, I have one general tip: break everything up into amounts and relationships. The amounts will become variables—make as few as possible! One of the amounts (variables) should give the answer to the question. The relationships will be equations.

Examples

[5.] Solve: $2 x - 7 = 4 (x - 5)$ .

Distribute: $2 x - 7 = 4 (x - 5) ⇒ 2 x - 7 = 4 x - 20$ . Combine like terms: $2 x - 7 = 4 x - 20 ⇒ 13 = 2 x$ . Divide: $\frac{13}{2} = x$ .

[6.] Solve for v: $G^{2} = 1 - \frac{v}{c}$ .

Subtract the 1: $G^{2} = 1 - \frac{v}{c} ⇒ G^{2} - 1 = - \frac{v}{c}$ . Multiply by -c: $G^{2} - 1 = - \frac{v}{c} ⇒ (G^{2} - 1) (- c) = v$ . Simplify: $(G^{2} - 1) (- c) = v ⇒ c - c G^{2} = v$ .

[7.] Solve for x, and state any restrictions: $\frac{x + y}{x - y} = 2$ .

Multiply both sides by $(x - z)$ : $\frac{x + y}{x - y} = 2 ⇒ x + y = 2 (x - z)$ . Distribute: $x + y = 2 (x - z) ⇒ x + y = 2 x - 2 z$ . Collect like terms (with respect to x): $x + y = 2 x - 2 z ⇒ y + 2 z = x$ . The only restriction is that $x \neq z$ .

[8.] The length of a rectangle is to be 2.5 times the width. The perimeter of the rectangle is to be 49 m. What are the dimensions of the rectangle?

Read this at least once through without writing anything down—then read it again. The first sentence suggests the equation $l = 2.5 w$ , where l is the length and w is the width. Remember that you must have as many equations as variables in order to solve; thus, there must be one more. Specifically, the second sentence suggests $P = 2 (l + w)$ , where P is the perimeter. However, we know P—so we have $49 = 2 (l + w)$ . That's two equations and two variables—let's solve! Start by replacing l with 2.5w: $49 = 2 (2.5 w + w)$ . Keep going: $49 = 2 (3.5 w) ⇒ 49 = 7 w ⇒ 7 = w$ . This doesn't answer the question—we need the dimensions. If $7 = w$ , then $l = 2.5 w = 2.5 (7) = 17.5$ . The rectangle has length 17.5m and width 7m.

1-4: Solving Inequalities

One Inequality

Solving inequalities is almost exactly like solving equations—in fact, there is only one difference: if you multiply or divide by a negative number, then you must change the direction of the inequality. Because you don't know whether a variable is positive or negative, that means that multiplying or dividing both sides of an inequality by a variable expression actually results in two answers—thus, we won't be doing that (for now).

The solution to an inequality is another inequality. This can be written algebraically, or it can be graphed on a number line. When graphing, strict inequalities (< and >) are terminated with open circles. Inclusive inequalities (≤ or ≥) are terminated with closed (filled-in) circles.

Compound Inequalities

Just as we sometimes have multiple equations that must be solved simultaneously (later!), we can also have multiple inequalities. Let's begin with just two. Two inequalities can be joined in one of two ways—with and, or with or. Inequalities joined with and mean that our solutions must work in both inequalities (make both true) at the same time. Inequalities joined with or have solutions that work in one (or both!).

Usually, the word (and, or) joins the two; however, sometimes an and is disguised by physically joining the two—instead of $2 < 3 x + 1$ and $3 x + 1 \leq 5$ , is written $2 < 3 x + 1 \leq 5$ . Whenever dealing with and-joined inequalities, turn them around so that the symbols point to the left (as in the previous example).

The solution to such a system of inequalities can be written algebraically or graphically. Algebraically, the solution to an and problem can also be reduced to a (fairly) simple inequality. Graphically, the solution will be a single region. The solution to an or problem will (almost) always be two pieces—algebraically or graphically. The exception is when the two pieces end up covering all real numbers—in which case, you can just answer "all real numbers."

Examples

[9.] Solve $2 x + 1 < 5 x + 3$ .

Collect like terms: $2 x + 1 < 5 x + 3 ⇒ -2 < 3 x$ . Divide: $-2 < 3 x ⇒ - \frac{2}{3} < x$ .

Graphical answer:

An Image

[10.] Solve $4 x - 2 \geq 2 (3 x + 5)$ .

Distribute: $4 x - 2 \geq 2 (3 x + 5) ⇒ 4 x - 2 \geq 6 x + 10$ . Collect like terms: $4 x - 2 \geq 6 x + 10 ⇒ -12 \geq 2 x$ . Divide: $-12 \geq 2 x ⇒ -6 \geq x$ .

Graphical answer:

An Image

[11.] Solve $9 x < 54$ and $-4 x < 12$ .

