Math Boot Camp

Math Boot CampLecture 1: Algebra

Matthew Morse

8/27/18

Math Boot Camp

Mathematics for the Political and Social Sciences

Mathematics provides a set of tools for manipulating andinterpreting quantitative (that is, numeric) data.

In order to use these tools effectively, we must have a clearunderstanding of the meaning and operation of mathematics.Mathematics is the language of quantitative data.

Math Boot Camp

1: Preliminaries

1.1: Variables and Constants

1.1: Variables

A variable is a symbol representing a quantity of interestassociated with a piece of data. The value of the variable changesdepending on the data.

Examples

x : per capita income in a city

y : proportion of vote for the Democratic candidate in a city

Mathematically, the value of a variable can be any real number,that is, positive or negative whole numbers, fractions, or irrationalnumbers. Depending on the context, some values may not makesense.

Math Boot Camp

1: Preliminaries

1.1: Variables and Constants

1.1: Constants

A constant is a symbol representing a quantity of interest whichdoes not change depending on the data.

Example

Suppose we propose a linear relationship between income and thevote. Then in the equation

y = ax + b

the variables x and y depend on the city, but the constants a andb are fixed and do not change.

Math Boot Camp

1: Preliminaries

1.1: Variables and Constants

1.1: Symbols for Variables and Constants

Generally we use lowercase or uppercase letters for variables orconstants, for example,

a,A, x ,X , . . .

If we have many variables, we may use subscripts.

x1, x2, x3, x4, . . .

From the previous example, x3 would be the per capita income ofthe third city in our data set.In addition to the standard Roman alphabet, we may also use theGreek alphabet.

α, β, γ, δ, . . .

Math Boot Camp

1: Preliminaries

1.6: Notation

1.6: Notation

Variables and constants are the nouns of the language ofmathematics. We combine these with other symbols intoexpressions to make mathematical statements. Interpretingmathematical statements requires understanding mathematicalnotation.

Examples

Addition a + bSubtraction a− bMultiplication a× b or a · b or abDivision a/b or a÷ b

Math Boot Camp

1: Preliminaries

1.6: Notation

1.6: Summation Notation

We use the Greek letter∑

(capital sigma) as a shorthand forsums. A typical use is

6∑i=3

xi

which is read, “the sum of xi from i = 3 through i = 6.”Below the summation symbol, we have the statement i = 3. Thevariable i is called the index of summation, and 3 is the startingvalue of the index.The number 6 appears above the summation symbol. This is theending value for the index. The index of summation takes on allvalues between the starting value and the ending value, so in thissum, i will be set to 3, 4, 5, and 6.

Math Boot Camp

1: Preliminaries

1.6: Notation

1.6: Summation Notation (Cont.)

Finally, we sum the expression to the right of the summationsymbol, xi , by replacing the index i by each of its values andadding the resulting terms. This gives us our result,

6∑i=3

xi = x3 + x4 + x5 + x6

Math Boot Camp

2: Algebra

2.1: Basic Properties of Algebra

2.1: Basic Properties of Algebra

Let a, b, and c be any real numbers. The following properties arealways true.

Property Addition Multiplication

Associative (a + b) + c = a + (b + c) (a · b) · c = a · (b · c)Commutative a + b = b + a a · b = b · aIdentity a + 0 = a a · 1 = aInverse (−a) + a = 0 (1/a) · a = 1 (a 6= 0)Distributive a · (b + c) = a · b + a · cEquivalence a = b is equivalent to a + c = b + c .

a = b is equivalent to a · c = b · c if c 6= 0.Subtraction a− b = a + (−b) = a + (−1) · bDivision a/b = a · (1/b) (b 6= 0)

Math Boot Camp

2: Algebra

2.1: Basic Properties of Algebra

2.1.1: Order of Operations

When evaluating an expression, computations must be performedin the standard Order of Operations: PEMDAS or “Please ExcuseMy Dear Aunt Sally”

1 Parentheses: Always evaluate terms inside parentheses first,working from the innermost set outward.

2 Exponents: Compute exponents and roots, working from leftto right.

3 Multiplication/Division: Perform any multiplications ordivisions in the order in which they occur, working from left toright.

4 Addition/Subtraction: Perform any additions or subtractionsin the order in which they occur, working from left to right.

Math Boot Camp

2: Algebra

2.1: Basic Properties of Algebra

2.1.1: Order of Operations - Example

Example: Use the Order of Operations to evaluate the expression ifx = −2, y = 3, and z = 4.

