Algebra
Chapter 1: Expressions and Algebraic Operations
Algebraic Expressions
An algebraic expression is a mathematical statement that consists of numbers, variables, and arithmetic operations between them.
For example,
\(3x + 10\) is an expression where \(3x\) and \(10\) are the terms and \(x\) is the variable. They are separated by the arithmetic sign +
Variable: A letter or symbol (e.g., \(x, n\)) representing an unknown.
Coefficient: The numerical factor multiplying the variable (e.g., in \(4x\), “4” is the coefficient).
Constant: A standalone number (e.g., 7 in \(3x + 7\)).
Creating Expressions
For translating word statements into algebraic expressions:
First, identify the variables (symbols that represent unknowns).
Next, look for the keywords that indicate mathematical operations:
- “Sum,” “plus,” “added to” → Addition (+)
- “Difference,” “minus,” “less than” → Subtraction (−)
- “Times,” “product of,” “multiplied by” → Multiplication (×)
- “Divide by,” “quotient of,” “over” → Division (÷)
Example:
“Five more than a number \(x” \rightarrow x + 5.\)
“Two times a number \(n” \rightarrow 2n\).
Equivalent Expressions
Equivalent expressions are expressions that have the same value or worth but do not look the same.
For example, \(3x (x + 2)\) and \(3x^2 + 3x\) are equivalent expressions.
To check if two expressions are equivalent, simplify both to the simplest form by combining like terms and applying the distributive property.
You can combine like terms by adding or subtracting their coefficients. For example,
\(3x^2 + x + 2x + x^2\) and \(5x^2 – x – x^2 + 4x\) equivalent expressions. They both simplify to \(4x^2 + 3x\).
The Distributive Property, which says \(a(b + c) = ab + ac\) lets you expand an expression or factor out a common factor. For example
\(3(y + 2)\) and \(3y + 6\) are equivalent expressions.
Evaluation through Substitution
To evaluate an expression at a certain value of the variable:
- Substitute the given value for the variable.
- Simplify by following the order of operations (PEMDAS).
Example:
Evaluate \(2x + 5\) at \(x = 4\).
\(2(4) + 5 = 8 + 5 = 13\)
Algebraic Operations
Adding & Subtracting Polynomials
Combine like terms (same variable, same exponent).
Example:
\((3x^22 + 2x) + (4x^2 – x) = 3x^2 + 4x^2 + 2x – x = 7x^2 + x\).
Multiplying Polynomials
Use distributive property or FOIL: F (First), O (Outside), I (Inside), L (Last).
Example
\((x + 3)(x + 5)\)
F: \(x \times x = x^2\)
O: \(x \times 5 = 5x\)
I: \(3 \times x = 3x\)
L: \(3 \times 5 = 15\)
Combining the results:
\(x^2 + 5x + 3x + 15\)\(=x^2 + 8x + 15\)
Chapter 2: Linear Equations
Creating Equations
When a word problem about finding an unknown quantity is given, set up the equation and solve it:
- Identify the main unknown quantity
- Assign one variable (e.g., \(x\)) to represent the main unknown.
- Translate each part of the sentence step by step into math operations.
Solving Equations
The goal is to isolate the variable that represents the main unknown on one side of the equation using the following steps:
- Use inverse operations: Subtract to cancel addition, add to cancel subtraction, divide to cancel multiplication, and multiply to cancel division.
- Balance the equation: Perform the same operation on both sides.
Example 1:
\(2x + 6 = 10\)
Here, \(x\) is the unknown variable. Isolate it on the left-hand side first:
\(2x + 6 – 6 = 10 – 6\)\(2x = 4\)
\(2x \div 2 = 4 \div 2\)
\(x = 2\)
Example 2:
\(3a − 5 = a + 7\)
Here \(a\) is the unknown variable. Isolate it on the left-hand side first:
\(3a – a − 5 = a – a + 7\)\(2a − 5 = 7\)
\(2a − 5 + 5 = 7 + 5 \)
\(2a = 12\)
\(2a \div 2 = 12 \div 2\)
\(a = 6\)
Example 3:
A movie theater charges $8 per ticket. You have $40 to spend. How many tickets can you buy?
Let x = number of tickets.
Each ticket is $8, total cost = \(8x\). You have $40,
\(8x = 40\)\(8x \div 8 = 40 \div 8\)
\(x = 5\)
Therefore, you can buy 5 tickets.
Chapter 3: Inequalities
Creating Inequalities
Use the same strategies as in creating equations, but remember the following additional information when setting up and solving inequalities.
Symbols involved in inequalities are > (Greater than), < (Less than), ≥ (Greater than or equal to), and ≤ (Less than or equal to).
