EPPS Math and Coding Camp

Algebra review

Instructor: Taowen Hu

2.1 BASIC PROPERTIES OF ARITHMETIC

  • The associative properties state that \((a+b)+c = a+(b+c)\) and \((a \times b) \times c = a \times (b \times c)\).
  • The commutative properties state that \(a+b=b+a\) and \(a \times b=b \times a\).
  • The distributive property states that \(a(b+c)=ab+ac\).
  • The identity properties state that there exists a zero such that \(x+0=x\) and that there exists a one such that \(x \times 1 =x\).
  • The inverse property states that there exists a \(-x\) such that \((-x)+x=0\). We might also consider an inverse under multiplication, \(x^{-1}\), such that \((x^{-1} \times x) = 1\).
  • The existence of this inverse is a property of the real numbers (and the rational numbers), but not the integers.

Ratios, Proportions, and Percentages

  • The ratio of two quantities is one divided by the other \(\frac{x}{y}\) is the ratio of x to y. Ratios are also written as \(x:y\).
  • Though a ratio may be negative, we typically consider ratio variables that range from \(0\) to \(\infty\).
  • The proportion of two variables, on the other hand, is the amount one variable represents of the sum of itself and a second variable: \(|\frac{x}{x+y}|\). A proportion ranges from a minimum of \(0\) to a maximum of \(1\).
  • The proportion of expenditures that is spent in a given category is often interested.
  • The percentage one variable represents of a total is the proportion represented over the range from \(0\) to \(100\): \(|\frac{x}{x+y}| \times 100 \%\).

2.2 ALGEBRA REVIEW

2.2.1 Fractions

\[ \frac{Numerator}{Denominator} \]

2.2.2 Factoring

  • Factoring involves rearranging the terms in an equation to make further manipulation possible or to reveal something of interest.

Example:
\[ \delta+\delta^2+4\delta-6\delta^2+18\delta^3\]

In this case we combine all the terms that have the same exponent, which gives us \[ 18\delta^3-5\delta^2+5\delta\]

2.2.2.1 Factoring Quadratic Polynomials

Quadratic polynomials are composed of a constant and a variable that is both squared and raised to the power of one: \(x^2-2x+3\), or \(7-12x+6x^2\). Quadratic polynomials can be factored into the product of two terms: \((x \pm ?) \times (x \pm ?)\) where you need to determine whether the sign is \(+\) or \(-\), and then replace the question marks with the proper values.

2.2.2.2 Factoring and Fractions

Consider the fraction \(\frac{x^2-1}{x-1}\)

We can factor the numerator \(x^2-1=(x+1)(x-1)\). We can thus rewrite the fraction as follows:

\[ \frac{x^2-1}{x-1}= \frac{(x+1)(x-1)}{x-1} \]

The term \(x-1\) is in both the numerator and the denominator and thus (as long as \(x\neq 1\)) cancels out, leaving \(x+1\) for \(x \neq 1\)

2.2.3 Expansion: The FOIL Method

\[ (\delta + \gamma)^2 \]

\[=(\delta + \gamma)(\delta + \gamma)\]

\[ =\delta^2 + 2\delta\gamma + \gamma^2 \]

F: Multiply the first terms: \((\underline{2\pi}+7) (\underline{4}+3\pi)=2 \pi \times 4=8 \pi\)

O: Multiply the outer terms: \((\underline{2\pi}+7) (4+\underline{3\pi})=2 \pi \times 3 \pi=6 \pi^2\)

I: Multiply the inner terms: \((2\pi+\underline{7}) (\underline{4}+3\pi)=4 \times 7=28\)

L: Multiply the last terms: \((2\pi+\underline{7}) (4+\underline{3\pi})=7 \times 3 \pi=21 \pi\)

Add terms to get: \(8\pi+6\pi^2+28+21\pi\)

Finally, group like terms to get: \(6\pi^2+29\pi+28\)

To test yourself, factor the final expression and show it yields the simplified expression with which we started. This is one way to check your work for any mistakes.