−4 · x2 − 5 · y + 4 · (z − 1)

Use parentheses when replacing letters with numbers:

−4 · x2 − 5 · y + 4 · (z − 1) = −4 · (−2)2 − 5 · (3) + 4 · ([4]− 1)

= −4 · (4)− 5 · (3) + 4 · (3)

= −16− 15 + 12

= −19

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.1: Fractions

Fractions are another way of writing division. The numerator of afraction is the number (or expression) on the top. Thedenominator is the number or expression on the bottom.

Example

x + 4

n= (x + 4)/n

x + 4 is the numerator. n is the denominator.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.1.1: Simplifying Fractions: Cancellation

a · cb · c

=a

b(b, c 6= 0)

Proof:

a · cb · c

=a · cb · c

· 1 (Identity)

=

[(a · c) · 1

b · c

]·(b · 1

b

)(Division and Inverse)

=

(a · 1

b

)·[

(b · c) · 1

b · c

](Associative and Commutative)

=a

b· 1 (Division and Inverse)

=a

b(Identity)

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.1.1: Fraction Cancellation Examples

12 · x8

=4 · 3 · x

4 · 2=

3 · x2

12 + x

8=

12

8+

x

8(Distributive)

=3

2+

x

8

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.1.2: Adding Fractions

a

b+

c

d=

a · d + b · cb · d

(b, d 6= 0)

Proof:

a

b+

c

d=

a · db · d

+b · cb · d

(Fraction Cancellation)

= (a · d) · 1

b · d+ (b · c) · 1

b · d(Division)

= (a · d + b · c) · 1

b · d(Distributive)

=a · d + b · c

b · d(Division)

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.1.2: Adding Fractions Example

4

3+

1

6=

4 · 63 · 6

+3 · 13 · 6

=24 + 3

18

=27

18

=3

2

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.2: Factoring

Factoring is the rewriting of a mathematical expression as aproduct. We do this to isolate an important term or to simplify anexpression.

Examples

Factor each expression:

12p − 18q = 6 · (2p − 3q)

15x3 + 9x2 − 12x = 3x · (5x2 + 3x − 4)

5(4α− 3)3 + 2(4α− 3)2 = (4α− 3)2 · [5(4α− 3) + 2]

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.2.2: Factoring and Fractions

We can often simplify fractions by factoring.

Example

Simplify:

3θ3 + 6θ2

9θ2 + 18θ=

3θ · (θ2 + 2θ)

3θ · (3θ + 6)

=θ2 + 2θ

3θ + 6(θ 6= 0)

=θ · (θ + 2)

3 · (θ + 2)

=θ

3(θ + 2 6= 0)

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.3: Expansion

Use the distributive property to expand expressions by multiplyingthem out. For an expression like (2x − 5)(3x + 4) we can use themnemonic FOIL to remember the expansion.

1 First: Multiply the first terms: 2x · 3x = 6x2

2 Outer: Multiply the outer terms: 2x · 4 = 8x

3 Inner: Multiply the inner terms: −5 · 3x = −15x

4 Last: Multiply the last terms: −5 · 4 = −20

5 Sum the products: 6x2 + 8x − 15x − 20

6 Combine like terms: 6x2 + (8− 15)x − 20 = 6x2 − 7x − 20

This is really just using the distributive property twice:

(2x − 5)(3x + 4) = 2x(3x + 4)− 5(3x + 4)

= 6x2 + 8x − 15x − 20

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4: Solving Equations

To solve an equation, use the rules of algebra to rewrite theequation so the variable of interest appears by itself on the lefthand side (LHS) and does not appear on the right hand side(RHS).This generally involves expanding terms, followed by adding orsubtracting from both sides of the equation to eliminate terms,and then multiplying or dividing from both sides of the equation.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4: Solving Equations Example

Solve for t: t + 3(t − 4) = 2(t − 4)

1 Distribute:t + 3t − 12 = 2t − 8

2 Add to both sides of the equation to eliminate the variablefrom the RHS and eliminate everything else from the LHS:

4t − 12− 2t + 12 = 2t − 8− 2t + 12

2t = 4

3 Multiply on both sides of the equation to eliminate thecoefficient from the variable on the LHS:

1

2· 2t =

1

2· 4

t = 2

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4: Applied Equation Solving

There are many examples of applications of equation solving in areal-world setting. Regardless of the application, the process ofsolving an applied problem should follow a similar process:

1 Identify all given information, as well as any unknownquantities. Draw a picture or table, if appropriate.

2 Identify the unknown quantity of interest, and denote it as avariable.

3 Write an equation that expresses the relationship between thequantities given in the problem and the variables. Note thatsometimes you may need more than one equation to expressthese relationships.