In addition to these, the word phrases “at least” “minimum” indicate that the inequality being set up is “less than or equal to”, and the phrases “No more than,” “maximum,” “up to” indicate that the inequality involved is “greater than or equal to”
Example:
Which inequality represents the phrase “A number is at least 5”?
Let the number be represented by \(x\).
The phrase “at least” indicates that the inequality involved is “greater than or equal to”
Therefore, the inequality is \(x \ge 5\).
Solving Inequalities
Use the same strategies as those used for linear equations:
First, identify the unknown variable.
Next, add, subtract, multiply, or divide on both sides to isolate the variable.
Crucial rule:
If you multiply or divide an inequality by a negative number, you must reverse the inequality symbol.
Example 1:
\(2x − 4 \gt 8\)
Here x is the unknown variable. Isolate it on the left-hand side first:
\(2x − 4 + 4 \gt 8 + 4\)\(2x \gt 12\)
\(2x \div 2 \gt 12 2\)
\(x \gt 6\)
Example 2:
\(– 3x \ – 15 \le 9\)
Here x is the unknown variable. Isolate it on the left-hand side first:
\(-3x \ – 15 + 15 \le 9 + 15\)\(-3x \le 24\)
\(-3x \div 2 \le 24 \div 2\)
\(-x \le 8\)
Multiplying both sides by -1 to isolate xwill result in the sign flipping.
\(-x \times -1 \ge 8 \ge -1\)\(x \ge -8\)
Example 3
Jose earns $20 per hour. He wants to earn at least $300 this week. How many hours \(h\) must Jose work to achieve his goal?
Setting up the equation:
Since Jose works \(h\) hours, the amount he earns in \(h\) hours is \(20h\)
This should be at least $300. Therefore, the inequality involved is greater than or equal to.
\(20h \ge 300\)\(20h \div 20 \ge 300 \div 20\)
\(h \ge 10\)
Jose must work 15 or more hours to save $300 or more.
Chapter 4: Quadratic Equations
A quadratic equation is typically written as:
\(ax^2 + bx + c = 0\)
where \(a, b,\) and \(c\) are real numbers and \(a≠0\).
Factorizing and Solving Quadratic Equations
To factorize and solve quadratic equations:
- Rewrite the equation to the \(ax^2 + bx + c = 0\) form.
- Factor the expression if possible.
- Set each factor to 0 and solve for \(x\).
For a simple quadratic equation where \(a = 1\) and \(b\) and \(c\) are integers \((x^2 + bx + c = 0)\), we look for two numbers \(m\) and \(n\) whose product is \(c\) and whose sum is \(b\).
Then we rewrite \(x^2 + bx + c = 0\) as
\(x^2 + (m + n) x + (mn) = 0\)
This is equivalent to:
\((x + m) (x + n)\)
Thus, the factors of \(x^2 + bx + c = 0\) are \((x + m)\) and \((x + n)\).
And, the solutions to \(x^2 + bx + c = 0\) are \(x = -m\) and \(x = -n\)
Example
Factorize: \(x^2 + 5x + 6 = 0\)
We need two numbers that multiply to 6 and add to 5: 2 and 3.
\(x^2 + 5x + 6 = x^2+ (2 + 3)x + (2 \times 3)\)\(x^2 + 5x + 6 = (x+2)(x+3)\)
Factors: \((x + 2)\) and \((x + 3)\)
Solutions: \(x = -2\) and \(x = -3\)
Quadratic Formula
When a quadratic equation cannot be easily factored, we use the Quadratic Formula to find the solutions.
The solutions of \(ax^2 + bx + c = 0\) are given by the formula:
\(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)
This yields two solutions for the equation,
\(x = \dfrac{-b + \sqrt{b^2 – 4ac}}{2a}\) and \(x = \dfrac{-b \sqrt{-b^2 – 4ac}}{2a}\)
Example:
\(3x^2 \ – 24x \ – 99 = 0\)
Here,
\(a = 3, b = -24,\) and \(c = -99\)
Substituting these in the quadratic formula:
\(x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)
\(x = \dfrac{-(-24) \pm \sqrt{(-24)^2 – 4(3)(-99)}}{2(3)}\)
\(x = \dfrac{24 \pm \sqrt{576+4(3)(99)}}{2(3)}\)
\(x = \dfrac{24 \pm \sqrt{576+1188}}{6}\)
\(x = \dfrac{24 \pm \sqrt{1764}}{6}\)
\(x = \dfrac{24 \pm 42}{6}\)
So,
\(x = \dfrac{24+42}{6}\) or \(x = \dfrac{24-42}{6}\)
\(x = \dfrac{66}{6}\) or \(x = \dfrac{-18}{6}\)
\(x = 11\) or \(x = -3\)