In both cases, just divide! $9 x < 54 ⇒ x < 6$ and $-4 x < 12 ⇒ x > -3$ . This can be collapsed to $-3 < x < 6$ . Graphical answer:

An Image

[12.] Solve $6 (x + 2) \geq 24$ or $5 x + 10 < 15$ .

Solve each of these separately. For the first, we'll begin with division: $6 (x + 2) \geq 24 ⇒ x + 2 \geq 4$ . Now subtract: $x + 2 \geq 4 ⇒ x \geq 2$ . For the second, do the reverse! $5 x + 10 < 15 ⇒ 5 x < 5 ⇒ x < 1$ . The answer is $x < 1$ or $x \geq 2$ . Graphical answer:

An Image

1-5: Absolute Value Equations and Inequalities

The Definition(s) of Absolute Value

The absolute value of a number is the distance from that number to the origin. If you think of absolute value this way, you'll rarely go wrong.

It is likely that you think of absolute value as something that strips negative signs off of numbers. For that viewpoint, there is a complicated definition:

$| x | = {\binom{x if x \geq 0}{- x if x < 0}}$

This is called a piecewise-defined function. More on those later!

From the previous function, you can see that an absolute value reduces by either disappearing completely, or by disappearing with a negative sign in front. This is the key to working with absolute value in equation and inequalities: it turns into two things! One with a + (or nothing), and one with a -.

Absolute Value Equations

To solve an absolute value equation, make two equations—one with the bars missing; the other with the bars missing and a negative sign in front.

Note that if an absolute value is equal to a negative number, then there is no solution. If you're not paying attention, and you go ahead and solve this as normal, then you'll create an extraneous solution—one that doesn't actually solve the equation! To check for extraneous solutions, you must plug your solutions into the original problem, and evaluate to see if it works.

Absolute Value Inequalities

These begin in the same way—split into two, with a + and a -. However…should these be connected with and? Should they be connected with or?

The answer can be determined by looking to see which way the inequality is pointing. If it points towards the absolute value, then use and. If the symbol points away from the absolute value, then use or.

Remember—there's no way an absolute value can be less than a negative number!

Examples

[13.] Solve $| 2 x + 3 | = 21$ .

Divide and conquer! $| 2 x + 3 | = 21$ ⇒ $2 x + 3 = 21$ and $- (2 x + 3) = 21$ . Solve each individually: $2 x + 3 = 21 ⇒ 2 x = 18 ⇒ x = 9$ ; $- (2 x + 3) = 21 ⇒ 2 x + 3 = -21 ⇒ 2 x = -24 ⇒ x = -12$ . Now, check: $| 2 (9) + 3 | = | 18 + 3 | = | 21 | = 21$ (OK), and $| 2 (-12) + 3 | = | -24 + 3 | = | -21 | = 21$ (OK). Both solutions are actual solutions.

[14.] Solve $| 2 x - 1 | = 4 x + 3$ .

$| 2 x - 1 | = 4 x + 3$ ⇒ $2 x - 1 = 4 x + 3$ and $- (2 x - 1) = 4 x + 3$ . $2 x - 1 = 4 x + 3 ⇒ -4 = 2 x ⇒ -2 = x$ , and $- (2 x - 1) = 4 x + 3 ⇒ 2 x - 1 = -4 x - 3 ⇒ 6 x = -2 ⇒ x = - \frac{1}{3}$ . Check: $| 2 (-2) - 1 | \overset{?}{=} 4 (-2) + 3 ⇒ | -4 - 1 | \overset{?}{=} -8 + 3 ⇒ | -5 | \overset{?}{=} -5$ . This is obviously false; $x = -2$ is an extraneous solution. Let's check the other one: $| 2 (- \frac{1}{3}) - 1 | \overset{?}{=} 4 (- \frac{1}{3}) + 3 ⇒ | - \frac{2}{3} - 1 | \overset{?}{=} - \frac{4}{3} + 3 ⇒ | - \frac{5}{3} | \overset{?}{=} \frac{5}{3} ⇒ \frac{5}{3} = \frac{5}{3}$ . Hurrah! The solution is $x = - \frac{1}{3}$ .

[15.] Solve $| 3 x - 5 | \leq 28$ .

Divide this into two parts, connected with and: $3 x - 5 \leq 28$ and $- (3 x - 5) \leq 28$ . Now solve: $3 x - 5 \leq 28 ⇒ 3 x \leq 33 ⇒ x \leq 11$ and $- (3 x - 5) \leq 28 ⇒ 3 x - 5 \geq -28 ⇒ 3 x \geq -23 ⇒ x \geq - \frac{23}{3}$ . The answer can be reduced to $- \frac{23}{3} \leq x \leq 11$ . Graphically:

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[16.] Solve $| x + 1 | > 5$ .

Divide this into $x + 1 > 5$ or $- (x + 1) > 5$ . Solve each: $x + 1 > 5 ⇒ x > 4$ or $- (x + 1) > 5 ⇒ x + 1 < -5 ⇒ x < -6$ . The solution is $x < -6$ or $x > 4$ . Graphically:

An Image

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