

# Variables and their role in algebraic equations



Let's consider a simple algebraic equation:

$$x + 3 = 7$$

In this equation, $x$ is a variable. We want to find the value of $x$ that satisfies this equation.

Variables can also be used in more complex algebraic expressions. For example, consider the expression:

$$(x + 2)(x - 3)$$

In this expression, $x$ is a variable, and we can evaluate it by substituting its value into the expression.

## Exercise
Evaluate the expression $(x + 2)(x - 3)$ when $x = 1$.



# The operations of addition, subtraction, multiplication, and division in algebraic expressions



Let's consider two algebraic expressions:

$$a = 2x + 3$$
$$b = 4x - 1$$

We can add these expressions as follows:

$$a + b = (2x + 3) + (4x - 1) = 6x + 2$$

Similarly, we can subtract, multiply, and divide algebraic expressions using the rules of arithmetic.

## Exercise
Find the product of the following expressions:

$$a = 3x^2 - 2x$$
$$b = 2x^2 + 4x + 1$$



# Simplifying algebraic expressions



Consider the algebraic expression:

$$(2x + 3) - (x + 2)$$

We can simplify this expression by subtracting the terms with the same variable:

$$(2x + 3) - (x + 2) = x + 1$$



# Solving algebraic equations using various methods

- Substitution
- Manipulation
- Graphical methods
- Numerical methods

Let's consider the algebraic equation:

$$2x^2 - 5x + 3 = 0$$

We can solve this equation using the substitution method by setting the expression equal to zero and isolating the variable $x$:

$$2x^2 - 5x = -3$$
$$x(2x - 5) = -3$$

Now, we can find the value of $x$ that satisfies this equation.



# Systems of linear equations



Consider the following system of linear equations:

$$\begin{cases}
x + y = 5 \\
2x - y = 1
\end{cases}$$

We can solve this system using the substitution method by substituting the value of $x$ from the first equation into the second equation:

$$2(5 - y) - y = 1$$
$$10 - 2y - y = 1$$
$$y = 3$$

Now, we can substitute the value of $y$ into the first equation to find the value of $x$:

$$x = 5 - 3 = 2$$



# Quadratic equations

A quadratic equation is an equation of the form:

$$ax^2 + bx + c = 0$$

In this section, we will learn how to solve quadratic equations using various methods, such as factoring, completing the square, and the quadratic formula.

Let's consider the quadratic equation:

$$2x^2 - 5x + 3 = 0$$

We can solve this equation using factoring:

$$(2x - 3)(x - 1) = 0$$

Now, we can find the values of $x$ that satisfy either factor.



# Inequalities



Consider the inequality:

$$x^2 - 4 > 0$$

We can solve this inequality by graphing the function $y = x^2 - 4$ and finding the intervals where the function is greater than zero:

$$x^2 - 4 > 0$$
$$x^2 > 4$$

Now, we can find the values of $x$ that satisfy this inequality.



# Polynomial expressions



Let's consider the polynomial expression:

$$(x + 1)(x - 2)$$

We can evaluate this expression by multiplying the terms:

$$(x + 1)(x - 2) = x^2 - x - 2$$

Now, we can factor this expression:

$$x^2 - x - 2 = (x - 2)(x + 1)$$



# Factoring and prime factorization



Let's factor the polynomial expression:

$$x^2 - 4$$

We can factor this expression by finding the values of $x$ that satisfy the equation:

$$x^2 - 4 = 0$$

Now, we can find the factors of the expression.



# Rational expressions



Let's consider the rational expression:

$$\frac{3x^2 - 2x}{x^2 + 1}$$

We can evaluate this expression by substituting the value of $x$ and simplifying the fraction:

$$\frac{3x^2 - 2x}{x^2 + 1} = \frac{3x(x - 1)}{x(x + 1)}$$

Now, we can simplify this expression.

In conclusion, this textbook has covered the basics of algebraic expressions and equations. We have explored variables, operations, simplification, solving methods, systems of linear equations, quadratic equations, inequalities, polynomial expressions, factoring, prime factorization, and rational expressions. By applying these concepts, you will be able to tackle a wide range of algebraic problems.