2.2.4.1 Solving Quadratics

Example:

\[ (x-3)^2=4 \]

\[ x-3=\pm 2 \]

\[ x=5 \text{ or } x=1 \]

Note that this quadratic equation will have two solutions in the real numbers, or zero, but not one. In other words, the cardinality of the solution set for a quadratic equation will be zero or two.

Example with no solutions in the real numbers: \(x^2+1=0\)

Another example: \[ x^2+8x+6=0 \]

The general form of a quadratic equation:
\[ ax^2+bx+c=0 \]

The general solutions to this equation: \[ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \]

2.2.5 Inequalities

  • All pairs of real numbers have exactly one of the following relations: \(x=y\), \(x>y\), or \(x<y\).
  • Adding any number to each side of these relations will not change them; this includes the inequalities.
    That is, inequalities have the same addition and subtraction properties as equalities such that: if \(x>y\), then \(x+a>y+a\) and \(x-a>y-a\)
  • You flip the \(<\) or \(>\) sign when multiplying or dividing by a negative. Multiplying or dividing an inequality by zero is not allowed.
  • For multiplication, if \(a\) is positive and \(x>y\), then \(ax>ay\).
    If \(a\) is negative and \(x>y\), then \(ax<ay\).
  • For division, if \(a\) is positive and \(x>y\), then \(\frac{x}{a}>\frac{y}{a}\).
    If \(a\) is negative and \(x>y\), then \(\frac{x}{a}<\frac{y}{a}\).

Example:
Solve for \(y\):
\[ -4y>2x+12 \]

\[ y<-\frac{x}{2}-3 \]

Exercises:

Exercise 1

Solve for x:

\[ 1.4x^2+3.7x+1.1=0 \]

Answer: \[ a=1.4, b=3.7, c=1.1 \]

\[ x=\frac{-3.7 \pm \sqrt{3.7^2-4 \times 1.4 \times 1.1}}{2.8} \]

Using a calculator to solve, we find that \(x=-.341 \text{ or } x=-2.301\)

Exercise 2

Simplify this expression as much as possible: \(\frac{2x^2 + 20x + 50}{2x^2 - 50}\)

Answer:

Factor numerator and denominator:

Numerator:
\[ 2x^2 + 20x + 50 = 2(x^2 + 10x + 25) = 2(x + 5)^2 \]

Denominator:
\[ 2x^2 - 50 = 2(x^2 - 25) = 2(x - 5)(x + 5) \]

Now simplify: \[ \frac{2(x + 5)^2}{2(x - 5)(x + 5)} = \frac{(x + 5)}{(x - 5)} \]

Final Answer:
\[ \frac{x + 5}{x - 5} \]

Exercise 3

FOIL: \((2x - 3)(5x + 7)\)

Answer:

Apply FOIL (First, Outer, Inner, Last):

\[ = 2x \cdot 5x + 2x \cdot 7 - 3 \cdot 5x - 3 \cdot 7 \]

\[ = 10x^2 + 14x - 15x - 21 \]

\[ = 10x^2 - x - 21 \]

Final Answer:
\[ 10x^2 - x - 21 \]

Exercise 4

Complete the square and solve for y: \(\frac{1}{3}y^2 + \frac{2}{3}y - 16 = 0\)

Multiply through by 3 to eliminate fractions: \[ y^2 + 2y - 48 = 0 \]

Move constant to the right: \[ y^2 + 2y = 48 \]

Complete the square: \[ y^2 + 2y + 1 = 49 \quad \Rightarrow \quad (y + 1)^2 = 49 \]

Solve: \[ y + 1 = \pm 7 \Rightarrow y = -1 \pm 7 \Rightarrow \boxed{y = 6 \text{ or } -8} \]

Exercise 5

Solve:

\[ -\delta > \frac{\delta + 4}{7} \]

Multiply both sides by 7 to eliminate denominator (note: 7 > 0, so inequality direction unchanged): \[ -7\delta > \delta + 4 \]

Move all terms to one side: \[ -7\delta - \delta > 4 \Rightarrow -8\delta > 4 \]

Divide both sides by -8 (reverse inequality): \[ \delta < -\frac{1}{2} \]

Final Answer:
\[ \delta < -\frac{1}{2} \]

Any Questions?

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