GED Math Study Guide: Algebra

In this section, you’ll learn how to use algebraic thinking to analyze relationships, work with variables, and solve equations step by step. These skills are essential not only for the GED test but also for everyday problem-solving and higher-level math. Topics include:

• Expressions & Algebraic Operations
• Linear Equations
• Inequalities
• Quadratic Equations

Practice Quiz

Expressions & Algebraic Operations

Algebraic Expressions

An algebraic expression is a mathematical statement that consists of numbers and variables and the 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 (\(+\)).

Note
There are a few key terms you need to know when working with algebraic expressions:
 
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 (÷)

Examples:
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 \(3x^2 + 6x\) 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 \(\; 5x^2 − x − x^2 + 4x\) are 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:

1. Substitute the given value for the variable.
2. 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

To solve these problems, simply combine like terms (same variable, same exponent).

Example:
\((3x^2 + 2x) + (4x^2 − x) \) \( = 3x^2 + 4x^2 + 2x − x \) \( = 7x^2 + x\)

Multiplying Polynomials
Use the 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:

\(x^2 + 5x + 3x + 15\)

 

\(=x^2 + 8x + 15\)

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.

Example:

Sarah wants to calculate how many cakes her bakery must sell to profit $500. Each cake is sold for $25, but they cost $9 in ingredients to produce. How many cakes must be sold to hit Sarah’s profit goal?

We don’t know how many cakes must be sold, so we can represent that with \(x\).

We know we want the equation to equal $500, so we know the equation will begin with \(500 =\).

We also know that each cake nets a profit of $16, because the sale price minus the cost to produce is \(25 – 9 = 16\).

Since each cake adds $16 to Sarah’s profits, we can find how many cakes will get us to $500 with \(500 = 16x\).

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:

\(2x + 6 − 6 = 10 − 6\)

 

\(2x = 4\)

 

\(\dfrac{2x}{2} = \dfrac{4}{2}\)

 

\(x = 2\)

 

 

Example 2:
\(3a − 5 = a + 7\)

Here, \(a\) is the unknown variable. Isolate it on the left-hand side first:

\(3a − a − 5 = a − a + 7\)

 

\(2a − 5 = 7\)

 

\(2a − 5 + 5 = 7 + 5 \)

 

\(2a = 12\)

 

\(\dfrac{2a}{2} = \dfrac{12}{2}\)

 

\(a = 6\)

 

 

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.

As each ticket is $8, the total cost = 8\(x\). Since you have $40:

\(8x = 40\)

 

\(\dfrac{8x}{8} = \dfrac{40}{8}\)

 

\(x = 5\)

 

Therefore, you can buy 5 tickets.

Inequalities

Creating Inequalities

To create an inequality, use the same strategies as in creating equations above, 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, the phrases “at least” and “minimum” indicate that the inequality being set up is “less than or equal to”, and the phrases “no more than,” “maximum,” and “up to” indicate that the inequality involved is “greater than or equal to.”

 

Example:
What 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

To solve, 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.

Note
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:

\(2x − 4 + 4 \gt 8 + 4\)

 

\(2x \gt 12\)

 

\(\dfrac{2x}{2} \gt \dfrac{12}{2}\)

 

\(x \gt 6\)

 

 

Example 2:
\(– 3x \ – 15 \le 9\)

Here, \(x\) is the unknown variable. Isolate it on the left-hand side first:

\(-3x \ − 15 + 15 \le 9 + 15\)

 

\(-3x \le 24\)

 

Now we divide both sides by –3, but remember: when dividing both sides of an inequality by a negative number, you must flip the inequality sign:

\(\dfrac{-3x}{−3} \ge \dfrac{24}{−3}\)

 

\(x \ge −8\)

 

 

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 20\(h\).

This needs to make at least $300 for Jose. Therefore, the inequality involved is “greater than or equal to.”

\(20h \ge 300\)

 

\(\dfrac{20h}{20} \ge \dfrac{300}{20}\)

 

\(h \ge 15\)

 

Jose must work 15 or more hours to earn $300 or more.

Quadratic Equations

A quadratic equation is typically written as:

\(ax^2 + bx + c = 0\)

 

where \(a, \ b,\) and \(c\) are real numbers and \(a≠0\).

Factoring and Solving Quadratic Equations

To factor and solve quadratic equations:

1. Rewrite the equation into \(ax^2 + bx + c = 0\) form.
2. Factor the expression (if possible).
3. 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 \((x + n)\).

Further, the solutions to \(x^2 + bx + c = 0\) are \(x = -m\) and \(x = -n\).

Example
Factor: \(x^2 + 5x + 6 = 0\)

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:

\(x = \dfrac{-b \pm \sqrt{b^2 − 4ac}}{2a}\)

 

 

\(x = \dfrac{-(-24) \pm \sqrt{(-24)^2 − 4(3)(-99)}}{2(3)}\)

 

 

\(x = \dfrac{24 \pm \sqrt{576+4(3)(99)}}{2(3)}\)

 

 

\(x = \dfrac{24 \pm \sqrt{576+1188}}{6}\)

 

 

\(x = \dfrac{24 \pm \sqrt{1764}}{6}\)

 

 

\(x = \dfrac{24 \pm 42}{6}\)

 

 

Thus, the solutions are:

 

\(x = \dfrac{24+42}{6} \ \) or \( \ x = \dfrac{24-42}{6}\)

 

\(x = \dfrac{66}{6} \ \) or \( \ x = \dfrac{-18}{6}\)

 

\(x = 11 \ \) or \( \ x = -3\)

Once you have carefully reviewed all of this material, you should take the algebra review quiz. These questions will test you on all the topics above. Remember, you may consult the GED formula sheet at any time.

Algebra Review Quiz