4 Solve for the unknown quantity of interest using typicalproblem-solving strategies.

5 Interpret your answer within the context of the originalproblem.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4: Applied Equation Solving: Example

A car rental firm charges $45.56 a day with unlimited mileage. Adiscount firm offers a similar car for $20 a day plus 18 cents permile. How far must you drive in a day in order for the cost to bethe same at both firms?

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4: Applied Equation Solving: Solution

1 Firm 1: $45.56 totalFirm 2: $20 plus $0.18 per mile.Let x = # miles traveled in one day.

2 Want to know when Firm 1 = Firm 2.

3 Relation: 45.56 = 20 + 0.18x

4 Solve for x:

45.56− 20 = 0.18x

25.56

0.18= x

x = 142

5 You must drive 142 miles/day in order for the cost to be thesame at both firms.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Solving Quadratic Equations

A quadratic equation of one variable is an equation which can bewritten so the variable appears as a square or multiplied by aconstant, but not in any more complicated expressions. Quadraticequations appear in many applications.

Two techniques for solving quadratic equations are completing thesquare and the quadratic formula.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Completing the Square

The idea behind completing the square is that if we know that

x2 = b2

(where x is a variable and b is a constant), then there are twopossibilities. Either x = b, or x = −b.

Example

Suppose x2 = 9.Then because 32 = 9, either x = 3 or x = −3.Both values of x are solutions to the original equation, which wecan verify by substitution.

We can apply this idea to solve more general quadratic equations.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Completing the Square Example

Solve 2x2 + 8x − 24 = 0.

1 Eliminate the constant from the LHS by adding to both sidesand eliminate the coefficient on the x2 term by dividing onboth sides.

2x2 + 8x − 24 + 24 = 0 + 24

1

2·(2x2 + 8x

)=

1

2· 24

x2 + 4x = 12

2 Divide the coefficient of x by 2, square that number, and addthe resulting number to both sides of the equation.

(4/2)2 = 4

x2 + 4x + 4 = 12 + 4

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Complete the Square Example (Cont.)

3 The LHS is now a perfect square and can be factored in theform (x + a)2. The constant a is the coefficient of x dividedby 2.

x2 + 4x + 4 = (x + [4/2])2

4 Take square roots of both sides, keeping both the positive andnegative square roots.

(x + 2)2 = 16

x + 2 = ±4

5 Solve for x .

x = −2± 4 or x = 2, x = −6

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: The Quadratic Formula

Instead of completing the square, we can use the quadraticformula. For a quadratic equation of the form

ax2 + bx + c = 0,

the solution is given by

x =−b ±

√b2 − 4ac

2a.

This can be proved by solving the equation by completing thesquare.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Quadratic Formula Example

Solve 2x2 + 8x − 24 = 0.

Note that a = 2, b = 8, and c = −24. Plug in to the quadraticformula to get

x =−(8)±

√(8)2 − 4(2)(−24)

2(2)

=−8±

√64 + 192

4

=−8± 16

4

= −2± 4

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Applied Quadratic Equations

There are many examples of applications of quadratic equations ina real-world setting.

These problems follow the same process of solving an appliedproblem as before, with the only difference being the quadraticnature of the relationship found in step 3.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Applied Quadratic Equations: Example

The yearly amount y of food distributed by the Louisianafoodbank is given by the equation:

y = 0.027x2 − 0.37x + 7.16

where x > 0, x = 0 corresponds to the year 1990, and y ismeasured in millions of pounds. Find the year when 17,000,000pounds of food were distributed.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.4.1: Applied Quadratic Equations: Solution

1 x = # years since 1990y = amount of food (millions of lbs)

2 Want to know when y = 173 Relation (given in problem): y = 0.027x2 − 0.37x + 7.164 Solve for x:

17 = 0.027x2 − 0.37x + 7.16

0 = 0.027x2 − 0.37x + 7.16− 17

0 = 0.027x2 − 0.37x − 9.84

x =0.37±

√(−0.37)2 − 4(0.027)(−9.84)

2(0.027)

x = 19.7879, x = −18.4175

5 Since we are given that x > 0, the solution is x = 19.7879,which corresponds to approximately 2009.

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.5: Inequalities

So far, we have discussed equalities, where both sides of astatement have the same value. We should also review inequalities.

Symbol Type of Inequality> Greater Than≥ Greater Than or Equal< Less Than≤ Less Than or Equal

Math Boot Camp

2: Algebra

2.2: Algebra Techniques

2.2.5: Equivalent Inequalities

If a > b and c is any real number, then

a + c > b + c

If a > b and c is any positive number, then

a · c > b · c

If a > b and c is any negative number, then

a · c < b · c